NCERT Solutions · Class 12 Chemistry Chemical Kinetics

The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law, and the units of \(\displaystyle k\) always adjust so that both sides of the rate equation come out in mol L\(\displaystyle ^{-1}\) s\(\displaystyle ^{-1}\). Once you know the order \(\displaystyle n\), the dimensions of \(\displaystyle k\) follow from\[\text{Rate} = k[\text{conc.}]^{n} \quad\Rightarrow\quad k = \frac{\text{Rate}}{[\text{conc.}]^{n}} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{n}} \]Here Rate is always mol L\(\displaystyle ^{-1}\) s\(\displaystyle ^{-1}\) — 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) \(\displaystyle 3\text{NO}(g) \rightarrow \text{N}_2\text{O}(g)\), Rate \(\displaystyle = k[\text{NO}]^2\)The exponent on \(\displaystyle [\text{NO}]\) is $\displaystyle 2$, so this is a second-order reaction (order $\displaystyle 2$ in NO, and $\displaystyle 2$ 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 \(\displaystyle k\): \[k = \frac{\text{Rate}}{[\text{NO}]^{2}} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{2}} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol}^{2}\text{L}^{-2}} = \text{mol}^{-1}\,\text{L}\,\text{s}^{-1} \]So \(\displaystyle k\) has units \(\displaystyle \text{L mol}^{-1}\text{s}^{-1}\) (equivalently \(\displaystyle \text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1}\)).(ii) \(\displaystyle \text{H}_2\text{O}_2(aq) + 3\text{I}^-(aq) + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O}(l) + \text{I}_3^-\), Rate \(\displaystyle = k[\text{H}_2\text{O}_2][\text{I}^-]\)The exponents are $\displaystyle 1$ on \(\displaystyle [\text{H}_2\text{O}_2]\) and $\displaystyle 1$ on \(\displaystyle [\text{I}^-]\), so the order is \(\displaystyle 1+1 = 2\) overall (first order in each reactant, second order overall). \(\displaystyle \text{H}^+\) does not appear in the rate law at all, so it contributes nothing to the order.Dimensions of \(\displaystyle k\): \[k = \frac{\text{Rate}}{[\text{H}_2\text{O}_2][\text{I}^-]} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})(\text{mol L}^{-1})} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol}^{2}\text{L}^{-2}} = \text{mol}^{-1}\,\text{L}\,\text{s}^{-1} \]So \(\displaystyle k\) again has units \(\displaystyle \text{L mol}^{-1}\,\text{s}^{-1}\).(iii) \(\displaystyle \text{CH}_3\text{CHO}(g) \rightarrow \text{CH}_4(g) + \text{CO}(g)\), Rate \(\displaystyle = k[\text{CH}_3\text{CHO}]^{3/2}\)The exponent is \(\displaystyle 3/2\), so this is a reaction of order \(\displaystyle 1.5\) (three-halves order) — orders are not required to be whole numbers; they are read off the experimental rate law exactly as written.Dimensions of \(\displaystyle k\): \[k = \frac{\text{Rate}}{[\text{CH}_3\text{CHO}]^{3/2}} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{3/2}} = \text{mol}^{1-\frac{3}{2}}\,\text{L}^{-1+\frac{3}{2}}\,\text{s}^{-1} = \text{mol}^{-1/2}\,\text{L}^{1/2}\,\text{s}^{-1} \]So \(\displaystyle k\) has units \(\displaystyle \text{L}^{1/2}\,\text{mol}^{-1/2}\,\text{s}^{-1}\).(iv) \(\displaystyle \text{C}_2\text{H}_5\text{Cl}(g) \rightarrow \text{C}_2\text{H}_4(g) + \text{HCl}(g)\), Rate \(\displaystyle = k[\text{C}_2\text{H}_5\text{Cl}]\)The exponent on the single concentration term is $\displaystyle 1$, so the reaction is first order overall.Dimensions of \(\displaystyle k\): \[k = \frac{\text{Rate}}{[\text{C}_2\text{H}_5\text{Cl}]} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1} \]So \(\displaystyle k\) has units \(\displaystyle \text{s}^{-1}\) — a plain reciprocal time, the hallmark of first-order kinetics (this is why first-order \(\displaystyle k\) never carries a concentration unit, unlike every other order above).Answer: (i) order $\displaystyle 2$, \(\displaystyle k\) in L mol\(\displaystyle ^{-1}\) s\(\displaystyle ^{-1}\); (ii) order $\displaystyle 2$, \(\displaystyle k\) in L mol\(\displaystyle ^{-1}\) s\(\displaystyle ^{-1}\); (iii) order $\displaystyle 3$/$\displaystyle 2$, \(\displaystyle k\) in L\(\displaystyle ^{1/2}\) mol\(\displaystyle ^{-1/2}\) s\(\displaystyle ^{-1}\); (iv) order $\displaystyle 1$, \(\displaystyle k\) in s\(\displaystyle ^{-1}\).

