Question 1

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N 2O (g) Rate = k[NO]2 (ii) H 2O 2 (aq) + 3I^- (aq) + 2H^+ → 2H 2O (l) + 3I^- Rate = k[ H 2 O 2 ][ I^- ] (iii) CH 3CHO (g) → CH 4 (g) + CO(g) Rate = k [CH 3CHO] 3 /2 (iv) C 2H 5Cl (g) → C 2H 4 (g) + HCl (g) Rate = k [C 2H 5Cl]

Accepted Answer

The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law, and the units of k always adjust so that both sides of the rate equation come out in mol L ^-1 s ^-1. Once you know the order n, the dimensions of k follow from Rate = k[ conc. ]^n ⇒ k = ( Rate )/([ conc. ]^n) = ( mol L^-1 s^-1)/(( mol L^-1)^n) Here Rate is always mol L ^-1 s ^-1 — concentration disappearing (or appearing) per unit time — regardless of how the reaction is written; the stoichiometric coefficients in the balanced equation do not decide the order, only the exponents actually written in the rate law do. (i) 3 NO (g) → N 2 O (g), Rate = k[ NO ]^2 The exponent on [ NO ] is 2, so this is a second-order reaction (order 2 in NO, and 2 overall) — even though the stoichiometric coefficient of NO in the balanced equation is 3. That coefficient-vs-exponent mismatch is exactly the trap here: order comes only from the rate law, never from the balanced equation. Dimensions of k: k = ( Rate )/([ NO ]^2) = ( mol L^-1 s^-1)/(( mol L^-1)^2) = ( mol L^-1 s^-1)/( mol^2 L^-2) = mol^-1 L s^-1 So k has units L mol^-1 s^-1 (equivalently dm^3 mol^-1 s^-1 ). (ii) H 2 O 2(aq) + 3 I^-(aq) + 2 H^+ → 2 H 2 O (l) + I 3^-, Rate = k[ H 2 O 2][ I^-] The exponents are 1 on [ H 2 O 2] and 1 on [ I^-], so the order is 1+1 = 2 overall (first order in each reactant, second order overall). H^+ does not appear in the rate law at all, so it contributes nothing to the order. Dimensions of k: k = ( Rate )/([ H 2 O 2][ I^-]) = ( mol L^-1 s^-1)/(( mol L^-1)( mol L^-1)) = ( mol L^-1 s^-1)/( mol^2 L^-2) = mol^-1 L s^-1 So k again has units L mol^-1 s^-1. (iii) CH 3 CHO (g) → CH 4(g) + CO (g), Rate = k[ CH 3 CHO ]^3/2 The exponent is 3/2, so this is a reaction of order 1.5 (three-halves order) — orders are not required to be whole numbers; they are read off the experimental rate law exactly as written. Dimensions of k: k = ( Rate )/([ CH 3 CHO ]^3/2) = ( mol L^-1 s^-1)/(( mol L^-1)^3/2) = mol^1-3/2 L^-1+3/2 s^-1 = mol^-1/2 L^1/2 s^-1 So k has units L^1/2 mol^-1/2 s^-1. (iv) C 2 H 5 Cl (g) → C 2 H 4(g) + HCl (g), Rate = k[ C 2 H 5 Cl ] The exponent on the single concentration term is 1, so the reaction is first order overall. Dimensions of k: k = ( Rate )/([ C 2 H 5 Cl ]) = ( mol L^-1 s^-1)/( mol L^-1) = s^-1 So k has units s^-1 — a plain reciprocal time, the hallmark of first-order kinetics (this is why first-order k never carries a concentration unit, unlike every other order above). Answer: (i) order 2, k in L mol ^-1 s ^-1; (ii) order 2, k in L mol ^-1 s ^-1; (iii) order 3/2, k in L ^1/2 mol ^-1/2 s ^-1; (iv) order 1, k in s ^-1.

Question 2

For the reaction: 2A + B → A 2 B the rate = k[A][B]2 with k = 2.0 × 10^-6 mol^-2 L^2 s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.

Accepted Answer

(i) 8.0 × 10 -9 mol L -1 s -1; 3.89 × 10 -9 mol L -1 s. Rate law tells you the rate at any instant once you know the concentrations at that instant — and stoichiometry tells you how those concentrations move together. For 2A + B → A 2B, the rate law is given as Rate = k[A][B]^2 where k = 2.0 × 10^-6 mol^-2 L^2 s^-1 is the rate constant, and [A], [B] are the molar concentrations of A and B at the instant the rate is measured. Step 1 — Initial rate Substitute the given initial concentrations, [A] = 0.1 mol L^-1 and [B] = 0.2 mol L^-1: Rate 1 = (2.0 × 10^-6 mol^-2 L^2 s^-1) × (0.1 mol L^-1) × (0.2 mol L^-1)^2 Rate 1 = (2.0 × 10^-6) × (0.1) × (0.04) mol L^-1 s^-1 = 8.0 × 10^-9 mol L^-1 s^-1 Step 2 — Find how much B has been used up when A drops to 0.06 mol L⁻¹ This is the step people skip: you are not told [B] at the later instant directly — you have to get it from the stoichiometry of the balanced equation, not just plug in [A]=0.06 and leave [B] unchanged. The amount of A consumed is Δ[A] = 0.1 - 0.06 = 0.04 mol L^-1 From the equation 2A + B → A 2B, 2 mol of A react with every 1 mol of B, so B is consumed at half the rate A is consumed: Δ[B] = 1/2Δ[A] = 1/2(0.04 mol L^-1) = 0.02 mol L^-1 So the new concentration of B is [B] new = 0.2 - 0.02 = 0.18 mol L^-1 Step 3 — Rate at the new concentrations Rate 2 = k[A] new [B] new^2 = (2.0 × 10^-6) × (0.06) × (0.18)^2 mol L^-1 s^-1 (0.18)^2 = 0.0324 Rate 2 = (2.0 × 10^-6) × (0.06) × (0.0324) mol L^-1 s^-1 = 3.888 × 10^-9 mol L^-1 s^-1 Rounding to three significant figures, consistent with the data given: Rate 2 ≈ 3.89 × 10^-9 mol L^-1 s^-1 Answer: Initial rate = 8.0 × 10^-9 mol L^-1 s^-1; rate when [A] = 0.06 mol L^-1 (so [B] = 0.18 mol L^-1 ) = 3.89 × 10^-9 mol L^-1 s^-1.

Question 3

The decomposition of NH 3 on platinum surface is zero order reaction. What are the rates of production of N 2 and H 2 if k = 2.5 × 10^-4 mol^-1 L s^-1?

Accepted Answer

For a zero-order reaction, the rate equals the rate constant k itself — it does not depend on the concentration of the reactant at all. The decomposition of ammonia on a platinum surface follows the balanced equation 2NH 3(g) Pt N 2(g) + 3H 2(g) For any reaction aA → bB + cC, the rate of reaction is defined so that it comes out the same number no matter which species you track, by dividing each rate of change by that species' stoichiometric coefficient: Rate = -1/a(d[A])/dt = 1/b(d[B])/dt = 1/c(d[C])/dt Here a = 2 (for NH 3 ), b = 1 (for N 2 ), c = 3 (for H 2 ). So Rate = -1/2(d[NH 3])/dt = (d[N 2])/dt = 1/3(d[H 2])/dt Because the reaction is zero order, the rate law is Rate = k[NH 3]^0 = k The units given for k, mol L^-1 s^-1, confirm this — a zero-order rate constant always carries the same units as the rate itself (this is the aside worth noting: the " mol^-1 " in the question is a printing slip; a rate constant with units of concentration/time, not concentration ^-1 /time, is what makes it zero order). Taking k = 2.5 × 10^-4 mol L^-1 s^-1 Rate of production of N 2: Since (d[N 2])/dt is exactly equal to the overall rate, (d[N 2])/dt = Rate = k = 2.5 × 10^-4 mol L^-1 s^-1 Rate of production of H 2: From the stoichiometric relation, (d[H 2])/dt = 3 × Rate, because 3 moles of H 2 form for every 1 mole of N 2 (this is the step people skip — the rate of formation of a product is not automatically equal to k; it is k multiplied by that product's own stoichiometric coefficient): (d[H 2])/dt = 3k = 3 × (2.5 × 10^-4 mol L^-1 s^-1) = 7.5 × 10^-4 mol L^-1 s^-1 Answer: Rate of production of N 2 = 2.5 × 10^-4 mol L^-1 s^-1; Rate of production of H 2 = 7.5 × 10^-4 mol L^-1 s^-1.

Question 4

The decomposition of dimethyl ether leads to the formation of CH 4, H 2 and CO and the reaction rate is given by Rate = k [CH 3OCH 3] 3 /2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., ( ) = p 3/2 Rate k CH OCH 3 3 If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Accepted Answer

bar -1/2s -1. When a reaction is followed by pressure change, "rate" is the rate of change of pressure with time — so its units come straight from the units you're given for pressure and time. The rate law is Rate = k (p CH 3OCH 3)^3/2 where p CH 3OCH 3 is the partial pressure of dimethyl ether and k is the rate constant for this pressure-based form of the rate law. Step 1: Units of Rate Rate is the change of a concentration-like quantity per unit time. Here concentration has been replaced by pressure, so Rate = -dp/dt Since pressure is measured in bar and time in minutes, the unit of Rate is simply unit of Rate = bar min^-1 Step 2: Units of the rate constant k Rearrange the rate law to isolate k: k = ( Rate )/((p CH 3OCH 3)^3/2) Substitute the units found above for Rate, and bar for pressure: unit of k = ( bar min^-1)/(( bar )^3/2) = bar^1-3/2 min^-1 = bar^-1/2 min^-1 This is the step people get wrong: the exponent on the concentration term in the rate law becomes the exponent on the pressure unit in k, and it must be subtracted correctly ( 1 - 3/2 = -1/2 ), not just copied or dropped. A fractional order (here 3/2 ) always gives a fractional power in the unit of k — that is expected, not a mistake. So the two units are: Rate: bar min^-1 k: bar^-1/2 min^-1 Answer: Rate has units of bar min⁻¹, and the rate constant k has units of bar⁻¹ᐟ² min⁻¹.

