Reactivity 1.4. Entropy and spontaneity (HL only)

The overall chemical equation for the electrolysis of dilute sulfuric acid using inert electrodes is shown below: 

2H₂O(l) → 2H₂(g) + O₂(g) 

Which row of the table gives the correct sign for each property of the reaction? 

Answer: D. 

What are the signs for the entropy changes associated with vaporization of water?

H₂O(l) → H₂O(g) 

Answer: B. 

Given that the enthalpy change of vaporization of water is +40.8 kJ mol^-1, what term gives the entropy change when 36.04 g of water boils to form water vapour?

A. `frac{40.8*10^3*36.04}{18.02*398}` J K^-1 

B. `-frac{40.8*36.04}{18.02*373}`J K^-1 

C. `frac{40.8*10^3*36.04}{18.02*373}`J K^-1 

D. `-frac{40.8*10^3*36.04}{18.02*373}`J K^-1 

Answer: C. `frac{40.8*10^3*36.04}{18.02*373}`J K^-1 

Formula for entropy change 

ΔS =`frac{q_"rev"}{T}`

For vaporization: 

 ΔS =`frac{ΔH_"vap"*"moles of water"}{T}`

moles of water = `frac{36.04}{18.02}`

⇒ ΔS = `frac{40.8*10^3*36.04}{18.02*373}`

Which of the following has the highest entropy? 

A. H₂O(g) at 150°C. 

B. H₂O(g) at 100°C. 

C. H₂O(l) at 100°C. 

D. H₂O(l) at 4°C (the temperature of maximum density). 

Answer: A. H₂O(g) at 150°C. 

A. Correct: Highest entropy – gases have much greater molecular disorder than liquids, and higher temperature further increases molecular motion and randomness.

B. Incorrect: Still gaseous, but cooler → less kinetic energy → slightly less entropy.

C. Incorrect: Much lower – liquids are far more ordered than gases; intermolecular  forces limit movement. 

D. Incorrect: Lowest entropy – cold liquid near maximum density has tightly packed,  ordered structure → minimal molecular freedom.

For systems at equilibrium, which of the following must always be true?

A. ΔS = 0. 

B. q = 0. 

C. ΔH = 0. 

D. ΔG = 0. 

Answer: D. ΔG = 0. 

The expression for the standard free energy change of a reaction is given by the Gibbs equation: 

ΔG° = ΔH° − TΔS° 

What are the signs for ΔH° and ΔS° for a reaction that is spontaneous at all temperatures? 

Answer: A.

The Gibbs equation 

ΔG° = ΔH° − TΔS° 

For a reaction to be spontaneous at all temperatures, it must always have ΔG° < 0 (negative). 

Effect of each term: 

   • ΔH° (enthalpy change): Negative (–) → exothermic reaction → favors  spontaneity. 

   • ΔS° (entropy change): Positive (+) → increase in disorder → also favors  spontaneity. 

Which statement is correct? 

A. If ΔH < 0, reaction is always spontaneous. 

B. If ΔH > 0, reaction is never spontaneous.

C. If ΔS < 0, reaction can be spontaneous if temperature is low enough.

D. If ΔS < 0, reaction can be spontaneous if temperature is high enough. 

Answer: C. If ΔS < 0, reaction can be spontaneous if temperature is low enough. Using the Gibbs Free Energy Equation 

ΔG = ΔH − TΔS 

A reaction is spontaneous when ΔG < 0. 

A. Incorrect: Exothermic (ΔH < 0) helps spontaneity, but if ΔS < 0 and temperature is high, TΔS can outweigh ΔH, making ΔG > 0. 

B. Incorrect: Endothermic reactions (ΔH > 0) can still be spontaneous at high temperature if ΔS > 0 (e.g. melting ice). 

C. Correct: When ΔS < 0, the –TΔS term is positive, so low T minimizes it. Spontaneous only if ΔH < 0 and T is small. 

D. Incorrect: Increasing T makes –TΔS more positive (since ΔS < 0), making ΔG less negative (less spontaneous). 

Magnesium carbonate decomposes on heating according to the equation shown below. The values of the standard enthalpy change and standard entropy change of the decomposition are provided. 