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 \(\displaystyle 2A + B \rightarrow A_2B\), the rate law is given as\[\text{Rate} = k[A][B]^{2} \]where \(\displaystyle k = 2.0 \times 10^{-6}\ \text{mol}^{-2}\,\text{L}^{2}\,\text{s}^{-1}\) is the rate constant, and \(\displaystyle [A]\), \(\displaystyle [B]\) are the molar concentrations of A and B at the instant the rate is measured.Step $\displaystyle 1$ — Initial rateSubstitute the given initial concentrations, \(\displaystyle [A] = 0.1\ \text{mol L}^{-1}\) and \(\displaystyle [B] = 0.2\ \text{mol L}^{-1}\):\[\text{Rate}_1 = (2.0 \times 10^{-6}\ \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}) \times (0.1\ \text{mol L}^{-1}) \times (0.2\ \text{mol L}^{-1})^{2} \]\[\text{Rate}_1 = (2.0 \times 10^{-6}) \times (0.1) \times (0.04)\ \text{mol L}^{-1}\text{s}^{-1} = 8.0 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1} \]Step $\displaystyle 2$ — Find how much B has been used up when A drops to $\displaystyle 0.06$ mol L⁻¹This is the step people skip: you are not told \(\displaystyle [B]\) at the later instant directly — you have to get it from the stoichiometry of the balanced equation, not just plug in \(\displaystyle [A]=0.06\) and leave \(\displaystyle [B]\) unchanged.The amount of A consumed is\[\Delta[A] = 0.1 - 0.06 = 0.04\ \text{mol L}^{-1} \]From the equation \(\displaystyle 2A + B \rightarrow A_2B\), $\displaystyle 2$ mol of A react with every $\displaystyle 1$ mol of B, so B is consumed at half the rate A is consumed:\[\Delta[B] = \frac{1}{2}\Delta[A] = \frac{1}{2}(0.04\ \text{mol L}^{-1}) = 0.02\ \text{mol L}^{-1} \]So the new concentration of B is\[[B]_{\text{new}} = 0.2 - 0.02 = 0.18\ \text{mol L}^{-1} \]Step $\displaystyle 3$ — Rate at the new concentrations\[\text{Rate}_2 = k[A]_{\text{new}}[B]_{\text{new}}^{2} = (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^{2}\ \text{mol L}^{-1}\text{s}^{-1} \]\[(0.18)^{2} = 0.0324 \]\[\text{Rate}_2 = (2.0 \times 10^{-6}) \times (0.06) \times (0.0324)\ \text{mol L}^{-1}\text{s}^{-1} = 3.888 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1} \]Rounding to three significant figures, consistent with the data given:\[\text{Rate}_2 \approx 3.89 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1} \]Answer: Initial rate = \(\displaystyle 8.0 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1}\); rate when \(\displaystyle [A] = 0.06\ \text{mol L}^{-1}\) (so \(\displaystyle [B] = 0.18\ \text{mol L}^{-1}\)) = \(\displaystyle 3.89 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1}\).