Question 5

Mention the factors that affect the rate of a chemical reaction.

Accepted Answer

The rate of a reaction depends on how often and how energetically the reacting particles collide — anything that changes collision frequency or collision energy changes the rate. Going through each factor and why it works this way: - Concentration of reactants. Rate is proportional to some power of the reactant concentrations (the rate law), because a higher concentration packs more particles into the same volume, so collisions between reactant molecules happen more often per unit time. Raising concentration almost always speeds up a reaction. - Temperature. Raising the temperature increases the average kinetic energy of the molecules, so a larger fraction of collisions now have energy at or above the activation energy E a — the minimum energy a collision needs before it can go on to form products. As a rough rule for many reactions, the rate roughly doubles for every 10 K rise in temperature. This is captured quantitatively by the Arrhenius equation, k = A e^-E a/RT where k is the rate constant, A is the frequency (pre-exponential) factor related to collision frequency and orientation, E a is the activation energy, R is the gas constant, and T is the absolute temperature (in kelvin — a common slip is to substitute the Celsius value here, which throws off the exponential term badly since T must be absolute). - Catalyst. A catalyst provides an alternative reaction pathway with a lower activation energy E a, without itself being consumed. Since more molecules now clear this lower energy barrier at a given temperature, the rate increases — the catalyst changes how the reaction happens, not the identity or amount of products formed. - Surface area of reactants (for heterogeneous/solid reactants). Breaking a solid into smaller pieces or a powder increases the surface area exposed to the other reactant. Since reaction can only occur where the reactants are in contact, more exposed surface means more collisions per second, so powdered or finely divided solids react faster than the same mass in a large lump. - Nature of the reactants. Both the physical state and the chemical nature of the substances matter. Reactions between ions in solution are typically very fast because no bonds need breaking before the reactive species meet; reactions requiring the breaking of strong covalent bonds are inherently slower. Similarly, gaseous and liquid reactants generally react faster than solids of the same chemical identity, because mixing (and hence collision opportunity) is far greater in a fluid phase. - Exposure to radiation (for photochemical reactions). For reactions that proceed through light absorption (e.g., the reaction between H 2 and Cl 2 ), the intensity and wavelength of light supplied controls the rate, since absorbed photons generate the reactive intermediates (like free radicals) that carry the reaction forward. Each of these factors works through the same underlying idea from collision theory: rate increases whenever collisions between reacting species become more frequent or more of those collisions carry enough energy (and the right orientation) to cross the activation-energy barrier. Answer: The rate of a chemical reaction is affected by (i) the concentration of reactants, (ii) temperature, (iii) presence of a catalyst, (iv) surface area of reactants (for solids), (v) the nature of the reactants (physical state and chemical nature), and (vi) exposure to radiation for photochemical reactions — all acting by changing either the frequency of collisions or the fraction of collisions with energy exceeding the activation energy E a.

Question 6

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Accepted Answer

(i) 4 times (ii) ¼ times. For a second-order reaction, the rate depends on the SQUARE of the concentration — doubling a concentration does not double the rate, it quadruples it. For a reaction that is second order in a reactant A, the rate law is Rate = k[A]^2 where k is the rate constant and [A] is the concentration of the reactant. Call the original rate r 1 = k[A]^2. (i) Concentration doubled Replace [A] with 2[A]: r 2 = k(2[A])^2 = k · 4[A]^2 = 4 (k[A]^2 ) = 4r 1 The squaring is the step that is easy to miss — doubling the concentration multiplies the rate by 2^2 = 4, not by 2. (ii) Concentration reduced to half Replace [A] with ([A])/2: r 3 = k(([A])/2)^2 = k · ([A]^2)/4 = 1/4 (k[A]^2 ) = r 1/4 Halving the concentration cuts the rate to (1/2)^2 = 1/4 of its original value, not to one-half. Answer: Doubling the concentration increases the rate 4 times (rate becomes 4× the original); reducing the concentration to half decreases the rate to 1/4 of the original.

Question 7

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Accepted Answer

A rate constant almost always increases as temperature rises — because more molecules cross the energy barrier, not because the mechanism changes. For most reactions, raising the temperature does not change the reaction pathway; it changes how many collisions happen with enough energy to react. The physical picture. At any temperature, molecular kinetic energies follow a spread (the Maxwell–Boltzmann distribution) — most molecules have energy near the average, and only a small fraction has energy equal to or above the activation energy E a (the minimum energy a collision needs to react). Raising the temperature shifts this distribution so a larger fraction of molecules clears the E a barrier. It is this growing fraction of "energetic enough" collisions — not a change in collision frequency alone — that makes the rate constant k rise sharply with T. A rough empirical rule (valid over the ordinary lab temperature range) is that the rate constant of a reaction nearly doubles for every 10 K (or 10^° C ) rise in temperature. This is captured by the temperature coefficient, Temperature coefficient = (k T+10)/(k T) ≈ 2 to 3 where k T and k T+10 are the rate constants at temperature T and T+10 K. This is only a rule of thumb — the exact relationship is quantitative and exponential, not linear. The quantitative relationship: the Arrhenius equation. k = A e^-E a/RT Here: - k is the rate constant of the reaction, - A is the pre-exponential factor (also called the frequency factor) — it accounts for the frequency of collisions and the fraction of collisions with the correct orientation; A has the same units as k, - E a is the activation energy in J mol^-1 (or kJ mol^-1 ), - R is the gas constant, 8.314 J K^-1 mol^-1, - T is the absolute temperature in kelvin. A step people get wrong here: T in this equation must be the absolute (kelvin) temperature, never degrees Celsius — the exponential blows up any Celsius value used by mistake. Why k rises with T follows directly from this equation. As T increases, the exponent -E a/RT becomes less negative, so e^-E a/RT increases, and hence k increases. The larger E a is, the more sensitive k is to a change in T — reactions with high activation energy speed up much more sharply on heating than those with low activation energy. A more convenient (linear) form. Taking the natural logarithm of both sides: ln k = ln A - E a/RT or, in base-10 form (more common for graphical work), log k = log A - E a/(2.303 RT) Since this last equation has the form y = c + mx, a plot of log k (y-axis) against 1/T (x-axis) gives a straight line with - slope = -E a/(2.303 R), so E a = -2.303 R × slope, and - intercept = log A. This is how E a and A are actually determined experimentally: measure k at several temperatures, plot log k vs. 1/T, and read off the slope and intercept. The two-temperature form. If the rate constants k 1 and k 2 are known at two temperatures T 1 and T 2, subtracting the log form of the Arrhenius equation at T 2 from that at T 1 eliminates A: log k 1 = log A - E a/(2.303 R T 1), log k 2 = log A - E a/(2.303 R T 2) log k 2 - log k 1 = E a/(2.303 R)(1/T 1 - 1/T 2) 2/k 1 = E a/(2.303 R)(1/T 1 - 1/T 2) This form is the one most often used in numerical problems: given E a and k at one temperature, it gives k at another temperature (or, given k at two temperatures, it gives E a ) without needing to know A separately. Answer: For most reactions, the rate constant k increases with rising temperature (roughly doubling for every 10 K /10^° C rise), because a larger fraction of molecules then possesses energy at or above the activation energy E a. This is represented quantitatively by the Arrhenius equation k = A e^-E a/RT, or in logarithmic form log k = log A - E a/(2.303 RT), whose two-temperature version is 2/k 1 = E a/(2.303 R)(1/T 1-1/T 2).

Question 8

In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L^-1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Accepted Answer

(i) 4.67 × 10 -3 mol L -1s -1 (ii) 1.98 × 10. Average rate is just -(Δ[A])/(Δ t) — the concentration lost divided by the time it took, no rate law needed yet. The data given are: t = 30 s, [A] = 0.31 mol L^-1 t = 60 s, [A] = 0.17 mol L^-1 For a reactant being consumed, the average rate over an interval is Average rate = -([A] 2 - [A] 1)/(t 2 - t 1) = -(Δ [A])/(Δ t) where [A] 1 and [A] 2 are the concentrations of A at the start and end of the interval, and the minus sign is there because [A] is falling as the reaction proceeds — without it the "rate" would come out negative, which a rate can never be. Substituting the values for the 30 s to 60 s window: Δ [A] = [A] 2 - [A] 1 = 0.17 - 0.31 = -0.14 mol L^-1 Δ t = 60 - 30 = 30 s Average rate = -((-0.14 mol L^-1))/(30 s ) = (0.14 mol L^-1)/(30 s ) Average rate = 4.666… × 10^-3 mol L^-1 s^-1 This is the average rate over that 30-second window, not the instantaneous rate at t = 30 s or t = 60 s — since the reaction is first order, the true (instantaneous) rate is actually falling continuously as [A] drops, so this number is really the mean of the (steeper) rate near 30 s and the (shallower) rate near 60 s. Rounding to three significant figures, matching the precision of the concentration data: Average rate ≈ 4.67 × 10^-3 mol L^-1 s^-1 Answer: The average rate of the reaction between t = 30 s and t = 60 s is 4.67 × 10^-3 mol L^-1 s^-1.

