Question 1

Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only.

Accepted Answer

(ii). A state function depends only on where the system is, not on how it got there. To pick the right option, check each one against what "independent of path" means. Internal energy U, enthalpy H, entropy S, pressure p, volume V, and temperature T are all state functions: once you fix the state of the system (its pressure, volume, temperature, and composition), each of these has one fixed value, no matter which route was taken to reach that state. Heat q and work w are different — they are path functions. The same overall change in a system (say, from state 1 to state 2) can be brought about reversibly or irreversibly, and q and w individually come out different for each route, even though their sum q + w = Δ U is always the same (this is the first law, and it works only because U is a state function). Now check the four options: (i) "used to determine heat changes" — heat itself is a path function, not a state function, so a quantity defined through heat changes is not automatically state-independent. This does not define a state function. (iii) "used to determine pressure–volume work" — pressure–volume work, w = -p ext Δ V, also depends on the path (for instance, whether the expansion is done in one step or reversibly in many infinitesimal steps against a continuously adjusted pressure gives different w for the same Δ V ). So this cannot be the defining property of a state function either. (iv) "whose value depends on temperature only" — this is too narrow. Enthalpy H, for example, depends on both pressure and composition as well as temperature, yet it is very much a state function. Depending on temperature alone is not what makes a quantity a state function — being path-independent is. (ii) "whose value is independent of path" — this is exactly the definition. A state function's change between two states, Δ X = X final - X initial, depends only on the initial and final states, never on the sequence of steps connecting them. That is the property shared by U, H, S, p, V, and T, and it is the one property that correctly separates state functions from path functions like q and w. Answer: (ii) — a thermodynamic state function is a quantity whose value is independent of path.

Question 2

For the process to occur under adiabatic conditions, the correct condition is: (i) ∆T = 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0

Accepted Answer

(iii). An adiabatic process is defined by zero heat exchange with the surroundings — not by zero temperature change, zero pressure change, or zero work. Start from the first law of thermodynamics, Δ U = q + w where Δ U is the change in internal energy of the system, q is the heat absorbed by the system from its surroundings, and w is the work done on the system. The word "adiabatic" refers specifically to the heat term: an adiabatic process is one carried out with the system thermally insulated from its surroundings, so no heat can enter or leave it. That condition is written as q = 0 Now test the other three options against this definition, one at a time. - Δ T = 0 is the defining condition of an isothermal process, not an adiabatic one. An adiabatic process is usually accompanied by a temperature change: with q=0, the first law reduces to Δ U = w, so if the system does work on the surroundings ( w negative for the system), its internal energy drops, and since internal energy is tied to temperature, the temperature falls. - Δ p = 0 is the defining condition of an isobaric process (constant pressure), which is unrelated to whether heat is exchanged. - w = 0 describes a process in which no work is done — this is characteristic of a rigid, constant-volume container, not of thermal insulation. Under adiabatic conditions, work is generally not zero: with q=0, Δ U = w, meaning it is precisely the work exchanged that accounts for the entire change in internal energy. The step students most often get wrong: "adiabatic" is a statement about heat ( q ), while "no work done" is a separate constraint tied to volume, and the two must not be swapped. A process can be adiabatic and still involve a large amount of work — that is exactly how an adiabatic expansion cools a gas. Only option (iii) matches the definition of an adiabatic process. Answer: (iii) q = 0 — an adiabatic process is one in which no heat is exchanged between the system and its surroundings.

Question 3

The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element

Accepted Answer

(ii). Standard enthalpy of formation of an element in its own standard state is fixed at zero — it is a reference point, not something measured or calculated for each element. Enthalpy H itself can never be measured in an absolute sense — only changes in enthalpy, Δ H, between two states can be measured. So to build a usable scale of enthalpies for compounds, chemists need a common reference (a "sea level") from which every other enthalpy is measured. That reference is the standard enthalpy of formation, Δ f H^: the enthalpy change when 1 mole of a compound is formed from its elements, with every element in its most stable form at the standard state (298 K, 1 bar). By this very definition, forming an element from itself involves no chemical change at all — nothing is transformed, nothing is broken or made — so there is no enthalpy change to record: Δ f H^ ( element, standard state ) = 0 This is a convention fixed by IUPAC, the same way sea level is arbitrarily set to 0 m in geography even though the ocean has a real depth beneath it. It does not mean elements contain "no energy" — it only means their enthalpy is the zero mark against which every compound's enthalpy of formation is measured. A common slip is to think this value should depend on the element (heavier atoms, more electrons, more energy stored), and pick option (iv). But the zero is not a statement about the physical energy content of the element — it is a bookkeeping choice, identical for every element, made purely so that Δ f H^ of compounds can be compared on one common scale. So of the four choices, "unity" (i) and "less than zero" (iii) are not the convention at all, and "different for each element" (iv) misunderstands why the reference is chosen — it must be the same value (zero) for every element to work as a common baseline. Answer: (ii) zero — the standard enthalpy of every element in its standard state is taken as zero by convention, since it forms the reference point from which all other standard enthalpies (of formation) are measured.

Question 4

∆U 0 of combustion of methane is – X kJ mol–1. The value of ∆ H 0 is (i) = ∆U 0 (ii) ∆U 0 (iii) < ∆U 0 (iv) = 0

Accepted Answer

(iii). At constant pressure, Δ H and Δ U differ by the work done by gas moles that appear or disappear — not by chance, but by a fixed formula. For a reaction run at constant temperature, the relation between enthalpy change and internal energy change is Δ H = Δ U + Δ n g RT where Δ n g is the change in the number of moles of gaseous species (moles of gaseous products − moles of gaseous reactants), R is the gas constant, and T is the absolute temperature. This formula comes from H = U + pV and treating the gases as ideal, so pV = n g RT. Write the balanced combustion equation and note the physical states — this is the step people skip. CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2O(l) Water is produced as a liquid at standard conditions, not a gas — only CH 4, O 2, and CO 2 count toward Δ n g. Gaseous moles on the product side: 1 (from CO 2 ) Gaseous moles on the reactant side: 1 + 2 = 3 (from CH 4 and O 2 ) Δ n g = 1 - 3 = -2 Substitute Δ n g into the formula. Δ H^0 = Δ U^0 + (-2)RT = Δ U^0 - 2RT Since R and T are both positive, 2RT is a positive quantity being subtracted. That means Δ H^0 < Δ U^0 regardless of the numerical value of X. (To see the size of the correction: at T = 298 K, 2RT = 2 × 8.314 × 10^-3 kJ mol^-1K^-1 × 298 K ≈ 4.96 kJ mol^-1 — a small but definite shift, always in the direction that makes Δ H^0 more negative than Δ U^0 here.) The combustion is exothermic, so Δ U^0 = -X kJ mol^-1 is already negative; subtracting a further positive number ( 2RT ) makes Δ H^0 even more negative, confirming Δ H^0 < Δ U^0. Answer: (iii) Δ H^0 < Δ U^0, because Δ n g = -2 for this combustion (3 mol gas → 1 mol gas, since water forms as a liquid), giving Δ H^0 = Δ U^0 - 2RT.