For a zero-order reaction, the rate equals the rate constant \(\displaystyle k\) 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\[2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g) \]For any reaction \(\displaystyle aA \rightarrow bB + cC\), 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:\[\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} \]Here \(\displaystyle a = 2\) (for \(\displaystyle NH_3\)), \(\displaystyle b = 1\) (for \(\displaystyle N_2\)), \(\displaystyle c = 3\) (for \(\displaystyle H_2\)). So\[\text{Rate} = -\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt} \]Because the reaction is zero order, the rate law is\[\text{Rate} = k[NH_3]^0 = k \]The units given for \(\displaystyle k\), \(\displaystyle mol\,L^{-1}\,s^{-1}\), confirm this — a zero-order rate constant always carries the same units as the rate itself (this is the aside worth noting: the "\(\displaystyle mol^{-1}\)" in the question is a printing slip; a rate constant with units of concentration/time, not concentration\(\displaystyle ^{-1}\)/time, is what makes it zero order). Taking\[k = 2.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1} \]Rate of production of \(\displaystyle N_2\):Since \(\displaystyle \dfrac{d[N_2]}{dt}\) is exactly equal to the overall rate,\[\frac{d[N_2]}{dt} = \text{Rate} = k = 2.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1} \]Rate of production of \(\displaystyle H_2\):From the stoichiometric relation, \(\displaystyle \dfrac{d[H_2]}{dt} = 3 \times \text{Rate}\), because $\displaystyle 3$ moles of \(\displaystyle H_2\) form for every $\displaystyle 1$ mole of \(\displaystyle N_2\) (this is the step people skip — the rate of formation of a product is not automatically equal to \(\displaystyle k\); it is \(\displaystyle k\) multiplied by that product's own stoichiometric coefficient):\[\frac{d[H_2]}{dt} = 3k = 3 \times (2.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1}) = 7.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1} \]Answer: Rate of production of \(\displaystyle N_2 = 2.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1}\); Rate of production of \(\displaystyle H_2 = 7.5 \times 10^{-4}\ mol\,L^{-1}\,s^{-1}\).

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 \[\text{Rate} = k\,(p_{CH_3OCH_3})^{3/2} \] where \(\displaystyle p_{CH_3OCH_3}\) is the partial pressure of dimethyl ether and \(\displaystyle k\) is the rate constant for this pressure-based form of the rate law.Step $\displaystyle 1$: Units of RateRate is the change of a concentration-like quantity per unit time. Here concentration has been replaced by pressure, so \[\text{Rate} = -\frac{dp}{dt} \] Since pressure is measured in bar and time in minutes, the unit of Rate is simply\[\text{unit of Rate} = \text{bar min}^{-1} \]Step $\displaystyle 2$: Units of the rate constant \(\displaystyle k\)Rearrange the rate law to isolate \(\displaystyle k\): \[k = \frac{\text{Rate}}{(p_{CH_3OCH_3})^{3/2}} \]Substitute the units found above for Rate, and bar for pressure: \[\text{unit of } k = \frac{\text{bar min}^{-1}}{(\text{bar})^{3/2}} = \text{bar}^{1-\frac{3}{2}}\,\text{min}^{-1} = \text{bar}^{-1/2}\,\text{min}^{-1} \]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 \(\displaystyle k\), and it must be subtracted correctly (\(\displaystyle 1 - \tfrac{3}{2} = -\tfrac{1}{2}\)), not just copied or dropped. A fractional order (here \(\displaystyle 3/2\)) always gives a fractional power in the unit of \(\displaystyle k\) — that is expected, not a mistake.So the two units are: \[\text{Rate: bar min}^{-1} \] \[k: \text{bar}^{-1/2}\,\text{min}^{-1} \]Answer: Rate has units of bar min⁻¹, and the rate constant k has units of bar⁻¹ᐟ² min⁻¹.

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: \[k = A\,e^{-E_a/RT} \] where \(\displaystyle k \) is the rate constant, \(\displaystyle A \) is the frequency (pre-exponential) factor related to collision frequency and orientation, \(\displaystyle E_a \) is the activation energy, \(\displaystyle R \) is the gas constant, and \(\displaystyle T \) is the absolute temperature (in kelvin — a common slip is to substitute the Celsius value here, which throws off the exponential term badly since \(\displaystyle T \) must be absolute).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 \(\displaystyle E_a \).