Question 9

A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?

Accepted Answer

(i) rate = k[A][B] (ii) 9 times. The order in each reactant tells you the power that concentration is raised to in the rate law — first order in A means power 1, second order in B means power 2. (i) The differential rate equation For a reaction of order 1 in A and order 2 in B, the rate law is written as Rate = -(d[R])/dt = k[A]^1[B]^2 where k is the rate constant, [A] and [B] are the molar concentrations of the two reactants, and the exponents (1 and 2) are the orders given in the problem, not the stoichiometric coefficients from a balanced equation. (ii) Effect of tripling [B] Start from the same rate law and replace [B] by 3[B], keeping [A] unchanged: Rate ' = k[A]^1(3[B])^2 = k[A](9[B]^2) = 9 (k[A][B]^2 ) = 9 × Rate The key point people miss here: because B appears squared in the rate law, tripling its concentration does not triple the rate — the factor of 3 gets squared too, giving 3^2 = 9. (iii) Effect of doubling both [A] and [B] Replace [A] by 2[A] and [B] by 2[B] simultaneously: Rate '' = k(2[A])^1(2[B])^2 = k(2[A])(4[B]^2) = 8 (k[A][B]^2 ) = 8 × Rate Here the factor of 2 from A contributes once (first order) and the factor of 2 from B contributes twice, since B is squared: 2^1 × 2^2 = 2 × 4 = 8. Doubling a concentration is not the same as doubling the rate unless that species is first order alone — with a mix of orders, each factor must be raised to its own power before multiplying. Answer: (i) Rate = k[A][B]^2; (ii) the rate increases 9-fold; (iii) the rate increases 8-fold.

Question 10

In a reaction between A and B, the initial rate of reaction ( r 0 ) was measured for different initial concentrations of A and B as given below: A/ mol L^-1 0.20 0.20 0.40 B/ mol L^-1 0.30 0.10 0.05 r 0 /mol L^-1 s^-1 5.07 × 10^-5 5.07 × 10^-5 1.43 × 10^-4 What is the order of the reaction with respect to A and B?

Accepted Answer

Orders with respect to A is 1.5 and order with respect to B is zero.. The rate law is r 0 = k[A]^x[B]^y, and each order is found by comparing two experiments where only ONE concentration changes — never by staring at the whole table at once. Take the three experiments as given: Experiment [A] /mol L ^-1 [B] /mol L ^-1 r 0 /mol L ^-1 s ^-1 --- --- --- --- 1 0.20 0.30 5.07×10^-5 2 0.20 0.10 5.07×10^-5 3 0.40 0.05 1.43×10^-4 Step 1 — order with respect to B. Experiments 1 and 2 hold [A] fixed at 0.20 mol L ^-1, so any change in rate between them must come from y, the order in B: (r 0,1)/(r 0,2) = (([B] 1)/([B] 2))^y Substituting, (5.07×10^-5)/(5.07×10^-5) = (0.30/0.10)^y ⇒ 1 = 3^y The only value that makes 3^y=1 is y = 0. Tripling [B] did nothing to the rate — the reaction is zero order in B, so B's concentration can be dropped from the rate law entirely. This is the step people skip: they try to fit A and B together over experiments 1 and 3, but experiment 3 has two things different from experiment 1 at once ([A] AND [B]), so that comparison alone can't isolate either order. You need a pair that isolates one variable first. Step 2 — order with respect to A. Now that B has zero order, its concentration cancels out of any comparison — experiments 2 and 3 can be compared for A even though their [B] values differ, because [B]^0=1 regardless of what [B] is: (r 0,3)/(r 0,2) = (([A] 3)/([A] 2))^x(([B] 3)/([B] 2))^0 = (([A] 3)/([A] 2))^x Substituting the numbers, (1.43×10^-4)/(5.07×10^-5) = (0.40/0.20)^x 2.821 = 2^x Taking logarithms of both sides, x = ( log (2.821))/( log (2)) = 0.4504/0.3010 = 1.496 ≈ 1.5 Checking this against a clean half-integer: 2^1.5 = 2√ 2 = 2.828, which reproduces the data ( 2.828 × 5.07×10^-5 = 1.434×10^-4, matching the given 1.43×10^-4 to the precision of the data). So x = 3/2. Step 3 — assemble the rate law. r 0 = k[A]^3/2[B]^0 = k[A]^3/2 The order in A is 1.5 (a fractional order — not unusual in kinetics, and a signal that the mechanism isn't a simple elementary step), the order in B is zero, and the overall order of the reaction is x+y = 1.5+0 = 1.5. Answer: order with respect to A = 1.5 (i.e., 3/2 ); order with respect to B = 0; overall order of reaction = 1.5.

Question 11

The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D Experiment [A]/mol L^-1 [B]/mol L^-1 Initial rate of formation of D/mol L^-1 min^-1 I 0.1 0.1 6.0 × 10^-3 II 0.3 0.2 7.2 × 10^-2 III 0.3 0.4 2.88 × 10^-1 IV 0.4 0.1 2.40 × 10^-2 Determine the rate law and the rate constant for the reaction.

Accepted Answer

rate law = k[A][B] 2; rate constant = 6.0 M. Rate law comes from comparing pairs of experiments where only ONE concentration changes — never guess the order from the stoichiometric coefficients (2 in front of A does not mean the reaction is second order in A). For the reaction 2A + B → C + D, write the rate law with unknown orders x and y: Rate = k[A]^x[B]^y Step 1: Find the order in B. Pick two experiments where [A] is held fixed and [B] changes — Experiments II and III both have [A] = 0.3 mol L^-1. ( Rate III )/( Rate II ) = (k(0.3)^x(0.4)^y)/(k(0.3)^x(0.2)^y) = (0.4/0.2)^y = (2)^y Substituting the rates: (2.88×10^-1)/(7.2×10^-2) = 4.0 = 2^y So y = 2: the reaction is second order in B. Step 2: Find the order in A. Now pick two experiments where [B] is held fixed and [A] changes — Experiments I and IV both have [B] = 0.1 mol L^-1. ( Rate IV )/( Rate I ) = (k(0.4)^x(0.1)^y)/(k(0.1)^x(0.1)^y) = (0.4/0.1)^x = (4)^x Substituting the rates: (2.40×10^-2)/(6.0×10^-3) = 4.0 = 4^x So x = 1: the reaction is first order in A. Step 3: Write the rate law. Rate = k[A][B]^2 The overall order is x + y = 1 + 2 = 3 (third order overall) — note this is different from the sum of the stoichiometric coefficients (3 as well here only by coincidence; in general the two need not match, which is exactly why the rate law must be found from data, not from the balanced equation). Step 4: Find the rate constant k. Rearrange the rate law and substitute Experiment I's data ( [A] = 0.1 mol L^-1, [B] = 0.1 mol L^-1, rate = 6.0×10^-3 mol L^-1 min^-1 ): k = ( Rate )/([A][B]^2) = (6.0×10^-3 mol L^-1 min^-1)/((0.1 mol L^-1)(0.1 mol L^-1)^2) k = (6.0×10^-3)/(0.1 × 0.01) mol L^-1 min^-1 · mol^-3 L^3 = (6.0×10^-3)/(1.0×10^-3) mol^-2 L^2 min^-1 k = 6.0 mol^-2 L^2 min^-1 Checking against the other three experiments confirms the same k — this cross-check is the whole point of using multiple experiments, not just a formality: Exp II: k = (7.2×10^-2)/((0.3)(0.2)^2) = (7.2×10^-2)/0.012 = 6.0 mol^-2 L^2 min^-1 Exp III: k = (2.88×10^-1)/((0.3)(0.4)^2) = (2.88×10^-1)/0.048 = 6.0 mol^-2 L^2 min^-1 Exp IV: k = (2.40×10^-2)/((0.4)(0.1)^2) = (2.40×10^-2)/0.004 = 6.0 mol^-2 L^2 min^-1 All four experiments give the identical value, which confirms the rate law is correct. The units of k are not arbitrary — for a rate law of overall order 3, k must carry units of mol^-2 L^2 min^-1 (equivalently L^2 mol^-2 min^-1 ) so that k[A][B]^2 comes out in mol L^-1 min^-1, the units of rate. Answer: Rate law is Rate = k[A][B]^2 (first order in A, second order in B, third order overall), with k = 6.0 L^2 mol^-2 min^-1.