Question 5

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH 4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.

Accepted Answer

(i). Hess's Law says enthalpy is a state function — so instead of measuring the formation reaction directly, you can build it by combining reactions you already have data for, as long as the reactants and products add up correctly. The target reaction is the formation of methane from its elements in their standard states: C(graphite) + 2 H 2(g) → CH 4(g), Δ fH^° =? You are given three combustion reactions instead: (1) CH 4(g) + 2 O 2(g) → CO 2(g) + 2 H 2 O (l), Δ cH 1 = -890.3 kJ mol^-1 (2) C(graphite) + O 2(g) → CO 2(g), Δ cH 2 = -393.5 kJ mol^-1 (3) H 2(g) + 1/2 O 2(g) → H 2 O (l), Δ cH 3 = -285.8 kJ mol^-1 The trick people miss: CH₄ is a product of the reaction you want, but a reactant in equation (1) — so equation (1) must be reversed (its sign flipped) when you add it in. Build the target by combining: reaction (2), plus twice reaction (3) (to get 2 H 2 instead of one), minus reaction (1) (to flip CH₄ from reactant to product): C + O 2 → CO 2 (2) + 2 ( H 2 + 1/2 O 2 → H 2 O ) (3) - ( CH 4 + 2 O 2 → CO 2 + 2 H 2 O ) (1) Add up the species on each side: CO 2 appears on both sides and cancels, 2 H 2 O appears on both sides and cancels, and the O 2 totals ( 1 + 2×1/2 = 2 on the left of the sum, 2 from reaction (1) moved to the left) cancel too. What survives is exactly C(graphite) + 2 H 2(g) → CH 4(g) which is the formation reaction you want. Since enthalpy changes add the same way the equations do, this fixes the formula: Δ fH^°( CH 4) = Δ cH 2 + 2Δ cH 3 - Δ cH 1 Substituting the values (all in kJ mol⁻¹, so the units carry straight through without conversion): Δ fH^°( CH 4) = (-393.5) + 2(-285.8) - (-890.3) Δ fH^°( CH 4) = -393.5 - 571.6 + 890.3 Δ fH^°( CH 4) = -74.8 kJ mol^-1 The result is negative, which makes physical sense: forming a stable molecule like methane from its elements releases energy, exactly as combustion (also exothermic) would suggest. Answer: Δ fH^°( CH 4) = -74.8 kJ mol^-1, option (i).

Question 6

A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature

Accepted Answer

(iv). Spontaneity is decided by Δ G, not by Δ S alone. The condition for a reaction to be spontaneous ("possible") at a given temperature is the Gibbs free energy equation Δ G = Δ H - TΔ S where Δ G is the free energy change, Δ H is the enthalpy change, T is the absolute temperature (in kelvin), and Δ S is the entropy change. A reaction runs on its own only when Δ G < 0. Read the equation for what q means. The reaction is written as A + B → C + D + q Heat q appears on the product side, which is the standard way of saying heat is released by the reaction — this is an exothermic reaction, so Δ H < 0 The step people skip is exactly this: a "+q" tacked onto the products is a statement about Δ H, and mistaking it for something about Δ S is what leads to the wrong option here. Now put both signs into the Gibbs equation. We are told separately that the entropy change is positive: Δ S 0 So -TΔ S is a negative quantity at every temperature T 0 (since T is always positive on the Kelvin scale). Substituting the two signs: Δ G = Δ H negative - TΔ S positive = ( negative ) + ( negative ) Both terms on the right push Δ G in the same direction — there is no temperature at which they can cancel to make Δ G positive. Hence Δ G < 0 for every value of T Contrast with the cases that do depend on temperature. A reaction is only temperature-sensitive (spontaneous at high T but not low T, or the reverse) when Δ H and Δ S have opposite signs, because then Δ H and -TΔ S compete. Here they don't compete — exothermic ( Δ H < 0 ) combined with increasing disorder ( Δ S 0 ) makes Δ G negative unconditionally, so raising or lowering T can never flip the sign. Answer: the reaction is possible at any temperature (exothermic with Δ S 0 gives Δ G = Δ H - TΔ S < 0 for all T, since both terms are negative).

Question 7

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Accepted Answer

q = + 701 J w = – 394 J, since work is done by the system ∆U = 307 J. Change in internal energy comes from the First Law: heat absorbed adds to it, work done BY the system subtracts from it. The First Law of Thermodynamics states Δ U = q + w where Δ U is the change in internal energy of the system, q is the heat absorbed by the system, and w is the work done on the system. The sign convention is the step people trip on: heat absorbed by the system is positive, and work done by the system (i.e., the system pushing on its surroundings) is negative, because that energy leaves the system. Here the system absorbs heat, so q = +701 J The system does work on the surroundings, so that energy is lost from the system: w = -394 J Substituting into the First Law: Δ U = q + w = (+701 J ) + (-394 J ) Δ U = 701 J - 394 J = 307 J Both given quantities (701 J, 394 J) are whole numbers of joules, so the result is reported to the same precision, with no decimal places. Answer: Δ U = 307 J (the internal energy of the system increases by 307 J).