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 \(\displaystyle A \), the rate law is\[\text{Rate} = k[A]^2 \]where \(\displaystyle k \) is the rate constant and \(\displaystyle [A] \) is the concentration of the reactant. Call the original rate \(\displaystyle r_1 = k[A]^2 \).(i) Concentration doubledReplace \(\displaystyle [A] \) with \(\displaystyle 2[A] \):\[r_2 = k(2[A])^2 = k \cdot 4[A]^2 = 4\big(k[A]^2\big) = 4r_1 \]The squaring is the step that is easy to miss — doubling the concentration multiplies the rate by \(\displaystyle 2^2 = 4 \), not by 2.(ii) Concentration reduced to halfReplace \(\displaystyle [A] \) with \(\displaystyle \dfrac{[A]}{2} \):\[r_3 = k\left(\frac{[A]}{2}\right)^2 = k \cdot \frac{[A]^2}{4} = \frac{1}{4}\big(k[A]^2\big) = \frac{r_1}{4} \]Halving the concentration cuts the rate to \(\displaystyle \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4} \) of its original value, not to one-half.Answer: Doubling the concentration increases the rate $\displaystyle 4$ times (rate becomes \(\displaystyle 4\times\) the original); reducing the concentration to half decreases the rate to \(\displaystyle \tfrac{1}{4}\) of the original.

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 \(\displaystyle E_a \) (the minimum energy a collision needs to react). Raising the temperature shifts this distribution so a larger fraction of molecules clears the \(\displaystyle E_a \) barrier. It is this growing fraction of "energetic enough" collisions — not a change in collision frequency alone — that makes the rate constant \(\displaystyle k \) rise sharply with \(\displaystyle T \).A rough empirical rule (valid over the ordinary lab temperature range) is that the rate constant of a reaction nearly doubles for every \(\displaystyle 10\ \text{K} \) (or \(\displaystyle 10\,^\circ\text{C} \)) rise in temperature. This is captured by the temperature coefficient, \[\text{Temperature coefficient} = \frac{k_{T+10}}{k_{T}} \approx 2 \text{ to } 3 \] where \(\displaystyle k_T \) and \(\displaystyle k_{T+10} \) are the rate constants at temperature \(\displaystyle T \) and \(\displaystyle T+10\,\text{K} \). 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: A step people get wrong here: \(\displaystyle T \) in this equation must be the absolute (kelvin) temperature, never degrees Celsius — the exponential blows up any Celsius value used by mistake.Why \(\displaystyle k \) rises with \(\displaystyle T \) follows directly from this equation. As \(\displaystyle T \) increases, the exponent \(\displaystyle -E_a/RT \) becomes less negative, so \(\displaystyle e^{-E_a/RT} \) increases, and hence \(\displaystyle k \) increases. The larger \(\displaystyle E_a \) is, the more sensitive \(\displaystyle k \) is to a change in \(\displaystyle T \) — 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: \[\ln k = \ln A - \frac{E_a}{RT} \] or, in base-$\displaystyle 10$ form (more common for graphical work), \[\log k = \log A - \frac{E_a}{2.303\,RT} \] Since this last equation has the form \(\displaystyle y = c + mx \), a plot of \(\displaystyle \log k \) (y-axis) against \(\displaystyle 1/T \) (x-axis) gives a straight line with This is how \(\displaystyle E_a \) and \(\displaystyle A \) are actually determined experimentally: measure \(\displaystyle k \) at several temperatures, plot \(\displaystyle \log k \) vs. \(\displaystyle 1/T \), and read off the slope and intercept.The two-temperature form. If the rate constants \(\displaystyle k_1 \) and \(\displaystyle k_2 \) are known at two temperatures \(\displaystyle T_1 \) and \(\displaystyle T_2 \), subtracting the log form of the Arrhenius equation at \(\displaystyle T_2 \) from that at \(\displaystyle T_1 \) eliminates \(\displaystyle A \): \[\log k_1 = \log A - \frac{E_a}{2.303\,R\,T_1}, \qquad \log k_2 = \log A - \frac{E_a}{2.303\,R\,T_2} \] \[\log k_2 - \log k_1 = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] \[\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] This form is the one most often used in numerical problems: given \(\displaystyle E_a \) and \(\displaystyle k \) at one temperature, it gives \(\displaystyle k \) at another temperature (or, given \(\displaystyle k \) at two temperatures, it gives \(\displaystyle E_a \)) without needing to know \(\displaystyle A \) separately.Answer: For most reactions, the rate constant \(\displaystyle k \) increases with rising temperature (roughly doubling for every \(\displaystyle 10\,\text{K}/10\,^\circ\text{C}\) rise), because a larger fraction of molecules then possesses energy at or above the activation energy \(\displaystyle E_a \). This is represented quantitatively by the Arrhenius equation \(\displaystyle k = A e^{-E_a/RT} \), or in logarithmic form \(\displaystyle \log k = \log A - \dfrac{E_a}{2.303\,RT} \), whose two-temperature version is \(\displaystyle \log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \).