MgCO₃(s) → MgO(s) + CO₂(g) 

ΔH^o = +117 kJ mol⁻¹ 

ΔS^o = +175 J mol⁻¹ K⁻¹ 

a. Explain why the entropy increases when magnesium carbonate decomposes.

b. Calculate the standard Gibbs free energy change, ΔG°, for the decomposition of  magnesium carbonate. Hence, comment on the spontaneity of the decomposition of magnesium carbonate under standard conditions.  

a. When magnesium carbonate decomposes, it produces a gas (CO₂) from a solid:

MgCO₃(s) → MgO(s) + CO₂(g) 

Gases have much higher entropy than solids because gas particles are more disordered  and occupy more microstates. Therefore, entropy increases because a gas is formed from  a solid.

ΔG^o = ΔH^o − TΔS^o

Given: 

ΔH^o = +117 kJ mol⁻¹  = 117000 J mol⁻¹ 

ΔS^o = +175 J mol⁻¹ K⁻¹ 

T = 298 K 

Substitute: 

ΔG^o = 117000 − (298)(175) 

ΔG^o = 117000 − 52150 = +64850 J mol⁻¹ = +64.9 kJ mol⁻¹

Interpretation: ΔG° is positive (+64.9 kJ mol⁻¹) 

→ The reaction is not spontaneous under standard conditions (298 K). It requires heating to become spontaneous (at higher temperatures, the TΔS term becomes larger and can  overcome the positive ΔH). 

At a pressure of 1.01 × 10⁵ Pa and a temperature of 188 K, the liquid and gaseous states  of HCl will be in equilibrium: 

HCl (l) ⇌ HCl (g) 

ΔG = 0.0 kJ mol⁻¹, ΔH = +16.8 kJ mol⁻¹ 

The enthalpy change for the vaporization process, the forward reaction as written, is shown. 

a. Calculate the entropy change, ΔS, for the vaporization, and explain the significance of its sign. 

b. Calculate ΔG for this process at a temperature of 298 K, and explain the significance of its sign. 

a. Calculate ΔS and explain its sign 

At equilibrium (given at 188 K): 

ΔG = 0 

From the Gibbs equation: 

ΔG = ΔH − TΔS 

So at equilibrium: 

0 = ΔH − TΔS 

ΔS = `frac{ΔH}{T}`

Given: 

ΔH = +16.8 kJ mol⁻¹ = 16,800 J mol⁻¹ 

T = 188 K 

ΔS = `frac{16.800}{188}`= 89.4 J mol⁻¹K⁻¹

Sign: ΔS is positive because vaporization increases disorder — gas molecules are more  randomly distributed than liquid molecules. 

b. Calculate ΔG at 298 K and explain its sign 

ΔG = ΔH − TΔS 

ΔG = 16,800 − (298)(89.4) 

ΔG = 16,800 − 26,641 = −9,841 J mol⁻¹ = −9.84 kJ mol⁻¹ 

Significance: ΔG < 0 → the vaporization process is spontaneous at 298 K. This is because the higher temperature increases the TΔS term, making the Gibbs energy negative. 

Solid mercury(II) sulfide, HgS, can exist in either the red or black form. The ΔH° for  the conversion from the red to black form is +4.2 kJ mol^-1.

HgS (red) → HgS (black); ΔH°= +4.2 kJ mol⁻¹ 

The standard entropy values, S°, for HgS (red) and HgS (black) are +77.8 J K^-1 mol^-1 and +88.3 J K^-1 mol^-1, respectively. 

a. Determine the minimum temperature HgS (red) must be heated to in order to change  it to HgS (black). 

b. Given that the equilibrium constant, K, of the reaction at 298 K is 2.97 × 10⁻¹³,  calculate the value of ΔG° of the reaction under standard conditions.

c. Comment on the sign of ΔG° with reference to the position of equilibrium of the reaction under standard conditions. 

a. Minimum temperature for HgS(red) → HgS(black) 

ΔS° = 88.3 − 77.8 = 10.5 J K^-1 mol^-1 = 0.0105 kJ K^-1 mol^-1

At the threshold of spontaneity, 

ΔG° = 0 = ΔH° − TΔS°

⇒ T_min = `frac{ΔH°}{ΔS°}`= `frac{4.2}{0.0105}`= 4.00 × 102 K (≈ 400 K, i.e. 127°C) 

b. ΔG° at 298 K from K = 2.97 × 10⁻¹³ 

ΔG° = −RTlnK

ΔG° = −(8.314 × 10⁻³ kJ K^-1 mol^-1)(298)ln (2.97 × 10⁻¹³) ≈ +71.4 kJ mol⁻¹

c. Significance of the sign 

   • ΔG° is positive, consistent with K ≪ 1. 