Question 8

The reaction of cyanamide, NH 2CN (s), with dioxygen was carried out in a bomb calorimeter, and Δ U was found to be –742.7 kJ mol^-1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH 2CN(g) + 3/2 O 2(g) → N 2(g) + CO 2(g) + H 2O(l)

Accepted Answer

–743.939 kJ. Δ H = Δ U + Δ n g RT, and this whole question is the sign of Δ n g — which counts GASES only, so a solid reactant and a liquid product contribute nothing. The bomb calorimeter measures at constant volume, so what it gives you is Δ U, not Δ H. Converting between them needs the change in the number of moles of GAS: Δ H = Δ U + (Δ n g) RT where Δ n g = (moles of gaseous products) − (moles of gaseous reactants), R = 8.314 J K^-1mol^-1 and T = 298 K, so RT = (8.314 × 298)/1000 = 2.478 kJ mol^-1 Counting the gases. On the product side: N 2(g) and CO 2(g) give 2 moles of gas. H 2O(l) is a LIQUID and contributes nothing — this is the step people lose the mark on. On the reactant side, 3/2 O 2(g) gives 1.5 moles of gas, and cyanamide contributes 1 mole if it is a gas and nothing if it is a solid. This exercise labels the cyanamide two different ways. Its prose calls it NH 2CN (s); the equation printed underneath writes NH 2CN(g). The two readings give opposite signs for Δ n g, so work both and know which you are being asked for. Reading it as the solid (which is what cyanamide is at 298 K — it melts at about 44 °C, and a bomb calorimeter burns a weighed solid pellet): Δ n g = 2 - 1.5 = +0.5 Δ H = -742.7 + (0.5)(2.478) = -742.7 + 1.239 = -741.5 kJ mol^-1 Reading it as the gas, as the printed equation writes it: Δ n g = 2 - (1 + 1.5) = -0.5 Δ H = -742.7 - 1.239 = -743.9 kJ mol^-1 NCERT's answer key prints -743.939 kJ, so the key follows the equation as printed rather than the prose. Nothing else in the calculation moves: only the 1.239 kJ term changes sign. In an exam, answer to the state symbols on the paper in front of you, and show the Δ n g line either way — that line is where the marks are, not the third decimal place. Answer: Δ H = -741.5 kJ mol^-1, taking NH 2CN as the solid it is at 298 K. NCERT's key prints -743.939 kJ, which is the same calculation with the cyanamide read as a gas, as the exercise's own printed equation writes it.

Question 9

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.

Accepted Answer

1.067 kJ. Heat absorbed at constant pressure is q = n C m Δ T, where n is the number of moles, C m is the molar heat capacity, and Δ T is the temperature change. You need moles of aluminium, not grams, because C m is defined per mole. Step 1: Convert mass to moles. The molar mass of aluminium is M = 27.0 g mol^-1. n = ( mass )/M = (60.0 g )/(27.0 g mol^-1) = 2.222 mol Step 2: Find the temperature change. Δ T = 55^° C - 35^° C = 20^° C = 20 K A change of 20^° C equals a change of 20 K — the two scales differ only by the fixed 273.15 offset, which cancels out in a difference. This is different from converting a single temperature reading, where you must add 273. Step 3: Apply q = n C m Δ T. Here C m = 24 J mol^-1 K^-1 is the molar heat capacity of Al (heat needed to raise 1 mole by 1 K). q = (2.222 mol ) × (24 J mol^-1 K^-1) × (20 K ) q = 2.222 × 24 × 20 J = 1066.7 J Step 4: Convert to kJ. q = 1066.7 J × (1 kJ )/(1000 J ) = 1.0667 kJ The given data (60.0 g, 35 °C, 55 °C, 24 J mol⁻¹ K⁻¹) each carry two to three significant figures, so the result is rounded once, at the end, to three significant figures. Answer: q ≈ 1.07 kJ

Question 10

Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C. Cp [H 2O(l)] = 75.3 J mol–1 K–1 Cp [H 2O(s)] = 36.8 J mol–1 K–1

Accepted Answer

∆H = –7.151 kJ mol–1. Enthalpy is a state function, so you can break an odd path (liquid at +10 °C → solid at –10 °C) into a chain of simple steps — cooling, then freezing, then cooling again — and just add up the Δ H of each step, even though water never actually stops at 0 °C on its way down. The direct process — water at 10.0 °C freezing to ice at –10.0 °C — is not something you have data for. But Hess's law says the total enthalpy change depends only on the start and end states, not the route. So build a three-step route between the same two states, using only the numbers you're given: Step 1: cool liquid water from 10.0 °C to 0.0 °C (no phase change) Step 2: freeze the liquid to ice, at 0.0 °C (the only temperature you have Δ fusH for) Step 3: cool the ice from 0.0 °C to –10.0 °C (no phase change) Step 1 — cooling the liquid For a temperature change with no phase change, use q p = n C p Δ T, where n is the number of moles, C p is the molar heat capacity at constant pressure, and Δ T = T final - T initial. Δ H 1 = n C p[ H 2 O (l)] Δ T = (1.0 mol )(75.3 J mol^-1 K^-1)(0.0^° C - 10.0^° C ) Since Δ T is a difference of temperatures, it's the same whether you read it in °C or in K — the 273.15 cancels out. So: Δ H 1 = (1.0)(75.3)(-10.0) J = -753 J = -0.753 kJ Step 2 — freezing at 0 °C You're given Δ fusH = 6.03 kJ mol^-1 — that's for melting (ice → liquid). Freezing is the reverse process, so its enthalpy change is the same size with the opposite sign. This is the step people skip: forgetting to flip the sign for the reverse of fusion gives an answer with the wrong direction of heat flow. Δ H 2 = -Δ fusH = -6.03 kJ mol^-1 × 1.0 mol = -6.03 kJ Step 3 — cooling the ice Same formula as Step 1, now with the solid's heat capacity: Δ H 3 = n C p[ H 2 O (s)] Δ T = (1.0 mol )(36.8 J mol^-1 K^-1)(-10.0^° C - 0.0^° C ) Δ H 3 = (1.0)(36.8)(-10.0) J = -368 J = -0.368 kJ Adding the three steps Because enthalpy is a state function, the sum of these three step-changes equals the enthalpy change for the actual, direct process: Δ H = Δ H 1 + Δ H 2 + Δ H 3 = (-0.753 kJ ) + (-6.03 kJ ) + (-0.368 kJ ) Δ H = -7.151 kJ Every piece of data here carries three significant figures (6.03, 75.3, 36.8, 10.0), so the final value is rounded to three significant figures too: Δ H ≈ -7.15 kJ mol^-1 The negative sign makes physical sense: freezing releases heat (it's exothermic), and cooling the water down before and after also releases heat, so all three steps push in the same direction. Answer: Δ H = -7.15 kJ mol^-1

Question 11

Enthalpy of combustion of carbon to CO 2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO 2 from carbon and dioxygen gas.