Question 12

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L^-1 [B]/ mol L^-1 Initial rate/ mol L^-1 min^-1 I 0.1 0.1 2.0 × 10^-2 II - 0.2 4.0 × 10^-2 III 0.4 0.4 - IV - 0.2 2.0 × 10^-2

Accepted Answer

When the rate law is Rate = k[A]^1[B]^0, the concentration of B never enters the rate — only [A] and the rate constant k matter. Zero order in B means changing [B] does nothing to the rate; every blank in this table is found from Experiment I's rate law once you have k. Step 1: Write the rate law and find k from Experiment I. Since the reaction is first order in A and zero order in B, Rate = k[A]^1[B]^0 = k[A] where k is the rate constant, [A] is the concentration of A, and [B] does not appear because its order is zero. Using Experiment I, where [A] = 0.1 mol L^-1 and Rate = 2.0×10^-2 mol L^-1 min^-1: 2.0×10^-2 mol L^-1 min^-1 = k × 0.1 mol L^-1 k = (2.0×10^-2 mol L^-1 min^-1)/(0.1 mol L^-1) = 0.2 min^-1 This k is fixed for the reaction (it does not change between experiments), so every other blank uses this same value. Step 2: Experiment II — find [A] from the given rate. Rate = 4.0×10^-2 mol L^-1 min^-1. The value of [B] = 0.2 mol L^-1 given alongside it is a distractor here — it plays no role in the rate law, since the order in B is zero. [A] = ( Rate )/k = (4.0×10^-2 mol L^-1 min^-1)/(0.2 min^-1) = 0.2 mol L^-1 Step 3: Experiment III — find the rate from the given [A]. [A] = 0.4 mol L^-1, and again [B] = 0.4 mol L^-1 is irrelevant to the calculation. Rate = k[A] = 0.2 min^-1 × 0.4 mol L^-1 = 8.0×10^-2 mol L^-1 min^-1 Step 4: Experiment IV — find [A] from the given rate. Rate = 2.0×10^-2 mol L^-1 min^-1; [B] = 0.2 mol L^-1 again does not enter the calculation. [A] = ( Rate )/k = (2.0×10^-2 mol L^-1 min^-1)/(0.2 min^-1) = 0.1 mol L^-1 Completed table: Experiment [A] /mol L ^-1 [B] /mol L ^-1 Initial rate/mol L ^-1 min ^-1 --- --- --- --- I 0.1 0.1 2.0×10^-2 II 0.2 0.2 4.0×10^-2 III 0.4 0.4 8.0×10^-2 IV 0.1 0.2 2.0×10^-2 Answer: k = 0.2 min^-1; Experiment II: [A] = 0.2 mol L^-1; Experiment III: Rate = 8.0×10^-2 mol L^-1 min^-1; Experiment IV: [A] = 0.1 mol L^-1.

Question 13

Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s^-1 (ii) 2 min^-1 (iii) 4 years^-1

Accepted Answer

(i) 3.47 x 10 -3 seconds. For a first-order reaction, the half-life does not depend on the starting concentration — it depends only on the rate constant k. The half-life formula comes from integrating the first-order rate law, ln ([R] 0)/([R]) = kt, and asking how long it takes for [R] to fall to [R] 0/2: t 1/2 = ( ln 2)/k = 0.693/k Here k is the rate constant (with units of inverse time, since first-order rate constants carry units of time^-1 ), and t 1/2 comes out in whatever time unit k is expressed in. This is the step people trip on: because t 1/2 = 0.693/k only, you never need a concentration value to answer this kind of question — just divide 0.693 by k, keeping the unit of time that k already carries. (i) k = 200 s^-1 t 1/2 = 0.693/(200 s^-1) = 3.465×10^-3 s Rounded to three significant figures: t 1/2 ≈ 3.47×10^-3 s. (ii) k = 2 min^-1 t 1/2 = 0.693/(2 min^-1) = 0.3465 min Rounded to three significant figures: t 1/2 ≈ 0.347 min. (iii) k = 4 years^-1 t 1/2 = 0.693/(4 years^-1) = 0.17325 years Rounded to three significant figures: t 1/2 ≈ 0.173 years. Notice the unit of t 1/2 simply matches the unit of time built into k in each case — there is no unit conversion to do, only the division 0.693/k. Answer: (i) t 1/2 ≈ 3.47×10^-3 s (ii) t 1/2 ≈ 0.347 min (iii) t 1/2 ≈ 0.173 years

Question 14

The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Accepted Answer

1845 years. Radioactive decay is always first order, so the fraction of the isotope left tells you the elapsed time directly through the first‑order rate law — you never need to know the actual number of atoms, only the ratio. Step 1 — Get the rate constant from the half-life. For a first-order process, k = 0.693/(t 1/2) where k is the decay (rate) constant and t 1/2 is the half-life. k = 0.693/(5730 year ) = 1.209×10^-4 year^-1 Step 2 — Write the integrated first-order rate law in terms of the remaining fraction. t = 2.303/k log ([R] 0)/([R]) Here [R] 0 is the amount of ^14 C in the living tree (taken as 100%) and [R] is the amount left in the artifact now. The artifact has 80% of the original ^14 C, so ([R])/([R] 0) = 0.80 ⇒ ([R] 0)/([R]) = 1/0.80 = 1.25 This is the step people invert by mistake — the ratio inside the log must be initial over remaining (a number bigger than 1), not remaining over initial, or the log comes out negative and the age comes out negative too. Step 3 — Substitute and solve for t. t = 2.303/(1.209×10^-4 year^-1)× log (1.25) log (1.25) = 0.0969 t = 1.9043×10^4 year × 0.0969 t ≈ 1845 year The half-life (5730 year, 4 significant figures) sets the precision here, so the answer is kept to four significant figures. Answer: The archaeological sample is about 1845 years old.

Question 15

The experimental data for decomposition of N 2O 5 [ 2N 2 O 5 → 4NO 2 + O 2 ] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 10^2 × [N 2O 5] / 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L^-1 (i) Plot [N 2O 5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[ N 2 O 5 ] and t. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

Accepted Answer

A straight line on a plot of log [N 2O 5] against t is the signature of first-order kinetics — its slope hands you the rate constant directly, which is why part (iii) asks for that particular graph and not just the raw concentration curve. (i) Plotting [N 2O 5] against t The raw data are t /s 0 400 800 1200 1600 2000 2400 2800 3200 --- --- --- --- --- --- --- --- --- --- [N 2O 5] /mol L ^-1 0.0163 0.0136 0.0114 0.00930 0.00780 0.00640 0.00530 0.00430 0.00350 Plotted on ordinary axes these nine points trace a smoothly falling, concave-up curve: steep near t=0 and flattening out as [N 2O 5] gets small — the shape any exponential decay produces. A curve like this cannot be measured with a ruler, which is exactly the problem part (iii) fixes. (ii) Half-life read from the plot Half of the starting concentration is ([N 2O 5] 0)/2=0.0163/2=0.00815 mol L^-1. This value sits between the readings at t=1200 s ( 0.00930 ) and t=1600 s ( 0.00780 ). Interpolating linearly along the curve between those two points: t 1/2=1200+400×0.00930-0.00815/0.00930-0.00780=1200+400×0.00115/0.00150=1200+307=1507 s. So, reading the graph, t 1/2≈ 1.51×10^3 s. (iii) Plotting log [N 2O 5] against t Taking log 10 of each concentration: t /s 0 400 800 1200 1600 2000 2400 2800 3200 --- --- --- --- --- --- --- --- --- --- log [N 2O 5] -1.788 -1.866 -1.943 -2.032 -2.108 -2.194 -2.276 -2.367 -2.456 These nine points fall on a single straight line of negative slope. Turning the curved decay in part (i) into a straight line here is the whole point of the log transform — it lets you read a slope, not just eyeball a curve. (iv) The rate law A straight line for log [N 2O 5] versus t is only possible if the integrated first-order law holds: log [N 2O 5]= log [N 2O 5] 0-k/2.303 t. This is the equation of a line with slope -k/2.303, and the data obey it (part iii). That means the reaction is first order in N 2O 5, so the rate law is Rate =k[N 2O 5]^1. This is consistent with the decomposition being effectively a unimolecular step for each N 2O 5 molecule, even though the balanced equation 2N 2O 5→ 4NO 2+O 2 has a stoichiometric coefficient of 2 — order is an experimental fact, not something you read off the balanced equation. (v) The rate constant, from the slope For a line y=mx+c, the slope is m=(Δ y)/(Δ x). Using the first and last points of the log table, (0, -1.788) and (3200, -2.456): m=(-2.456-(-1.788))/3200-0=-0.668/3200=-2.088×10^-4 s^-1. Since m=-k/2.303: k=-m×2.303=(2.088×10^-4)×2.303=4.81×10^-4 s^-1. Fitting the best straight line through all nine points (rather than just the two ends) gives the same slope to three figures, k=4.80×10^-4 s^-1 — a useful check that the "first order" call in part (iv) isn't riding on two convenient data points. A first-order rate constant carries units of s ^-1 only — no concentration in sight — which is a quick way to catch an order mistake before it reaches the rate law. (vi) Half-life calculated from k, and comparison For a first-order reaction the half-life does not depend on the starting concentration: t 1/2=0.693/k=0.693/(4.80×10^-4 s^-1)=1444 s ≈1.44×10^3 s. This is close to the graphical estimate of 1.51×10^3 s from part (ii) — the two differ by only about 5%, which is well inside the uncertainty of interpolating a curve by eye. Getting essentially the same half-life two different ways (once from the raw decay curve, once from the rate constant) is the real confirmation that the reaction is first order, not just the straight log-plot on its own. Answer: First order in N 2O 5, Rate =k[N 2O 5]; k≈4.80×10^-4 s^-1; t 1/2 (from graph) ≈1.51×10^3 s, t 1/2 (from k ) ≈1.44×10^3 s — in close agreement.

Question 16

The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/ 16^th value?

