CBSE Class 11 Chemistry HOTs Thermodynamics Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): The standard entropy of an element is not zero, whereas its standard enthalpy of formation is taken as zero.
Reason (R): Entropy is zero only for a perfectly ordered crystal at 0 K, while the enthalpy of formation of an element in its reference state is zero by definition.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Entropy has an absolute scale from the third law: only a perfect crystal at 0 K has S = 0, so at 298 K an element has a positive standard entropy.
Δ_fH° measures the heat change in forming a substance from its elements. For an element in its reference state, no change occurs, so it is set to zero by definition. R explains A.
In simple words: Entropy has a true zero at 0 K; enthalpy of formation uses a chosen reference.

Teacher's Note:
1. Link the third law to the reason.
2. Remember that Δ_fH° is zero only for the reference state of the element.

Question 2. The equilibrium constant of a reaction is greater than or equal to one if its Δ_rG° is less than zero. Explain.
Answer:
Δ_rG° = −RT ln K, so ln K = −Δ_rG°/RT.
If Δ_rG° < 0, the right-hand side is positive, so ln K > 0 and K > 1. If Δ_rG° = 0, K = 1.
A large K means the products dominate at equilibrium.
In simple words: A negative standard Gibbs energy change means the products are favoured at equilibrium.

Teacher's Note:
1. Start from Δ_rG° = −RT ln K.
2. State the condition for K > 1 clearly.

Question 3. Many thermodynamically feasible reactions do not occur under ordinary conditions. Why?
Answer:
A negative ΔG only shows that a reaction is possible; it says nothing about the rate.
Under ordinary conditions the reacting molecules may not have enough energy to cross the activation energy barrier, so the reaction does not start.
Example: the reaction of H₂ and O₂ to form water has a large negative ΔG, but a mixture of the gases stays unchanged at room temperature until a spark or catalyst is supplied.
In simple words: Thermodynamics tells whether a reaction can happen; kinetics tells whether it happens quickly.

Teacher's Note:
1. Mention activation energy.
2. Give the H₂ and O₂ example.

Question 4. Which of the following statements are correct? (more than one may be correct)
(a) The heat exchanged at constant volume equals the change in internal energy
(b) Work is a state function because it is a form of energy
(c) The heat exchanged at constant pressure equals the change in enthalpy
(d) ΔH = ΔU for every chemical reaction
Answer:
Ans. (a) and (c) At constant volume w = 0, so q_V = ΔU. At constant pressure q_p = ΔH.
Work is a path function, so (b) is wrong. ΔH = ΔU + Δn_gRT, so they are equal only when Δn_g = 0, which makes (d) wrong.
In simple words: Heat at constant volume gives ΔU and heat at constant pressure gives ΔH.

Teacher's Note:
1. Test each statement against the definitions of q_V and q_p.

Question 5. One mole of an ideal gas is taken round the cycle A→B→C→D→A shown in the p–V diagram: A to B at constant pressure 2p, B to C at constant volume 2v, C to D at constant pressure p and D to A at constant volume v. Find the total work done by the system and the heat absorbed in the complete cycle.
Find the total work done and the heat absorbed in the cycle.
Answer:
A→B (isobaric expansion): w = −2p(2v − v) = −2pv.
B→C (isochoric): w = 0.
C→D (isobaric compression): w = −p(v − 2v) = +pv.
D→A (isochoric): w = 0.
Total w = −2pv + pv = −pv, so the system does net work of magnitude pv, which is the area of the rectangle ABCD.
The cycle is complete, so ΔU = 0 and q = −w = +pv.
In simple words: The loop turns pv of absorbed heat into pv of net work, because the gas returns to the same state.

Teacher's Note:
1. The net work in a cyclic process equals the enclosed area.
2. A clockwise cycle on a p–V diagram means net work done by the system.

Question 6. One mole of an ideal gas at 300 K, initially in a 2.00 L insulated cylinder, is allowed to expand against vacuum to 8.00 L. Determine w, q, ΔU, ΔH, the final temperature, and predict the signs of ΔS(system), ΔS(surroundings), ΔS(universe) and ΔG.
Answer:
The external pressure is zero (free expansion), so w = −p_exΔV = 0. The cylinder is insulated, so q = 0.
ΔU = q + w = 0. For an ideal gas U depends only on T, so the temperature is unchanged: T_final = 300 K, and ΔH = ΔU + Δ(pV) = Δ(nRT) = 0.
The gas occupies a larger volume, so it is more disordered: ΔS(system) > 0. The surroundings exchange no heat, so ΔS(surroundings) = 0, and ΔS(universe) > 0.
ΔG = −TΔS(universe) < 0, so the expansion is spontaneous (and irreversible).
In simple words: The gas spreads out on its own with no energy change; only the disorder increases.

Teacher's Note:
1. List all eight quantities, even those that are zero.
2. Free expansion is the standard example of a spontaneous process with ΔU = 0.

Question 7. Calculate the work done and ΔU for each case. (i) One mole of an ideal gas in a 10 L bulb is allowed to enter an evacuated 100 L bulb. (ii) One mole of gas expands from 1 L to 5 L against a constant external pressure of 1 atm, in an adiabatic process. (1 L atm = 101.3 J)
Answer:
(i) The gas expands into a vacuum, so p_ex = 0 and w = 0. The process is adiabatic, so q = 0 and ΔU = q + w = 0.
(ii) ΔV = 5 − 1 = 4 L. w = −p_exΔV = −1 × 4 = −4 L atm = −405.2 J (work done by the gas).
For an adiabatic change q = 0, so ΔU = w = −405.2 J. The internal energy and temperature fall.
In simple words: A gas that expands without heat flowing in must pay for the work with its own internal energy.