Accepted Answer

– 314.8 kJ. Enthalpy of combustion tells you the heat released per mole of substance formed — not per gram. To use it for 35.2 g of CO 2, first convert that mass into moles. The reaction is C(s) + O 2( g ) → CO 2( g ), Δ cH = -393.5 kJ mol^-1 The negative sign means the reaction is exothermic — 393.5 kJ of heat is released for every 1 mole of CO 2 formed. Step 1: Convert mass of CO 2 to moles. Number of moles, n = ( given mass )/( molar mass ), where the molar mass of CO 2 is M( CO 2) = 12 + 2(16) = 44 g mol^-1 This is the step people rush past: you cannot plug 35.2 g directly into an enthalpy given "per mole" — the unit mismatch has to be fixed first. n( CO 2) = (35.2 g )/(44 g mol^-1) = 0.8 mol Step 2: Scale the enthalpy of combustion to this many moles. Heat released, q = n × Δ cH, where n is the moles of CO 2 formed and Δ cH is the enthalpy of combustion per mole. q = 0.8 mol × 393.5 kJ mol^-1 q = 314.8 kJ The data (393.5, four significant figures; 35.2 g, three significant figures) justifies keeping three significant figures in the final answer, so q = 315 kJ. Since combustion is exothermic, this heat is released (given out), consistent with the negative sign on Δ cH. Answer: 315 kJ (314.8 kJ) of heat is released.

Question 12

Enthalpies of formation of CO(g), CO 2(g), N 2O(g) and N 2O 4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction: N 2O 4(g) + 3CO(g) → N 2O(g) + 3CO 2(g)

Accepted Answer

∆rH = –778 kJ. Use Hess's law: Δ r H = Σ Δ f H( products ) - Σ Δ f H( reactants ), each term weighted by its coefficient in the balanced equation. Here Δ f H is the standard enthalpy of formation of each substance — the enthalpy change when 1 mole of that compound forms from its elements in their standard states. This law works because enthalpy is a state function: it doesn't matter whether the reactants go straight to products or are imagined to pass through their elements first, so building products from elements and tearing reactants down to elements (in the reverse direction) must sum to the same Δ r H as the direct reaction. The reaction is N 2 O 4(g) + 3 CO (g) → N 2 O (g) + 3 CO 2(g) The given formation enthalpies are Δ f H[ CO (g)] = -110 kJ mol^-1, Δ f H[ CO 2(g)] = -393 kJ mol^-1 Δ f H[ N 2 O (g)] = 81 kJ mol^-1, Δ f H[ N 2 O 4(g)] = 9.7 kJ mol^-1 The step people get wrong here: every formation enthalpy must be multiplied by its own stoichiometric coefficient from the balanced equation — CO and CO₂ each carry a factor of 3 — before you subtract. Skipping this scaling is the most common way to get this type of question wrong. Applying Hess's law: Δ r H = [1 × Δ f H( N 2 O ) + 3 × Δ f H( CO 2) ] - [1 × Δ f H( N 2 O 4) + 3 × Δ f H( CO ) ] Substitute the values, keeping units of kJ mol⁻¹ attached throughout: Δ r H = [(81 kJ mol^-1) + 3 × (-393 kJ mol^-1) ] - [(9.7 kJ mol^-1) + 3 × (-110 kJ mol^-1) ] Work out each bracket separately. Products bracket: 81 kJ mol^-1 + (-1179 kJ mol^-1) = -1098 kJ mol^-1 Reactants bracket: 9.7 kJ mol^-1 + (-330 kJ mol^-1) = -320.3 kJ mol^-1 Now subtract reactants from products: Δ r H = -1098 kJ mol^-1 - (-320.3 kJ mol^-1) = -1098 kJ mol^-1 + 320.3 kJ mol^-1 Δ r H = -777.7 kJ mol^-1 The data are given to one decimal place at best (9.7 kJ mol⁻¹), so the result is reported to one decimal place, matching the input precision — no further rounding is needed since the arithmetic above is exact. The negative sign shows the reaction releases heat: N₂O₄ and CO together are higher in enthalpy than the N₂O and CO₂ they form, so energy comes out as the bonds rearrange. Answer: Δ r H = -777.7 kJ mol^-1

Question 13

Given N 2(g) + 3H 2(g) → 2NH 3(g); ∆rH0 = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH 3 gas?

Accepted Answer

– 46.2 kJ mol–1. Standard enthalpy of formation is defined per ONE mole of the compound formed from its elements in their standard states — but the equation given here makes 2 mol of NH 3, so you must divide before you can call the number a formation enthalpy. The given thermochemical equation is N 2(g) + 3 H 2(g) → 2 NH 3(g), Δ r H^0 = -92.4 kJ mol^-1 Here Δ r H^0 is the enthalpy change for the reaction exactly as written — that is, for the formation of 2 mol of NH 3(g) from 1 mol N 2(g) and 3 mol H 2(g), both elements already in their standard states. The standard enthalpy of formation, Δ f H^0, of a substance is the enthalpy change when 1 mol of that substance is formed from its elements in their standard states. So: Δ f H^0( NH 3) = (Δ r H^0)/( moles of NH 3 formed ) This is the step people skip: they copy -92.4 kJ mol^-1 straight in as the formation enthalpy, but that number belongs to the reaction producing 2 mol of product, not to 1 mol of NH 3. Substituting the values: Δ f H^0( NH 3) = (-92.4 kJ )/(2 mol ) Δ f H^0( NH 3) = -46.2 kJ mol^-1 The data ( -92.4 ) is given to 3 significant figures, so the halved result is kept to the same precision: -46.2 kJ mol^-1, with no further rounding needed since the division is exact. Answer: Δ f H^0( NH 3) = -46.2 kJ mol^-1

Question 14

Calculate the standard enthalpy of formation of CH 3OH(l) from the following data: CH 3OH (l) + 3/2 O 2(g) → CO 2(g) + 2H 2O(l); Δ rH^ = -726 kJ mol^-1 C(graphite) + O 2(g) → CO 2(g); Δ cH^ = -393 kJ mol^-1 H 2(g) + 1/2 O 2(g) → H 2O(l); Δ fH^ = -286 kJ mol^-1.