Accepted Answer

4.6 × 10 -2 s. A first-order reaction has a fixed integrated rate law linking time to how much the concentration has fallen — you don't need to know the actual starting concentration, only the ratio by which it has dropped. For a first-order reaction, the integrated rate law is k = 2.303/t log ([A] 0)/([A]) where k is the rate constant, t is the elapsed time, [A] 0 is the initial concentration, and [A] is the concentration at time t. Rearranging for t: t = 2.303/k log ([A] 0)/([A]) Set up the concentration ratio. The reactant falls to 1/16 th of its initial value, so ([A] 0)/([A]) = ([A] 0)/([A] 0/16) = 16 A step people rush past: 16 = 2^4, so log 16 = 4 log 2 = 4(0.3010) = 1.2041. Using this instead of trying to evaluate log 16 directly avoids arithmetic slips. Substitute k = 60 s^-1 and log ([A] 0)/([A]) = 1.2041: t = 2.303/(60 s^-1) × 1.2041 t = (2.303 × 1.2041)/60 s t = 2.7731/60 s t = 0.04622 s Another place to be careful: the units work out to seconds only because k is given in s^-1; the 2.303 and log terms are pure numbers, so t inherits its unit directly from 1/k. Rounding to three significant figures (matching the precision of log 2 = 0.3010 ): t = 4.62 × 10^-2 s Answer: t = 4.62 × 10^-2 s (about 0.0462 s, i.e. 46.2 ms).

Question 17

During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Accepted Answer

0.7814 μg and 0.227 μg.. Radioactive decay is a first-order process, so the amount left after any time t is fixed once you know the half-life — you don't halve it in chunks, you use the exponential decay law. Step 1 — Get the rate constant from the half-life. For a first-order reaction, k = 0.693/(t 1/2) where t 1/2 is the half-life and k is the decay constant. Substituting t 1/2 = 28.1 years: k = 0.693/(28.1 year ) = 0.02466 year^-1 Step 2 — Apply the first-order decay law. N = N 0 e^-kt where N 0 is the amount present at t = 0 (here N 0 = 1 μ g ) and N is the amount remaining after time t. This is the step people skip: because 10 years and 60 years are not whole multiples of the 28.1-year half-life, you cannot just say "less than half is gone" — you have to put kt into the exponential directly. After 10 years: kt = 0.02466 year^-1 × 10 year = 0.2466 N = 1 μ g × e^-0.2466 = 1 μ g × 0.7814 = 0.7814 μ g After 60 years: kt = 0.02466 year^-1 × 60 year = 1.480 N = 1 μ g × e^-1.480 = 1 μ g × 0.2277 = 0.2277 μ g So of the original 1 μg of ^90 Sr, about 78.1% is still in the bone after 10 years, and only about 22.8% is left after 60 years — the decay looks slow at first only because 10 years is a small fraction of the 28.1-year half-life; by 60 years (a bit over two half-lives) most of it is gone. Answer: about 0.7814 μg remains after 10 years, and about 0.2277 μg remains after 60 years.

Question 18

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Accepted Answer

A first-order reaction has a fixed integrated rate law, and that law converts "% completion" into a log term — comparing two percentages is just comparing two logarithms. For a first-order reaction, the integrated rate equation is k = 2.303/t log ([A] 0)/([A]) where [A] 0 is the initial concentration, [A] is the concentration left at time t, and k is the rate constant (the same fixed number throughout the reaction — this is the property that makes the comparison work). Rearranging for t: t = 2.303/k log ([A] 0)/([A]) Time for 90% completion, t 90. If 90% of [A] 0 has reacted, 10% remains, so [A] = 0.10 [A] 0, which gives ([A] 0)/([A]) = ([A] 0)/(0.10 [A] 0) = 10 Substituting: t 90 = 2.303/k log 10 = 2.303/k(1) = 2.303/k ( log 10 = 1 because log here is log 10, not ln — the 2.303 already absorbs the conversion from natural log, so no further base change is needed.) Time for 99% completion, t 99. If 99% has reacted, 1% remains, so [A] = 0.01 [A] 0, which gives ([A] 0)/([A]) = ([A] 0)/(0.01 [A] 0) = 100 Substituting: t 99 = 2.303/k log 100 = 2.303/k(2) = 2 × 2.303/k ( log 100 = log 10^2 = 2 — this is the step that produces the factor of 2; it is easy to instead write log 100 as some decimal and lose the exact relationship.) Comparing the two. Since t 90 = 2.303/k, the expression for t 99 is just t 99 = 2 × 2.303/k = 2 t 90 The rate constant k is a constant for a given reaction at a given temperature, so it cancels out of the comparison entirely — the factor of 2 comes purely from log 100 = 2 log 10, which is a property of first-order kinetics, not of any particular reaction. This is why the result holds for every first-order reaction, regardless of its actual k value: doubling how far a first-order reaction has progressed (in this log sense, from 90% left-to-go being 10× down to 99% being 100× down) exactly doubles the time. Answer: t 99 % = 2 t 90 %, since t 90 = 2.303/k log 10 = 2.303/k and t 99 = 2.303/k log 100 = (2× 2.303)/k = 2 t 90.

Question 19

A first order reaction takes 40 min for 30% decomposition. Calculate t 1/2.

Accepted Answer

77.7 minutes. For a first-order reaction, the rate constant depends only on the ratio of initial to remaining concentration — not on the actual amounts — so "30% decomposed" is enough information by itself. The integrated first-order rate law is k = 2.303/t log ([A] 0)/([A]) where k is the rate constant, t is the elapsed time, [A] 0 is the concentration at the start, and [A] is the concentration remaining at time t. Setting up the concentrations. Take [A] 0 = 100 (any convenient starting value works, since only the ratio matters). If 30% has decomposed in t = 40 min, then 70% remains: [A] 0 = 100, [A] = 100 - 30 = 70 Substituting into the rate law. k = 2.303/(40 min ) log 100/70 Evaluate the logarithm first: log 100/70 = log (1.4286) = 0.1549 So k = 2.303/(40 min ) × 0.1549 = 0.35674/40 min^-1 = 8.918×10^-3 min^-1 Getting the half-life. For a first-order reaction the half-life is independent of concentration and given by t 1/2 = 0.693/k Here 0.693 = ln 2, and this formula holds only for first-order kinetics — for any other order t 1/2 would depend on the starting concentration, which is exactly why this problem could be solved without knowing an actual concentration in moles per litre. Substituting the value of k: t 1/2 = 0.693/(8.918×10^-3 min^-1) = 77.70 min Rounding to three significant figures (matching the precision of the given data, 40 min and 30%): Answer: t 1/2 ≈ 77.7 min

Question 20

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.

Accepted Answer

2.20 × 10 -3 s -1. The trick here is that pressure is being used as a stand-in for concentration — you never get to see [A] directly, only the total pressure of the gas mixture, so the first job is to dig the partial pressure of azoisopropane back out of that total. The decomposition is (CH 3) 2 CHN=NCH(CH 3) 2(g) → N 2(g) + C 6 H 14(g) One mole of azoisopropane (call it A ) gives one mole of N 2 and one mole of hexane — two moles of gas for every one that reacts. That is why the total pressure rises above the starting value of 35.0 mm Hg as the reaction proceeds. Setting up pressures with an ICE-type table Let P 0 be the initial pressure of A (35.0 mm Hg), and let x be the drop in the partial pressure of A after time t: The total pressure at time t is the sum of all three partial pressures: P t = (P 0-x) + x + x = P 0 + x So the extra pressure above P 0 tells you exactly how much A has reacted: x = P t - P 0 ⇒ P A = P 0 - x = 2P 0 - P t Getting P A at each recorded time Using P 0 = 35.0 mm Hg: - At t = 360 s: P A = 2(35.0) - 54.0 = 70.0 - 54.0 = 16.0 mm Hg - At t = 720 s: P A = 2(35.0) - 63.0 = 70.0 - 63.0 = 7.0 mm Hg Applying the first-order integrated rate law Since pressure is directly proportional to concentration at fixed temperature and volume, the first-order equation can be written directly in pressures: k = 2.303/t log 10 (P 0/P A) Here k is the rate constant, t is the elapsed time, P 0 is the initial pressure of A, and P A is the pressure of unreacted A remaining at that time. Using t = 360 s: k 1 = 2.303/(360 s ) log 10 (35.0/16.0) = 2.303/(360 s ) log 10(2.1875) log 10(2.1875) = 0.3400 k 1 = (2.303 × 0.3400)/(360 s ) = 0.78302/(360 s ) = 2.175 × 10^-3 s^-1 Using t = 720 s: k 2 = 2.303/(720 s ) log 10 (35.0/7.0) = 2.303/(720 s ) log 10(5.0) log 10(5.0) = 0.6990 k 2 = (2.303 × 0.6990)/(720 s ) = 1.60979/(720 s ) = 2.236 × 10^-3 s^-1 Checking consistency, then averaging k 1 and k 2 come out close to each other (2.175 and 2.236, both × 10^-3 s^-1 ), which confirms the reaction really is first order — a constant k from two independent data points is the whole point of testing it this way, not an assumption made in advance. The reported rate constant is their average: k = (k 1 + k 2)/2 = ((2.175 + 2.236)× 10^-3)/2 s^-1 = 2.206 × 10^-3 s^-1 Rounding to three significant figures, consistent with the precision of the pressure data: Answer: k ≈ 2.20 × 10^-3 s^-1

Question 21

The following data were obtained during the first order thermal decomposition of SO 2Cl 2 at a constant volume. SO Cl () g → SO () g + Cl () g 2 2 2 2 Experiment Time/ s^-1 Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.