Teacher's Note:
1. Free expansion against zero pressure does no work.
2. Adiabatic means q = 0, so ΔU = w.

Question 8. Five moles of an ideal gas at 300 K expand isothermally from 500 Pa to 100 Pa. Calculate w, q, ΔU and ΔH if the expansion is (i) against a constant external pressure of 100 Pa (irreversible) and (ii) reversible. Which gives more work? (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
For an isothermal process of an ideal gas, ΔU = ΔH = 0 in both cases, so q = −w. nRT = 5 × 8.314 × 300 = 12 471 J.
(i) Irreversible: V₁ = nRT/p₁ = 12 471/500 = 24.94 m³ and V₂ = 12 471/100 = 124.7 m³. w = −p_ex(V₂ − V₁) = −100 × 99.77 = −9977 J, q = +9977 J.
(ii) Reversible: w = −2.303 nRT log (p₁/p₂) = −2.303 × 12 471 × log 5 = −20 071 J, q = +20 071 J.
The reversible expansion gives about twice as much work as the irreversible one (20.07 kJ against 9.98 kJ), because in a reversible process the external pressure is always almost equal to the gas pressure.
In simple words: A reversible path squeezes out the maximum possible work.

Teacher's Note:
1. For an isothermal expansion the gas pressure ratio equals the inverse volume ratio.
2. Reversible work is the maximum work.

Question 9. The enthalpy of combustion of methane is given by CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = −890.3 kJ mol⁻¹. (a) How many grams of methane are needed to produce 445.15 kJ of heat? (b) What mass of CO₂ forms when 445.15 kJ is evolved? (c) What volume of O₂ at STP is used?
Answer:
Heat 445.15 kJ is exactly half of 890.3 kJ, so half a mole of reaction occurs.
(a) 0.5 mol of CH₄ = 0.5 × 16 = 8 g.
(b) 0.5 mol of CO₂ = 0.5 × 44 = 22 g.
(c) O₂ used = 0.5 × 2 mol = 1 mol, and at STP 1 mol occupies 22.4 L.
In simple words: Heat released scales with the amount of methane burnt.

Teacher's Note:
1. Find the fraction of the thermochemical equation that reacts.
2. Use molar masses and 22.4 L per mole at STP.

Question 10. The thermochemical equations for two rocket fuels are 2Al(s) + 3/2 O₂(g) → Al₂O₃(s); ΔH = −1667.8 kJ and H₂(g) + ½O₂(g) → H₂O(l); ΔH = −285.9 kJ. (a) If equal masses of aluminium and hydrogen are used, which is the better fuel? (b) Find ΔH for Al₂O₃(s) → 2Al(s) + 3/2 O₂(g).
Answer:
(a) 2 mol Al = 54 g gives 1667.8 kJ, so 1 g of Al gives 1667.8/54 = 30.9 kJ.
1 mol H₂ = 2 g gives 285.9 kJ, so 1 g of H₂ gives 285.9/2 = 142.95 kJ. Hydrogen releases more heat per gram, so H₂ is the better fuel.
(b) The required reaction is the reverse of the first, so ΔH = +1667.8 kJ.
In simple words: Compare the heat released for one gram of each fuel.

Teacher's Note:
1. Convert moles to grams using molar masses.
2. Reversing a reaction changes the sign of ΔH.

Question 11. Can sodium reduce aluminium oxide to aluminium at 298 K? Δ_fG°(Al₂O₃) = −1582 kJ mol⁻¹ and Δ_fG°(Na₂O) = −377 kJ mol⁻¹.
Answer:
The reaction is Al₂O₃(s) + 6Na(s) → 3Na₂O(s) + 2Al(s).
Δ_rG° = 3Δ_fG°(Na₂O) − Δ_fG°(Al₂O₃) = 3(−377) − (−1582) = −1131 + 1582 = +451 kJ (the elements Na and Al have zero Δ_fG°).
Δ_rG° is positive, so the reaction is non-spontaneous at 298 K. Sodium cannot reduce Al₂O₃ under these conditions.
In simple words: Al₂O₃ is more stable than Na₂O, so sodium is unable to take its oxygen.

Teacher's Note:
1. Standard Gibbs energy of formation of an element is zero.
2. Positive Δ_rG° means non-spontaneous.

Question 12. Assertion (A): A reaction with ΔH < 0 and ΔS > 0 is spontaneous at all temperatures.
Reason (R): ΔG = ΔH − TΔS is then negative at every temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) With ΔH negative and ΔS positive, ΔH is negative and −TΔS is also negative for every T > 0. So ΔG is always negative and the reaction is spontaneous at all temperatures. R explains A.
In simple words: Both terms of ΔG push it below zero, whatever the temperature.

Teacher's Note:
1. Check the sign of each term in ΔG = ΔH − TΔS.

Question 13. Assertion (A): The enthalpy of combustion of a substance is always negative.
Reason (R): All combustion reactions are exothermic.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Combustion is the complete burning of a substance in oxygen, which always releases heat, so ΔH is always negative. R explains A.
Enthalpies of solution and formation, by contrast, can be positive or negative.
In simple words: Burning always releases heat.

Teacher's Note:
1. Compare with formation and solution enthalpies, which may have either sign.

Question 14. Assertion (A): For PCl₅(g) → PCl₃(g) + Cl₂(g), ΔH is greater than ΔU at a given temperature.
Reason (R): The number of moles of gas increases in the reaction, so Δn_g is positive.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) ΔH = ΔU + Δn_gRT. Here Δn_g = 2 − 1 = +1, so ΔH = ΔU + RT, which is larger than ΔU. R explains A.
In simple words: More gas moles in the products means ΔH exceeds ΔU by Δn_gRT.

Teacher's Note:
1. Count only gaseous species when finding Δn_g.