Accepted Answer

– 239 kJ mol–1. Hess's Law: enthalpy change depends only on the initial and final states, not on the path. So the enthalpy of formation of CH 3OH(l) can be built by combining the three given equations, even though none of them is that formation reaction by itself. The target reaction — the standard enthalpy of formation of methanol — is the one where 1 mole of CH 3OH(l) is formed from its elements in their standard states: C(graphite) + 2H 2(g) + 1/2O 2(g) → CH 3OH(l), Δ f H^0 =? The three data equations are: (i) CH 3OH(l) + 3/2O 2(g) → CO 2(g) + 2H 2O(l), Δ r H^0 (i) = -726 kJ mol^-1 (ii) C(graphite) + O 2(g) → CO 2(g), Δ c H^0 (ii) = -393 kJ mol^-1 (iii) H 2(g) + 1/2O 2(g) → H 2O(l), Δ f H^0 (iii) = -286 kJ mol^-1 Finding the combination is bookkeeping on atoms, done before touching a single number. The target has CH 3OH(l) as a product, but in (i) it is a reactant — so (i) must be reversed (flip the sign of its Δ H ). The target needs 2 mol of H 2, and (iii) supplies 1 mol of H 2 — so (iii) is used twice. Equation (ii) already matches the carbon term as written, used once. Check this combination reproduces the target: reverse (i), add (ii), add 2× (iii): CO 2(g) + 2H 2O(l) → CH 3OH(l) + 3/2O 2(g) + C(graphite) + O 2(g) → CO 2(g) + 2H 2(g) + O 2(g) → 2H 2O(l) Adding these three, CO 2(g) cancels (appears once on each side), 2H 2O(l) cancels (appears once on each side), and the oxygen on the right totals 3/2 + 1 + 1 = 7/2 mol while the oxygen on the left totals 1 + 1 = 2 mol, leaving 7/2 - 2 = 3/2 mol O 2 on the right side net. So the sum is exactly: C(graphite) + 2H 2(g) + 1/2O 2(g) → CH 3OH(l) which is the target reaction. Good — the same combination applies to the enthalpies. The step people skip: enthalpy is additive only when the equations are added exactly this way — reverse a reaction and you must reverse the sign of Δ H too; scale a reaction by a factor and you must scale Δ H by the same factor. Δ f H^0( CH 3OH, l ) = -Δ r H^0 (i) + Δ c H^0 (ii) + 2 Δ f H^0 (iii) Substituting the values (all in kJ mol ^-1, so no unit conversion is needed): Δ f H^0( CH 3OH, l ) = -(-726) + (-393) + 2(-286) = 726 - 393 - 572 = -239 kJ mol^-1 The negative sign shows the formation of methanol from graphite, hydrogen gas, and oxygen gas is exothermic, releasing 239 kJ of heat per mole of CH 3OH(l) formed. Answer: Δ f H^0( CH 3OH, l ) = -239 kJ mol ^-1

Question 15

Calculate the enthalpy change for the process CCl 4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ∆vapH0(CCl4) = 30.5 kJ mol–1. ∆fH0 (CCl 4) = –135.5 kJ mol–1. ∆aH0 (C) = 715.0 kJ mol–1, where ∆aH0 is enthalpy of atomisation ∆aH0 (Cl 2) = 242 kJ mol–1

Accepted Answer

326 kJ mol–1. Hess's Law lets you build an unknown enthalpy change from known ones — by inserting a path through the elements, even though the actual reaction never goes that way. The reaction to solve is the complete atomization of gaseous carbon tetrachloride: CCl 4(g) → C (g) + 4 Cl (g), Δ H =? None of the given data describes this step directly. What's given are four separate thermochemical quantities, each for a different transformation: - Δ f H^0( CCl 4) = -135.5 kJ mol^-1 — enthalpy of forming liquid CCl 4(l) from graphite and chlorine gas: C(graphite) + 2 Cl 2(g) → CCl 4(l) - Δ vap H^0( CCl 4) = 30.5 kJ mol^-1 — enthalpy to turn liquid CCl 4(l) into gas: CCl 4(l) → CCl 4(g) - Δ a H^0( C ) = 715.0 kJ mol^-1 — enthalpy of atomisation of carbon: C(graphite) → C (g) - Δ a H^0( Cl 2) = 242 kJ mol^-1 — enthalpy of atomisation of chlorine, per mole of Cl 2: Cl 2(g) → 2 Cl (g) The trick people miss: you must build a path from CCl 4(g) back down to the elements, then back up to atoms — which means reversing the formation and vaporisation steps (flipping their signs). Route from CCl 4(g) to C (g) + 4 Cl (g), going via the elements: Step 1 — condense the gas back to liquid (reverse of vaporisation): CCl 4(g) → CCl 4(l), Δ H 1 = -Δ vap H^0 = -30.5 kJ mol^-1 Step 2 — decompose the liquid into its elements (reverse of formation): CCl 4(l) → C(graphite) + 2 Cl 2(g), Δ H 2 = -Δ f H^0 = -(-135.5) = +135.5 kJ mol^-1 Step 3 — atomise the carbon (used as given): C(graphite) → C (g), Δ H 3 = Δ a H^0( C ) = +715.0 kJ mol^-1 Step 4 — atomise the chlorine. Here's the step that trips people up: the data is per mole of Cl 2, but Step 2 produced 2 mol of Cl 2, so the atomisation enthalpy must be doubled: 2 Cl 2(g) → 4 Cl (g), Δ H 4 = 2 × Δ a H^0( Cl 2) = 2 × 242 = +484.0 kJ mol^-1 By Hess's Law, since Steps 1–4 carry CCl 4(g) to exactly the same final state ( C (g) + 4 Cl (g) ) as the direct reaction, their enthalpy changes simply add: Δ H = Δ H 1 + Δ H 2 + Δ H 3 + Δ H 4 Δ H = (-30.5) + (135.5) + (715.0) + (484.0) kJ mol^-1 Δ H = 1304.0 kJ mol^-1 This is the energy needed to break all four C–Cl bonds in one mole of CCl 4(g), turning it completely into isolated gaseous atoms. Bond enthalpy is an average, per-bond quantity — divide the total by the number of bonds broken, not by the number of atoms or moles of compound. Since the four C–Cl bonds are taken as identical, the mean bond enthalpy is: Bond enthalpy (C–Cl) = (Δ H)/4 = (1304.0 kJ mol^-1)/4 Bond enthalpy (C–Cl) = 326.0 kJ mol^-1 Answer: The enthalpy change for CCl 4(g) → C (g) + 4 Cl (g) is Δ H = 1304.0 kJ mol^-1, and the bond enthalpy of C–Cl in CCl 4 is 326.0 kJ mol^-1.