Accepted Answer

2.23 × 10 -3 s -1, 7.8 ×10 -4 atm s -1. For a gas-phase reaction run at constant volume and temperature, pressure is directly proportional to moles — so you can track the reaction using partial pressures exactly as you would use concentrations. The reaction is SO 2Cl 2(g) → SO 2(g) + Cl 2(g) Let the initial pressure of SO 2Cl 2 be p 0, and let x be the drop in its pressure by time t. Since one mole of SO 2Cl 2 gives one mole each of SO 2 and Cl 2: SO 2Cl 2 SO 2 Cl 2 --- --- --- --- at t=0 p 0 0 0 at t p 0-x x x The total pressure at time t is the sum of all three partial pressures: p t = (p 0-x) + x + x = p 0 + x ⇒ x = p t - p 0 So the partial pressure of the reactant that is actually left is p(SO 2Cl 2) = p 0 - x = p 0 - (p t-p 0) = 2p 0 - p t This is the step people skip — the total pressure given in the table is not the reactant's pressure; you have to subtract off what the products have contributed. Step 1: find k from the first data point. Here p 0 = 0.5 atm (the pressure at t=0, Experiment 1). For a first-order reaction, using pressure in place of concentration: k = 2.303/t 0/(p(SO 2Cl 2)) = 2.303/t 0/2p 0-p t From Experiment 2, t = 100 s, p t = 0.6 atm: p(SO 2Cl 2) = 2(0.5) - 0.6 = 0.4 atm k = 2.303/(100 s ) log 0.5/0.4 = 2.303/(100 s ) log (1.25) = (2.303 × 0.0969)/(100 s ) k = 2.232×10^-3 s^-1 Step 2: find the reactant's pressure when the total pressure is 0.65 atm. Using the same relation p(SO 2Cl 2) = 2p 0 - p t with p t = 0.65 atm: p(SO 2Cl 2) = 2(0.5) - 0.65 = 0.35 atm Step 3: get the rate from the first-order rate law. For a first-order reaction, Rate = k × [ reactant ]; with pressure standing in for concentration, Rate = k × p(SO 2Cl 2) Rate = (2.232×10^-3 s^-1)×(0.35 atm ) Rate = 7.81×10^-4 atm s^-1 Because concentration here is expressed as a partial pressure (atm) rather than mol L⁻¹, the rate correctly carries units of atm s⁻¹, not mol L⁻¹ s⁻¹ — don't relabel the unit out of habit. Answer: Rate ≈ 7.81×10^-4 atm s^-1 (with k ≈ 2.23×10^-3 s^-1 ).

Question 22

The rate constant for the decomposition of N 2O 5 at various temperatures is given below: T/°C 0 20 40 60 80 10^5 × k/ s^-1 0.0787 1.70 25.7 178 2140 Draw a graph between ln k and 1/T and calculate the values of A and E a. Predict the rate constant at 30° and 50°C.

Accepted Answer

The Arrhenius equation turns into a straight line only after you take logs — plot ln k (y-axis) against 1/T (x-axis), and the slope gives E a, the intercept gives A. Starting from the Arrhenius equation k = A e^-E a/RT taking the natural log of both sides gives ln k = ln A - E a/R·1/T This is a straight line y = mx + c with y= ln k, x = 1/T, slope m = -E a/R, and intercept c = ln A. So a graph of ln k vs 1/T lets you read off E a and A without ever plugging numbers into the exponential form directly. Before plotting anything, fix the two traps in this data: temperature must be in kelvin (not °C), and the tabulated numbers are 10^5× k, not k itself. T/ K = T/^° C + 273, and k = ( tabulated value )×10^-5 s^-1. Working these out, and 1/T and ln k alongside them: T/^° C T/ K k/ s^-1 10^3×(1/T)/ K^-1 ln k --- --- --- --- --- 0 273 7.87×10^-7 3.663 -14.055 20 293 1.70×10^-5 3.413 -10.982 40 313 2.57×10^-4 3.195 -8.266 60 333 1.78×10^-3 3.003 -6.331 80 353 2.14×10^-2 2.833 -3.844 Plotting these five (1/T, ln k) points gives a straight line falling from upper left (low T, very negative ln k ) to lower right (high T, less negative ln k ) — a negative slope, exactly as the Arrhenius equation predicts, since k grows as T rises. The slope of the best-fit line through those points is what you're actually reading off the graph — here it's obtained from the least-squares slope formula, m = (Σ i (x i- x)(y i- y))/(Σ i (x i- x)^2), x i = 1 T i, y i= ln k i The means of the five points are x = 3.2213×10^-3 K^-1 and y = -8.6959. Carrying these through the sums gives m = -1.2103×10^4 K and the intercept c = y - m x = -8.6959 - (-1.2103×10^4)(3.2213×10^-3) = 30.293 Reading E a off the slope: m=-E a/R, so E a=-mR. With R = 8.314 J mol^-1K^-1, E a = -m R = (1.2103×10^4 K )(8.314 J mol^-1K^-1 ) = 1.006×10^5 J mol^-1 E a ≈ 101 kJ mol^-1 (This lands close to the accepted experimental activation energy for N 2O 5 decomposition, which is a useful sanity check that the line was read correctly.) Reading A off the intercept: c = ln A, so A = e^c. A = e^30.293 = 1.43×10^13 s^-1 A pre-exponential factor of this order ( 10^13 s^-1 ) is typical for a unimolecular gas-phase decomposition — it's roughly the frequency of a bond vibration, which is exactly the physical picture behind A for this kind of reaction. Predicting k at a temperature not in the table just means reading a point on the same line — plug T into the fitted equation, don't re-derive anything. ln k = c + m· 1T = 30.293 - (1.2103×10^4)· 1T At 30^° C = 303 K: 1/T = 3.3003×10^-3 K^-1 ln k = 30.293 - (1.2103×10^4)(3.3003×10^-3) = -9.652 k = e^-9.652 = 6.43×10^-5 s^-1 At 50^° C = 323 K: 1/T = 3.0960×10^-3 K^-1 ln k = 30.293 - (1.2103×10^4)(3.0960×10^-3) = -7.178 k = e^-7.178 = 7.63×10^-4 s^-1 Both predictions sit exactly where they should: the value at 30^° C falls between the measured k at 20^° C ( 1.70×10^-5 ) and 40^° C ( 2.57×10^-4 ), and the value at 50^° C falls between 40^° C ( 2.57×10^-4 ) and 60^° C ( 1.78×10^-3 ) — confirming the line was fitted correctly. Answer: E a ≈ 101 kJ mol^-1 and A ≈ 1.43×10^13 s^-1; at 30^° C, k ≈ 6.43×10^-5 s^-1, and at 50^° C, k ≈ 7.63×10^-4 s^-1.

Question 23

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Accepted Answer

3.9 × 10 12 s -1. The Arrhenius equation is the bridge between the rate constant you measure and the pre-exponential factor you're asked for — solve it for A instead of k. The Arrhenius equation is k = A e^-E a/RT where k is the rate constant, A is the pre-exponential (frequency) factor, E a is the activation energy, R = 8.314 J K^-1 mol^-1 is the gas constant, and T is the absolute temperature. Taking natural logs of both sides and switching to base-10 logarithms (dividing by ln 10 = 2.303 ) puts it in the form you actually calculate with: ln k = ln A - E a/RT ⇒ log A = log k + E a/(2.303 RT) The activation energy is given in kJ/mol but R is in J — convert before you divide, or the exponent is off by a factor of 1000. E a = 179.9 kJ mol^-1 = 1.799×10^5 J mol^-1 The remaining data: k = 2.418×10^-5 s^-1, T = 546 K. Step 1 — evaluate 2.303RT: 2.303 × 8.314 J K^-1 mol^-1 × 546 K = 1.0454×10^4 J mol^-1 Step 2 — evaluate E a/(2.303RT): (1.799×10^5 J mol^-1)/(1.0454×10^4 J mol^-1) = 17.208 Notice the units of energy cancel top and bottom, leaving a pure number — exactly what a logarithm needs to be. Step 3 — evaluate log k: log (2.418×10^-5) = log (2.418) + log (10^-5) = 0.3835 - 5 = -4.6165 Step 4 — add them to get log A: log A = -4.6165 + 17.208 = 12.592 Step 5 — undo the logarithm. Since A = 10^ log A, A = 10^12.592 = 10^12×10^0.592 = 10^12× 3.905 = 3.905×10^12 A carries the same units as k (the Arrhenius equation only multiplies A by a dimensionless exponential), so here A is in s^-1, the same as this first-order rate constant — not "per mole" or any other unit, since nothing in the exponential has units. Rounding to three significant figures, matching the precision of the given k and E a: Answer: A ≈ 3.91×10^12 s^-1

Question 24

Consider a certain reaction A → Products with k = 2.0 × 10 - 2s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.

