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Chemistry Thermodynamics and Thermochemistry General Single Correct MCQ
Published on: August 12, 2026

Dependence of spontaneity on temperature :

For a process to be spontaneous, at constant temperature and pressure, there must be decrease in free energy of the system in the direction of the process, i.e. Δ G P, T < 0. Δ G P, T = 0 implies the equilibrium condition and Δ G P, T > 0 corresponds to non-spontaneity.

Gibbs-Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : = Δ H – T Δ S ...

A
The magnitude of Δ H does not change much with the change in temperature but the entropy factor T Δ S changes appreciably. Thus, spontaneity of a process depends very much on temperature. For endothermic process, both Δ H and Δ S are positive. The energy factor, the first factor of equation, opposes the spontaneity whereas entropy factor favours it. At low temperature the favourable factor T Δ S will be small and may be less than Δ H, Δ G will have positive value indicating the nonspontaneity of the process. On raising temperature, the factor T Δ S increases appreciably and when it exceeds Δ H, Δ G would become negative and the process would be spontaneous. For an exothermic process, both Δ H and Δ S would be negative. In this case the first factor of
eq. 1 favours the spontaneity whereas the second factor opposes it. At high temperature, when T Δ S > Δ H, Δ G will have positive value, showing thereby the non-spontaneity of the process. However, on decreasing temperature, the factor T Δ S decreases rapidly and when T Δ S < Δ H, Δ G becomes negative and the process occurs spontaneously. Thus, an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. (i) When CaCO 3 is heated to a high temperature, it undergoes decomposition into CaO and CO 2 whereas it is quite stable at room temperature. The most likely explanation of it, is The enthalpy of reaction ( Δ H) overweighs the term T Δ S at high temperature. spontaneous 1000 K 0, – 0.1 cal K –1 – 12.84 kcal mol –1 , spontaneous
B
The term T Δ S overweighs the enthalpy of reaction at high temperature. non-spontaneous 1500 K 0,100 cal K –1 12.84 kcal mol –1 , non-spontaneous
C
At high temperature, both enthalpy of reaction and entropy change become negative. at equilibrium 2000 K – 10 kcal, – 100 cal K –1 – 17.16 kcal mol –1 , spontaneous
D
None of these. (ii) For the reaction at 25°C, X 2 O 4 ( λ ) ⎯→ 2XO 2 (g) Δ H = 2.1 Kcal and Δ S = 20 cal K –1 . The reaction would be unpredictable (iii) For the reaction at 298 K, 2A + B ⎯→ C Δ H = 100 kcal and Δ S = 0.050 kcal K –1 . If Δ H and Δ S are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous ? 2500 K (iv) A reaction has a value of Δ H = – 40 kcal at 400K. Above 400 K, the reaction is spontaneous, below this temperature, it is not. The values of Δ G and Δ S at 400 K are respectively 0, – 100 cal K –1 (v) The enthalpy change for a certain reaction at 300 K is – 15.0 K cal mol –1 . The entropy change under these conditions is – 7.2 cal K –1 mol –1 . The free energy change for the reaction and its spontaneous/non-spontaneous character will be None of these

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Text Solution

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The correct answer is:
C

(i) Δ G = Δ H – T Δ S

If T Δ S dominates Δ H at high temperature then Δ G < 0 hence CaCO 3 decomposes at high temperature.

(ii) Δ G = Δ H – T Δ S = 2.1 × 10 3 – 20 × 298 < 0.

(iii) Reaction to be spontaneous

Δ G < 0 ⇒ Δ H – T Δ S < 0 ⇒ Δ T > ⇒ T > ⇒ T > 2000 K.

(iv) Clearly at 400 K, reaction is in equilibrium Δ G = 0 Δ S = = – = – 100 cal K –1

(v) Use Δ G = Δ H – T Δ S

Δ G = – 15 – = – 12.84 Kcal mol –1

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