Question 16

For an isolated system, ∆U = 0, what will be ∆S?

Accepted Answer

∆S 0. An isolated system cannot exchange heat, work, or matter with its surroundings — so Δ U = 0 only tells you energy is conserved inside the system; it says nothing yet about entropy, which is governed by a separate law. Step 1: What "isolated" forces on the first law. For an isolated system there is no heat flow and no work exchanged with the surroundings, so q = 0 and w = 0. The first law of thermodynamics, Δ U = q + w, where Δ U is the change in internal energy, q is heat absorbed by the system, and w is work done on the system, then gives Δ U = 0 + 0 = 0. This is exactly the given condition — it is a restatement of energy conservation, not a statement about spontaneity. This is the step people trip on: seeing Δ U = 0 and assuming "nothing changes," so Δ S = 0 too. The first law (energy) and the second law (entropy) are independent statements — one being zero does not force the other to be zero. Step 2: Bring in the second law, which is what actually decides Δ S. The second law of thermodynamics says a process occurs spontaneously (on its own, without outside intervention) only if it increases the total entropy of the universe: Δ S total = Δ S system + Δ S surroundings 0 for a spontaneous process. For a system that is isolated, there is no separate surroundings to exchange entropy with — the system is the entire universe as far as this process is concerned, so Δ S surroundings = 0 and the whole criterion collapses onto the system itself: Δ S system 0 for a spontaneous change happening inside it. Step 3: Check this against the definition of entropy. Entropy change along a reversible path is defined by dS = δ q rev T, where δ q rev is heat added reversibly and T is the absolute temperature. If the change inside the isolated system could somehow happen reversibly, q = 0 would force dS = 0. But a change that occurs on its own — without any external agent driving it — is by definition irreversible, and the Clausius inequality for an irreversible step reads dS (δ q)/T = 0/T = 0. So the real, spontaneous change proceeding inside the isolated system must have Δ S 0; only in the limiting, idealized case of a perfectly reversible process (no net spontaneous change at all) would Δ S = 0. This is the same distinction as molarity vs. molality tripping people up elsewhere: two formulas that look like they should give the same sign of answer, but come from different physical premises. Here, "conserved" (first law, Δ U ) and "increasing" (second law, Δ S ) are not the same idea, and confusing them is the trap this question is built to catch. Answer: For an isolated system, Δ S 0 — entropy increases for the spontaneous process occurring within it (it equals zero only in the ideal limiting case of a reversible process).

Question 17

For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.

Accepted Answer

2000 K. A reaction becomes spontaneous when Δ G turns negative, and Δ G = Δ H - TΔ S is the equation that connects the three quantities you're given. Here Δ H is the enthalpy change, Δ S is the entropy change, and T is the absolute temperature in kelvin. For this reaction, Δ H is large and positive ( +400 kJ mol^-1 ) — the reaction is strongly endothermic, so at low temperature the Δ H term dominates and Δ G is positive (non-spontaneous). The -TΔ S term is what can pull Δ G negative, and it only grows in magnitude as T increases. So there is a threshold temperature above which spontaneity kicks in. Finding that threshold: set Δ G = 0. This is the boundary between non-spontaneous and spontaneous — the reaction is at equilibrium at exactly this temperature, and spontaneous ( Δ G < 0 ) for any temperature above it (since Δ S 0 here). 0 = Δ H - TΔ S Rearranging for T: T = (Δ H)/(Δ S) A step people trip on: Δ H is usually quoted in kJ but Δ S is usually quoted in J — mixing those units without converting gives an answer wrong by a factor of 1000. Here both are already given in kilojoules, so no conversion is needed: Δ H = 400 kJ mol^-1, Δ S = 0.2 kJ K^-1 mol^-1 Substituting: T = (400 kJ mol^-1)/(0.2 kJ K^-1 mol^-1) = 2000 K The mol ^-1 cancels top and bottom, and dividing kJ by (kJ K ^-1 ) leaves K — the unit comes out as a temperature, as it should. At T = 2000 K, Δ G = 0 exactly (the reaction sits at equilibrium). For any temperature above 2000 K, the -TΔ S term outweighs Δ H, making Δ G < 0, so the reaction becomes spontaneous. Answer: The reaction becomes spontaneous above T = 2000 K (with Δ G = 0 exactly at 2000 K).

Question 18

For the reaction, 2 Cl(g) → Cl 2(g), what are the signs of ∆H and ∆S?