Accepted Answer

0.135 M. For a first-order reaction, the concentration left after time t follows an exponential decay law — you cannot subtract the amount reacted linearly. The rate law for a first-order reaction A → Products is ln ([A] 0)/([A] t) = kt where [A] 0 is the initial concentration, [A] t is the concentration remaining at time t, and k is the rate constant. This form comes directly from integrating -(d[A])/dt = k[A], and it only holds because the reaction is first order — a key aside is that you must check the order before reaching for this formula, since a second-order reaction would need 1/([A] t) - 1/([A] 0) = kt instead. Substituting the given values Here [A] 0 = 1.0 mol L^-1, k = 2.0 × 10^-2 s^-1, and t = 100 s. ln ([A] 0)/([A] t) = kt = (2.0 × 10^-2 s^-1)(100 s ) = 2.0 Notice the units cancel cleanly: s^-1 × s = 1 (dimensionless), which is exactly what a logarithm requires. Solving for [A] t ([A] 0)/([A] t) = e^2.0 [A] t = [A] 0 e^-2.0 = (1.0 mol L^-1) × e^-2.0 Using e^-2.0 = 0.13534, [A] t = 1.0 mol L^-1 × 0.13534 = 0.13534 mol L^-1 A common slip here is to compute kt and then treat it as the fraction reacted rather than the exponent in e^-kt — the concentration remaining is the whole initial amount scaled by e^-kt, not [A] 0(1-kt). Rounding to three significant figures (matching the two sig figs of k extended slightly for the exponential): [A] t ≈ 0.135 mol L^-1 Answer: The concentration of A remaining after 100 s is 0.135 mol L^-1 (obtained from [A] t = [A] 0 e^-kt = 1.0 × e^-2.0 ).

Question 25

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t 1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Accepted Answer

0.158 M. A first-order rate law means the fraction of reactant left depends only on kt, never on the starting concentration — so you don't need to know how much sucrose you started with. Step 1: Get the rate constant from the half-life. For a first-order reaction the half-life is related to the rate constant k by t 1/2 = 0.693/k Rearranging for k, with t 1/2 = 3.00 h: k = 0.693/(t 1/2) = 0.693/(3.00 h ) = 0.231 h^-1 Step 2: Use the first-order integrated rate law to relate concentration to time. The integrated first-order rate law is k = 2.303/t log ([R] 0)/([R]) where [R] 0 is the initial concentration of sucrose, [R] is the concentration left after time t, and t is the elapsed time. Here t = 8 h. A common slip: plugging t 1/2 back into this formula instead of the actual elapsed time. The half-life was only used to find k in Step 1 — from here on, t = 8 h is the number that matters. Substituting k = 0.231 h^-1 and t = 8 h: 0.231 h^-1 = 2.303/(8 h ) log ([R] 0)/([R]) Step 3: Solve for the concentration ratio. log ([R] 0)/([R]) = (0.231 × 8)/2.303 = 1.848/2.303 = 0.8024 Taking the antilog: ([R] 0)/([R]) = 10^0.8024 = 6.346 Step 4: Invert to get the fraction remaining, not the fraction reacted. The question asks what fraction of sucrose remains, which is [R]/[R] 0, the reciprocal of what Step 3 gives: ([R])/([R] 0) = 1/6.346 = 0.1576 Rounding to three significant figures (matching the precision of the given t 1/2 ): ([R])/([R] 0) ≈ 0.158 So about 15.8% of the original sucrose is still present after 8 hours — consistent with the fact that 8 hours is a little under three half-lives (3 × 3.00 h = 9.00 h), during which the fraction remaining would fall to (1/2)^3 = 0.125; since 8 h is slightly less than 9 h, a fraction slightly above 0.125 is exactly what's expected. Answer: The fraction of sucrose remaining after 8 hours is about 0.158 (15.8%) of the original amount.

Question 26

The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11 s^-1 ) e^-28000 K^/ T Calculate E a.

Accepted Answer

232.79 kJ mol -1. Match the given rate law against the Arrhenius equation term by term — the number in the exponent IS E a/R, nothing more needs to be calculated separately. The Arrhenius equation is k = A e^-E a/RT where A is the pre-exponential (frequency) factor, E a is the activation energy, R is the gas constant, and T is the absolute temperature. The given equation is k = (4.5×10^11 s^-1) e^ -28000 K /T Comparing the two exponents directly, E a/RT = (28000 K )/T The T on both sides cancels, leaving E a/R = 28000 K This is the step people rush past: the "28000" is not E a itself — it is E a/R, so you still have to multiply by R to get an energy. Taking R = 8.314 J K^-1 mol^-1, E a = 28000 K × 8.314 J K^-1 mol^-1 E a = 232792 J mol^-1 Converting to kJ (dividing by 1000, since 1 kJ = 1000 J): E a = 232.792 kJ mol^-1 ≈ 232.79 kJ mol^-1 The pre-exponential factor 4.5×10^11 s^-1 is simply A — it plays no role in finding E a; it would only be needed if you were asked to calculate k at a given T. Answer: E a = 232.79 kJ mol^-1

Question 27

The rate constant for the first order decomposition of H 2O 2 is given by the following equation: log k = 14.34 - 1.25 × 10^4K /T Calculate E a for this reaction and at what temperature will its half-period be 256 minutes?

Accepted Answer

239.339 kJ mol -1. The Arrhenius equation becomes a straight line once you take logarithms — matching the given equation to that line hands you E a directly, and lets you solve backwards for T from any known k. The Arrhenius equation is k = A e^-E a/RT where k is the rate constant, A is the pre-exponential (frequency) factor, E a is the activation energy, R = 8.314 J K^-1 mol^-1 is the gas constant, and T is the absolute temperature in kelvin. Taking log 10 of both sides converts the exponential into a straight line: log k = log A - E a/(2.303 R)×1/T Compare this term by term with the equation given for H 2O 2 decomposition: log k = 14.34 - 1.25×10^4 K ×1/T Matching the two, the slope tells you the activation energy: E a/(2.303 R) = 1.25×10^4 K Step 1 — Find E a E a = 2.303 R × (1.25×10^4 K ) E a = 2.303 × 8.314 J K^-1 mol^-1 × 1.25×10^4 K E a = 19.147 J K^-1 mol^-1 × 1.25×10^4 K = 2.3934×10^5 J mol^-1 E a ≈ 239.3 kJ mol^-1 Step 2 — Turn the half-period into a rate constant, in the SAME time unit the data equation uses The pre-exponential term log A = 14.34 makes A ≈ 2.2×10^14, which is a per-second frequency factor — the kind of value quoted for a unimolecular gas-phase-type decomposition. So the k in this equation is in s^-1, not min^-1. The half-period given is in minutes, so it must be converted to seconds before it can be plugged in — mixing the two time units here is the step that silently wrecks the answer. t 1/2 = 256 min × 60 s min^-1 = 1.536×10^4 s For a first-order reaction, the half-life doesn't depend on concentration: t 1/2 = 0.693/k ⇒ k = 0.693/(t 1/2) k = 0.693/(1.536×10^4 s ) = 4.512×10^-5 s^-1 Step 3 — Feed this k back into the original equation and solve for T log (4.512×10^-5) = 14.34 - 1.25×10^4 K ×1/T -4.346 = 14.34 - (1.25×10^4 K )/T Rearranging to isolate the 1/T term: (1.25×10^4 K )/T = 14.34 - (-4.346) = 18.686 T = (1.25×10^4 K )/18.686 = 668.96 K Rounding to three significant figures (consistent with the precision of the given constants), T ≈ 669 K Answer: E a ≈ 239.3 kJ mol^-1; the half-period becomes 256 minutes at T ≈ 669 K (668.96 K).

Question 28

The decomposition of A into product has value of k as 4.5 × 10^3 s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10^4 s^-1?

Accepted Answer

24°C -1 MnO−. When k changes with temperature, use the two-point Arrhenius equation — it lets you find one unknown (here, T 2 ) without ever calculating the pre-exponential factor A. The Arrhenius equation is k = A e^-E a/RT where A is the frequency factor, E a is the activation energy, R is the gas constant, and T is the absolute temperature. Taking this at two temperatures and dividing eliminates A, giving the two-point form: 2/k 1 = E a/(2.303 R)(1/T 1-1/T 2) List what's given, and convert everything to consistent units first — this is the step most solutions get wrong, using °C directly or leaving E a in kJ. k 1 = 4.5×10^3 s^-1, T 1 = 10 + 273 = 283 K k 2 = 1.5×10^4 s^-1, E a = 60 kJ mol^-1 = 6.0×10^4 J mol^-1 R = 8.314 J K^-1 mol^-1 Compute the rate ratio. k 2/k 1 = (1.5×10^4)/(4.5×10^3) = 3.333 log (3.333) = 0.5229 Substitute into the Arrhenius expression and solve for T 2. 0.5229 = (6.0×10^4)/(2.303 × 8.314)(1/283-1/T 2) First evaluate the coefficient: E a/(2.303 R) = (6.0×10^4)/19.147 = 3133.6 K So 0.5229 = 3133.6(1/283-1/T 2) 1/283-1/T 2 = 0.5229/3133.6 = 1.6686×10^-4 K^-1 Now isolate 1/T 2: 1/T 2 = 1/283 - 1.6686×10^-4 = 3.5336×10^-3 - 0.16686×10^-3 = 3.3667×10^-3 K^-1 Invert to get T 2 — not 1/T 2, which is the value people mistakenly report. T 2 = 1/(3.3667×10^-3) = 297.0 K That's 297 - 273 = 24^° C — only a 14° rise, because the exponential dependence of k on T means even a small temperature increase multiplies the rate constant several-fold. Answer: T 2 ≈ 297 K (about 24^° C )

Question 29

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10^10 s^-1. Calculate k at 318K and E a.