Average rate is just \(\displaystyle -\dfrac{\Delta[A]}{\Delta t}\) — the concentration lost divided by the time it took, no rate law needed yet.The data given are:\[t = 30\ \text{s}, \quad [A] = 0.31\ \text{mol L}^{-1} \] \[t = 60\ \text{s}, \quad [A] = 0.17\ \text{mol L}^{-1} \]For a reactant being consumed, the average rate over an interval is\[\text{Average rate} = -\frac{[A]_2 - [A]_1}{t_2 - t_1} = -\frac{\Delta [A]}{\Delta t} \]where \(\displaystyle [A]_1\) and \(\displaystyle [A]_2\) are the concentrations of A at the start and end of the interval, and the minus sign is there because \(\displaystyle [A]\) 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 $\displaystyle 30$ s to $\displaystyle 60$ s window:\[\Delta [A] = [A]_2 - [A]_1 = 0.17 - 0.31 = -0.14\ \text{mol L}^{-1} \]\[\Delta t = 60 - 30 = 30\ \text{s} \]\[\text{Average rate} = -\frac{(-0.14\ \text{mol L}^{-1})}{30\ \text{s}} = \frac{0.14\ \text{mol L}^{-1}}{30\ \text{s}} \]\[\text{Average rate} = 4.666\ldots \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1} \]This is the average rate over that $\displaystyle 30$-second window, not the instantaneous rate at t = $\displaystyle 30$ s or t = $\displaystyle 60$ s — since the reaction is first order, the true (instantaneous) rate is actually falling continuously as \(\displaystyle [A]\) drops, so this number is really the mean of the (steeper) rate near $\displaystyle 30$ s and the (shallower) rate near $\displaystyle 60$ s.Rounding to three significant figures, matching the precision of the concentration data:\[\text{Average rate} \approx 4.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1} \]Answer: The average rate of the reaction between t = $\displaystyle 30$ s and t = $\displaystyle 60$ s is \(\displaystyle 4.67 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}\).

The order in each reactant tells you the power that concentration is raised to in the rate law — first order in A means power $\displaystyle 1$, second order in B means power 2.(i) The differential rate equationFor a reaction of order $\displaystyle 1$ in A and order $\displaystyle 2$ in B, the rate law is written as\[\text{Rate} = -\frac{d[R]}{dt} = k[A]^{1}[B]^{2} \]where \(\displaystyle k\) is the rate constant, \(\displaystyle [A]\) and \(\displaystyle [B]\) are the molar concentrations of the two reactants, and the exponents ($\displaystyle 1$ and $\displaystyle 2$) are the orders given in the problem, not the stoichiometric coefficients from a balanced equation.(ii) Effect of tripling \(\displaystyle [B]\)Start from the same rate law and replace \(\displaystyle [B]\) by \(\displaystyle 3[B]\), keeping \(\displaystyle [A]\) unchanged:\[\text{Rate}' = k[A]^{1}(3[B])^{2} = k[A](9[B]^{2}) = 9\big(k[A][B]^{2}\big) = 9 \times \text{Rate} \]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 $\displaystyle 3$ gets squared too, giving \(\displaystyle 3^2 = 9\).(iii) Effect of doubling both \(\displaystyle [A]\) and \(\displaystyle [B]\)Replace \(\displaystyle [A]\) by \(\displaystyle 2[A]\) and \(\displaystyle [B]\) by \(\displaystyle 2[B]\) simultaneously:\[\text{Rate}'' = k(2[A])^{1}(2[B])^{2} = k(2[A])(4[B]^{2}) = 8\big(k[A][B]^{2}\big) = 8 \times \text{Rate} \]Here the factor of $\displaystyle 2$ from A contributes once (first order) and the factor of $\displaystyle 2$ from B contributes twice, since B is squared: \(\displaystyle 2^1 \times 2^2 = 2 \times 4 = 8\). 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 \(\displaystyle = k[A][B]^2\); (ii) the rate increases $\displaystyle 9$-fold; (iii) the rate increases $\displaystyle 8$-fold.