Accepted Answer

∆H is negative (bond energy is released) and ∆S is negative (There is less randomness among the molecules than among the atoms). Forming a bond releases energy, and turning two free particles into one takes away freedom of motion — so here both Δ H and Δ S come out negative. Sign of Δ H (enthalpy change): In 2 Cl(g) → Cl 2 (g), two free chlorine atoms come together and form a single covalent bond, Cl - Cl. Making a bond is always an energy-releasing process — you have to put in energy to break a bond (bond dissociation enthalpy is defined as a positive quantity for that reason), so the reverse process, forming the bond, gives that same amount of energy back out. Since the system releases energy to the surroundings, the reaction is exothermic, and by the sign convention (heat given out by the system is negative): Δ H < 0 The step people get wrong: bond dissociation enthalpy (breaking a bond) is always quoted as positive; the moment a question is about forming that same bond, the sign flips to negative. Here the reaction is bond formation, not bond breaking, so Δ H is negative even though the numerical value used to compute it (the Cl–Cl bond enthalpy) is tabulated as a positive number. Sign of Δ S (entropy change): Entropy is a measure of the disorder, or the number of ways a system's particles can be arranged/move — more independent particles almost always means more disorder. On the left side there are 2 mol of separate Cl atoms, each free to move independently in the gas phase. On the right side these have combined into 1 mol of Cl 2 molecules — the same atoms now move together as a single, more constrained unit, and the total number of independently-moving gas particles has gone down (2 mol of particles → 1 mol of particles). Fewer independent particles means fewer ways to arrange them, i.e., lower disorder. So the entropy of the system decreases going from reactants to products: Δ S < 0 The step people get wrong: it is tempting to think "gas is involved, so entropy must increase" — but what matters is the change in the number of moles of gas (or more precisely, the change in the number of freely-moving particles), not merely the phase. Here the mole count of gas drops from 2 to 1, so entropy decreases, not increases. Answer: Δ H < 0 (negative, exothermic — a new Cl–Cl bond is formed) and Δ S < 0 (negative — 2 mol of gaseous atoms combine into 1 mol of gaseous molecules, decreasing disorder).

Question 19

For the reaction 2 A(g) + B(g) → 2D(g) ∆U 0 = –10.5 kJ and ∆ S 0 = –44.1 JK–1. Calculate ∆G0 for the reaction, and predict whether the reaction may occur spontaneously.

Accepted Answer

0.164 kJ, the reaction is not spontaneous.. Δ U^° is not the same thing as Δ H^° — for a gas-phase reaction where the mole count of gas changes, you must convert between them before you can use the Gibbs equation. The Gibbs equation needs Δ H^°, not Δ U^°: Δ G^° = Δ H^° - TΔ S^° where Δ H^° is the standard enthalpy change, T is the absolute temperature in kelvin, and Δ S^° is the standard entropy change. The question gives you Δ U^°, so the first job is to get Δ H^° from it. Step 1 — convert Δ U^° to Δ H^° For a reaction of ideal gases, Δ H^° = Δ U^° + Δ n g RT Here Δ n g is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants), R = 8.314 J K^-1 mol^-1 is the gas constant, and T is the temperature. This is the step it's easy to skip — treating Δ U^° as if it were Δ H^° — but whenever the number of gas moles changes, the two differ by a real, calculable amount. For 2A(g) + B(g) → 2D(g): Δ n g = 2 - (2+1) = -1 The problem gives no temperature, so use the standard reference temperature implied by the "°" superscripts, T = 298 K. Δ n g RT = (-1)(8.314 J K^-1 mol^-1)(298 K ) = -2477.6 J mol^-1 = -2.478 kJ mol^-1 Δ H^° = Δ U^° + Δ n g RT = (-10.5 kJ ) + (-2.478 kJ ) = -12.978 kJ Step 2 — apply the Gibbs equation Convert Δ S^° to kJ so the units match Δ H^°: Δ S^° = -44.1 J K^-1 = -0.0441 kJ K^-1 Mixing J and kJ here is the other easy slip — always bring both terms to the same unit before subtracting. TΔ S^° = (298 K )(-0.0441 kJ K^-1) = -13.142 kJ Δ G^° = Δ H^° - TΔ S^° = (-12.978 kJ ) - (-13.142 kJ ) = -12.978 kJ + 13.142 kJ = 0.164 kJ Step 3 — read the sign Δ G^° has come out positive. A positive Δ G^° means the reaction is not spontaneous in the forward direction under standard conditions at 298 K — even though Δ U^° itself was negative. The internal-energy change alone does not tell you about spontaneity; only Δ G^°, which weighs in the entropy term as well, does. Answer: Δ G^° ≈ +0.164 kJ (i.e. about +164 J ); since Δ G^° 0, the reaction is not spontaneous under standard conditions at 298 K.

Question 20

The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.

Accepted Answer

–5.744 kJ mol–1. The link between the equilibrium constant and Δ G^0 is Δ G^0 = -RT ln K, and this is exact — no approximation. Here Δ G^0 is the standard Gibbs energy change of the reaction, R is the gas constant, T is the absolute temperature in kelvin, and K is the equilibrium constant. This equation is what tells you whether a reaction favours products ( K 1, Δ G^0<0 ) or reactants ( K<1, Δ G^0 0 ) at equilibrium. A step people rush past: T must be in kelvin, not degrees Celsius — the formula only works with absolute temperature, and here it is already given as 300 K. Given: K = 10, R = 8.314 J K^-1 mol^-1, T = 300 K Substitute directly into Δ G^0 = -RT ln K: Δ G^0 = -(8.314 J K^-1 mol^-1)(300 K ) ln (10) A convenient rewrite people forget: ln x = 2.303 log 10 x, so ln (10) = 2.303 × log 10(10) = 2.303 × 1 = 2.303. This lets the whole calculation run on base-10 logs. Δ G^0 = -(8.314 J K^-1 mol^-1)(300 K )(2.303) Multiply the numbers in stages, keeping units attached: 8.314 J K^-1 mol^-1 × 300 K = 2494.2 J mol^-1 2494.2 J mol^-1 × 2.303 = 5744.1 J mol^-1 So: Δ G^0 = -5744.1 J mol^-1 The data ( K=10, T = 300 K) supports three significant figures, so this rounds to -5.74 × 10^3 J mol^-1, i.e. -5.74 kJ mol^-1. The negative sign is not incidental — it says something physical: because K 1, the reaction is product-favoured at equilibrium, and Δ G^0 must come out negative. A positive Δ G^0 would instead signal K < 1 (reactant-favoured), so the sign is a built-in check on the arithmetic, not just a label. Answer: Δ G^0 = -5.74 kJ mol^-1 ( -5744 J mol^-1 ).