Accepted Answer

E a = 76.750 kJ mol -1, k = 0.9965 × 10 -2 s. For a first-order reaction, k = 2.303/t /a-x^- and because the same time t is used for two different extents of reaction at two different temperatures, that t cancels out completely. You never need to know its actual value. Here k is the rate constant, t the elapsed time, a the initial amount taken as 100 (so percentages read off directly), and a-x the amount left unreacted. Step 1 — Write k at each temperature using the first-order law. At 298 K, 10% has reacted, so a=100, a-x = 90: k 298 = 2.303/t log 100/90 At 308 K, the same t gives 25% completion, so a-x = 75: k 308 = 2.303/t log 100/75 This pairing is the part to get right: 10% goes with the lower temperature and 25% with the higher one, because only that way can the same t produce both extents — a faster reaction (higher T ) covers more ground in the same time. Step 2 — Divide the two equations; t and 2.303 cancel. (k 298)/(k 308) = ( log (100/90))/( log (100/75)) = 0.04576/0.12494 = 0.3662 so (k 308)/(k 298) = 1/0.3662 = 2.730 Step 3 — Feed this ratio into the two-temperature Arrhenius equation to get E a. log (k 308)/(k 298) = E a/2.303R(1/T 1-1/T 2) with R = 8.314 J mol^-1 K^-1, T 1 = 298 K, T 2 = 308 K. log (2.730) = 0.4362, 1/298-1/308 = 308-298/(298×308) = 10/91784 = 1.090×10^-4 K^-1 Substituting: 0.4362 = E a/(2.303×8.314)×(1.090×10^-4) E a = (0.4362×2.303×8.314)/(1.090×10^-4) = 76664 J mol^-1 ≈ 76.7 kJ mol^-1 Step 4 — Use this E a with the given A in the Arrhenius equation to find k at 318 K. log k = log A - E a/2.303RT with A = 4×10^10 s^-1 and T = 318 K: log A = log (4×10^10) = 10.602 E a/2.303RT = 76664/(2.303×8.314×318) = 76664/6088.8 = 12.591 log k 318 = 10.602 - 12.591 = -1.989 k 318 = 10^-1.989 = 1.03×10^-2 s^-1 Since this is a first-order reaction, k carries units of s^-1 — the same units as A — with no concentration unit attached, unlike second-order rate constants. Answer: E a ≈ 76.7 kJ mol^-1, and k at 318 K ≈ 1.03×10^-2 s^-1.

Question 30

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Accepted Answer

52.8 kJ mol -1. The Arrhenius equation connects a change in rate constant to a change in temperature through the activation energy — this is the two-temperature (integrated) form, so you don't need the pre-exponential factor A at all. ln (k 2/k 1) = E a/R(1/T 1 - 1/T 2) where - k 1, k 2 are the rate constants at temperatures T 1 and T 2, - E a is the activation energy (in J mol⁻¹, to match R ), - R = 8.314 J K^-1 mol^-1 is the gas constant, - T 1, T 2 are absolute temperatures in kelvin. A step people rush past: "rate quadruples" means k 2/k 1 = 4, not that the rate itself is 4 — for a reaction whose rate law has the form rate = k[ reactant ], the rate ratio and the k ratio are the same number, so this substitution is valid. Given: T 1 = 293 K, T 2 = 313 K, k 2/k 1 = 4 Left side — evaluate the logarithm. ln (4) = 1.3863 Right side — the temperature term. Keep this as an exact fraction rather than subtracting two rounded decimals, since 1/293 and 1/313 are both small numbers whose difference otherwise loses precision: 1/T 1 - 1/T 2 = (T 2 - T 1)/(T 1 T 2) = (313 - 293)/(293 × 313) = 20/91709 K^-1 = 2.1808 × 10^-4 K^-1 Now solve for E a: E a = (R ln (k 2/k 1))/(1/T 1 - 1/T 2) Substituting: E a = ((8.314 J K^-1 mol^-1)(1.3863))/(2.1808 × 10^-4 K^-1) E a = (11.526 J K^-1 mol^-1)/(2.1808 × 10^-4 K^-1) = 52850 J mol^-1 Converting to kJ (the unit activation energies are conventionally reported in) and rounding once, to three significant figures — matching the precision of the given temperatures: E a = 52850 J mol^-1 = 52.9 kJ mol^-1 Answer: The activation energy of the reaction is E a ≈ 52.9 kJ mol^-1 ( ≈ 5.29 × 10^4 J mol^-1 ).

Experiment	\(\displaystyle [A]\)/mol L\(\displaystyle ^{-1}\)	\(\displaystyle [B]\)/mol L\(\displaystyle ^{-1}\)	\(\displaystyle r_{0}\)/mol L\(\displaystyle ^{-1}\)s\(\displaystyle ^{-1}\)
$\displaystyle 1$	$\displaystyle 0.20$	$\displaystyle 0.30$	\(\displaystyle 5.07\times10^{-5}\)
$\displaystyle 2$	$\displaystyle 0.20$	$\displaystyle 0.10$	\(\displaystyle 5.07\times10^{-5}\)
$\displaystyle 3$	$\displaystyle 0.40$	$\displaystyle 0.05$	\(\displaystyle 1.43\times10^{-4}\)

Experiment	\(\displaystyle [A]\)/mol L\(\displaystyle ^{-1}\)	\(\displaystyle [B]\)/mol L\(\displaystyle ^{-1}\)	Initial rate/mol L\(\displaystyle ^{-1}\)min\(\displaystyle ^{-1}\)
I	$\displaystyle 0.1$	$\displaystyle 0.1$	\(\displaystyle 2.0\times10^{-2}\)
II	$\displaystyle 0.2$	$\displaystyle 0.2$	\(\displaystyle 4.0\times10^{-2}\)
III	$\displaystyle 0.4$	$\displaystyle 0.4$	\(\displaystyle 8.0\times10^{-2}\)
IV	$\displaystyle 0.1$	$\displaystyle 0.2$	\(\displaystyle 2.0\times10^{-2}\)

\(\displaystyle t\)/s	$\displaystyle 0$	$\displaystyle 400$	$\displaystyle 800$	$\displaystyle 1200$	$\displaystyle 1600$	$\displaystyle 2000$	$\displaystyle 2400$	$\displaystyle 2800$	$\displaystyle 3200$
\(\displaystyle [N_2O_5]\)/mol L\(\displaystyle ^{-1}\)	$\displaystyle 0.0163$	$\displaystyle 0.0136$	$\displaystyle 0.0114$	$\displaystyle 0.00930$	$\displaystyle 0.00780$	$\displaystyle 0.00640$	$\displaystyle 0.00530$	$\displaystyle 0.00430$	$\displaystyle 0.00350$

\(\displaystyle t\)/s	$\displaystyle 0$	$\displaystyle 400$	$\displaystyle 800$	$\displaystyle 1200$	$\displaystyle 1600$	$\displaystyle 2000$	$\displaystyle 2400$	$\displaystyle 2800$	$\displaystyle 3200$
\(\displaystyle \log[N_2O_5]\)	\(\displaystyle -1.788\)	\(\displaystyle -1.866\)	\(\displaystyle -1.943\)	\(\displaystyle -2.032\)	\(\displaystyle -2.108\)	\(\displaystyle -2.194\)	\(\displaystyle -2.276\)	\(\displaystyle -2.367\)	\(\displaystyle -2.456\)

	\(\displaystyle SO_2Cl_2\)	\(\displaystyle SO_2\)	\(\displaystyle Cl_2\)
at \(\displaystyle t=0\)	\(\displaystyle p_0\)	$\displaystyle 0$	$\displaystyle 0$
at \(\displaystyle t\)	\(\displaystyle p_0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

NCERT Solutions · Class 12 Chemistry Chemical Kinetics

Exercises 3.1–3.10 (part 1 of 3)

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

Exercise 3.8

Exercise 3.9

Exercise 3.10

Exercises 3.11–3.20 (part 2 of 3)

Exercise 3.11

Exercise 3.12

Exercise 3.13

Exercise 3.14

Exercise 3.15

Exercise 3.16

Exercise 3.17

Exercise 3.18

Exercise 3.19

Exercise 3.20

Exercises 3.21–3.30 (part 3 of 3)

Exercise 3.21

Exercise 3.22

Exercise 3.23

Exercise 3.24

Exercise 3.25

Exercise 3.26

Exercise 3.27

Exercise 3.28

Exercise 3.29

Exercise 3.30

\(\displaystyle T/^{\circ}\mathrm{C}\)	\(\displaystyle T/\mathrm{K}\)	\(\displaystyle k/\mathrm{s^{-1}}\)	\(\displaystyle 10^{3}\times(1/T)/\mathrm{K^{-1}}\)	\(\displaystyle \ln k\)
$\displaystyle 0$	$\displaystyle 273$	\(\displaystyle 7.87\times10^{-7}\)	$\displaystyle 3.663$	\(\displaystyle -14.055\)
$\displaystyle 20$	$\displaystyle 293$	\(\displaystyle 1.70\times10^{-5}\)	$\displaystyle 3.413$	\(\displaystyle -10.982\)
$\displaystyle 40$	$\displaystyle 313$	\(\displaystyle 2.57\times10^{-4}\)	$\displaystyle 3.195$	\(\displaystyle -8.266\)
$\displaystyle 60$	$\displaystyle 333$	\(\displaystyle 1.78\times10^{-3}\)	$\displaystyle 3.003$	\(\displaystyle -6.331\)
$\displaystyle 80$	$\displaystyle 353$	\(\displaystyle 2.14\times10^{-2}\)	$\displaystyle 2.833$	\(\displaystyle -3.844\)