The rate law is \(\displaystyle r_{0} = k[A]^{x}[B]^{y} \), 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	\(\displaystyle [A]\)/mol L\(\displaystyle ^{-1}\)	\(\displaystyle [B]\)/mol L\(\displaystyle ^{-1}\)	\(\displaystyle r_{0}\)/mol L\(\displaystyle ^{-1}\)s\(\displaystyle ^{-1}\)
$\displaystyle 1$	$\displaystyle 0.20$	$\displaystyle 0.30$	\(\displaystyle 5.07\times10^{-5}\)
$\displaystyle 2$	$\displaystyle 0.20$	$\displaystyle 0.10$	\(\displaystyle 5.07\times10^{-5}\)
$\displaystyle 3$	$\displaystyle 0.40$	$\displaystyle 0.05$	\(\displaystyle 1.43\times10^{-4}\)

Step $\displaystyle 1$ — order with respect to B.Experiments $\displaystyle 1$ and $\displaystyle 2$ hold \(\displaystyle [A]\) fixed at $\displaystyle 0.20$ mol L\(\displaystyle ^{-1}\), so any change in rate between them must come from \(\displaystyle y\), the order in B:\[\frac{r_{0,1}}{r_{0,2}} = \left(\frac{[B]_1}{[B]_2}\right)^{y} \]Substituting,\[\frac{5.07\times10^{-5}}{5.07\times10^{-5}} = \left(\frac{0.30}{0.10}\right)^{y} \implies 1 = 3^{y} \]The only value that makes \(\displaystyle 3^{y}=1\) is \(\displaystyle y = 0\). Tripling \(\displaystyle [B]\) 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 $\displaystyle 1$ and $\displaystyle 3$, but experiment $\displaystyle 3$ has two things different from experiment $\displaystyle 1$ at once ([A] AND [B]), so that comparison alone can't isolate either order. You need a pair that isolates one variable first.Step $\displaystyle 2$ — order with respect to A.Now that B has zero order, its concentration cancels out of any comparison — experiments $\displaystyle 2$ and $\displaystyle 3$ can be compared for A even though their \(\displaystyle [B]\) values differ, because \(\displaystyle [B]^{0}=1\) regardless of what \(\displaystyle [B]\) is:\[\frac{r_{0,3}}{r_{0,2}} = \left(\frac{[A]_3}{[A]_2}\right)^{x}\left(\frac{[B]_3}{[B]_2}\right)^{0} = \left(\frac{[A]_3}{[A]_2}\right)^{x} \]Substituting the numbers,\[\frac{1.43\times10^{-4}}{5.07\times10^{-5}} = \left(\frac{0.40}{0.20}\right)^{x} \]\[2.821 = 2^{x} \]Taking logarithms of both sides,\[x = \frac{\log(2.821)}{\log(2)} = \frac{0.4504}{0.3010} = 1.496 \approx 1.5 \]Checking this against a clean half-integer: \(\displaystyle 2^{1.5} = 2\sqrt{2} = 2.828\), which reproduces the data (\(\displaystyle 2.828 \times 5.07\times10^{-5} = 1.434\times10^{-4}\), matching the given \(\displaystyle 1.43\times10^{-4}\) to the precision of the data). So \(\displaystyle x = \dfrac{3}{2}\).Step $\displaystyle 3$ — assemble the rate law.\[r_{0} = k[A]^{3/2}[B]^{0} = k[A]^{3/2} \]The order in A is \(\displaystyle 1.5\) (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 \(\displaystyle x+y = 1.5+0 = 1.5\).Answer: order with respect to A = $\displaystyle 1.5$ (i.e., \(\displaystyle 3/2\)); order with respect to B = $\displaystyle 0$; overall order of reaction = 1.5.