Question 21

Comment on the thermodynamic stability of NO(g), given 1/2 N 2(g) + 1/2 O 2(g) → NO(g); Δ rH^ = 90 kJ mol^-1 NO(g) + 1/2 O 2(g) → NO 2(g): Δ rH^ = -74 kJ mol^-1

Accepted Answer

NO(g) is unstable, but NO2(g) is formed.. The sign of Δ fH^° tells you where a compound sits on the energy scale relative to its own elements — positive means "uphill," and uphill compounds are thermodynamically unstable with respect to falling back to those elements. Balancing each equation so that exactly one mole of product is formed (that is what makes these enthalpies of formation, Δ fH^°, and not just reaction enthalpies for an arbitrary equation) fixes the fractional coefficients: (1) 1/2 N 2(g) + 1/2 O 2(g) → NO (g); Δ rH 1^° = +90 kJ mol^-1 (2) NO (g) + 1/2 O 2(g) → NO 2(g); Δ rH 2^° = -74 kJ mol^-1 The easy slip here is dropping the 1/2 in front of N 2 and O 2 — an enthalpy of formation is always defined per mole of the compound formed, and one mole of NO only needs half a mole of each diatomic element. Step 1 — read reaction (1) on its own. Reaction (1) is literally the formation reaction of NO from its elements in their standard states, so Δ fH^°( NO ) = Δ rH 1^° = +90 kJ mol^-1. This is strongly positive: making NO out of N 2 and O 2 costs 90 kJ for every mole formed. Equivalently, the reverse process — NO(g) breaking back down into N 2(g) and O 2(g) — releases 90 kJ mol⁻¹, an exothermic, energetically "downhill" path. A compound that can fall downhill in energy back to its elements like this is, by definition, thermodynamically unstable relative to those elements. Step 2 — use Hess's law to see what happens on the way to NO 2. Adding equation (1) and equation (2), the intermediate NO(g) cancels (one mole is produced in (1) and consumed in (2)): 1/2 N 2(g) + 1/2 O 2(g) + NO (g) + 1/2 O 2(g) → NO (g) + NO 2(g) ⇒ 1/2 N 2(g) + O 2(g) → NO 2(g) Because enthalpy is a state function, the enthalpy of this net reaction is just the sum of the two steps (Hess's Law: Δ fH^° = Δ rH 1^° + Δ rH 2^°, where each Δ rH^° is the heat absorbed or released by one step of the path): Δ fH^°( NO 2) = (+90 kJ mol^-1) + (-74 kJ mol^-1) = +16 kJ mol^-1 Step 3 — compare the two compounds. Both Δ fH^° values are positive, so both NO(g) and NO 2(g) sit above N 2(g) + O 2(g) on the energy scale — neither is stable relative to the free elements. But +90 kJ mol⁻¹ is a much steeper climb than +16 kJ mol⁻¹, so NO(g) is the less stable of the two: it sits higher up the hill, and rolling it further downhill by oxidising it to NO 2(g) releases a large 74 kJ mol⁻¹, which is exactly why reaction (2) runs in the exothermic direction as written. Answer: Δ fH^°( NO ) = +90 kJ mol^-1 is large and positive, so NO(g) is thermodynamically unstable with respect to decomposition into N 2(g) and O 2(g) (that reverse step is exothermic by 90 kJ mol⁻¹). Combining the two given steps by Hess's Law gives Δ fH^°( NO 2) = 90 + (-74) = +16 kJ mol^-1; since this is a much smaller positive value than for NO, NO 2(g) is comparatively more stable than NO(g), though still less stable than the free elements N 2(g) and O 2(g).

Question 22

Calculate the entropy change in surroundings when 1.00 mol of H 2O(l) is formed under standard conditions. ∆f H 0 = –286 kJ mol–1.

Accepted Answer

qsurr = + 286 kJ mol–1 ∆Ssurr = 959.73 J K–1. Heat leaving the system enters the surroundings — that's what changes the surroundings' entropy, not the reaction happening inside the system. When a reaction runs at constant temperature and pressure, the heat it releases (or absorbs) is transferred to the surroundings so slowly and evenly that the surroundings can be treated as absorbing it reversibly. That gives a formula that connects the system's enthalpy change directly to the surroundings' entropy change: Δ S surr = -Δ H system T Here Δ H system is the enthalpy change of the reaction happening in the system (what NCERT calls Δ f H^0 for a formation reaction), T is the absolute temperature in kelvin, and the minus sign is doing real work: heat leaving the system ( Δ H system < 0, exothermic) is heat entering the surroundings, so it makes Δ S surr positive. Set up the numbers. The reaction is the formation of liquid water, H 2(g) + 1/2O 2(g) → H 2O(l), Δ f H^0 = -286 kJ mol^-1 "Standard conditions" fixes the temperature at T = 298 K (this is the temperature standard thermodynamic data is tabulated at, unless a different one is stated). Before substituting, convert kilojoules to joules — mixing kJ into a formula whose answer you want in J K⁻¹ mol⁻¹ is a common slip: Δ f H^0 = -286 kJ mol^-1 = -286 × 10^3 J mol^-1 = -2.86×10^5 J mol^-1 Substitute. Δ S surr = (-(-2.86×10^5 J mol^-1))/(298 K ) = (2.86×10^5 J mol^-1)/(298 K ) The two negative signs (the minus in the formula, and the minus already inside Δ f H^0 ) cancel — that's expected, since releasing heat from an exothermic reaction should raise the surroundings' entropy, not lower it. Δ S surr = 959.73 J K^-1 mol^-1 The data (286 kJ mol⁻¹, three significant figures) justifies rounding the final value to three significant figures: Δ S surr ≈ 960 J K^-1 mol^-1 Answer: Δ S surroundings = (2.86×10^5 J mol^-1)/(298 K ) ≈ 960 J K^-1 mol^-1 (positive, because the exothermic reaction releases heat into the surroundings).

NCERT Solutions · Class 11 Chemistry Thermodynamics

Exercises 5.1–5.10 (part 1 of 2)

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Exercise 5.5

Exercise 5.6

Exercise 5.7

Exercise 5.8

Exercise 5.9

Exercise 5.10

Exercises 5.11–5.22 (part 2 of 2)

Exercise 5.11

Exercise 5.12

Exercise 5.13

Exercise 5.14

Exercise 5.15

Exercise 5.16

Exercise 5.17

Exercise 5.18

Exercise 5.19

Exercise 5.20

Exercise 5.21

Exercise 5.22