   • The equilibrium at 298 K lies far to the left (toward HgS red), so the conversion  to the black form is not spontaneous under standard conditions. 

   • Heating above ~400 K makes the black form favored. 

Hydrogen can be made from steam according to the following equation:

H₂O (g) + C (s) → H₂(g) + CO (g) 

The Gibbs free energy change of the reaction at two different temperatures are shown: ΔG₁ = +78 kJ mol⁻¹ at 378 K 

ΔG₂ = −58 kJ mol⁻¹ at 1300 K 

a. Deduce the correct signs of ΔH and ΔS for this reaction. 

b. Calculate the values of ΔH and ΔS. You can assume they are independent of temperature. 

a. Use ΔG = ΔH − TΔS 

ΔG₁ = +78 kJ mol⁻¹ at 378 K, ΔG₂ = −58 kJ mol⁻¹ at 1300 K 

As T increases, ΔG goes from positive to negative ⇒ the −TΔS term must drive ΔG  down. Therefore ΔS > 0. Since ΔG is still positive at low T, ΔH > 0 (endothermic). Signs: ΔH > 0, ΔS > 0 

b. Solve from the two temperatures (assume ΔH, ΔS constant) 

ΔG₁ = ΔH − T₁ΔS  = 78 

ΔG₂ = ΔH − T₂ΔS  = −58 

Subtract:

ΔS = `frac{ΔG_1-ΔG_2}{T_2-T_1}` = `frac{78-(-58)}{1300-378}`= `frac{136}{922}`= 0.147 kJ K⁻¹mol⁻¹ = +148 J K⁻¹mol⁻¹

Then: 

ΔH = ΔG₁ + T₁ΔS = 78 + 378(0.147) ≈ +134 kJ mol⁻¹ 

Consider the following reaction: 

2CH₃OH(g) + H₂(g) → C₂H₆(g) + 2H₂O(g) 

a. The standard enthalpy change of formation for CH₃OH(g) at 298 K is −201 kJ mol^-1, and for H₂O(g) is −242 kJ mol^-1. Using information from Table 11 of the Data Booklet,  determine the enthalpy change for this reaction.  

b. The standard entropy for CH₃OH(g) at 298 K is 238 J K^-1 mol^-1, for H₂(g) is 131 J K^-1 mol^-1, and for H₂O(g) is 189 J K^-1 mol^-1. Using information from Table 11 of the Data Booklet, determine the entropy change for this reaction.

c. Calculate the standard change in free energy, at 298 K, for the reaction and deduce whether the reaction is spontaneous or non-spontaneous. 

a. Enthalpy change, ΔH° 

Use Hess’s law with formation enthalpies: 

`""ΔH_"rxn"^°` = ∑v`"ΔH_f^°`(products) − ∑v`"ΔH_f^°`(reactants) 

Data: (`"ΔH_f^°` [kJ·mol^-1]) 

   • CH₃OH(g) = −201 

   • H₂O(g) = −242 

   • H₂(g) = 0 (element) 

   • C₂H₆(g) ≈ −84.7 

Calculation: 

Products: 

−84.7 + 2(−242) = −84.7 − 484 = −568.7 

Reactants: 

2(−201) + 0 = −402 

ΔH° = −568.7 − (−402) = −166.7 kJ (≈ −167 kJ) 

b. Entropy change, ΔS°

`""ΔS_"rxn"^°` = ∑vΔS°(products) − ∑vΔS°(reactants) 

Data: (S° [J·K^-1·mol^-1]) 

   • CH₃OH(g) = 238 

   • H₂(g) = 131 

   • H₂O(g) = 189 

   • C₂H₆(g) = 229.5 

Calculation: 

Products: 

229.5 + 2(189) = 229.5 + 378 = 607.5 

Reactants: 

2(238) + 131 = 476 + 131 = 607 

ΔS° = 607.5 − 607 = +0.5 J K^-1·mol^-1

(≈ 0; essentially no net disorder change) 

c. Free energy change, ΔG° at 298 K 

ΔG° = ΔH° − TΔS°

Convert: 

ΔS° = 0.5 J K^-1·mol^-1 = 0.0005 kJ K^-1·mol^-1

TΔS° = 298 × 0.0005 = 0.149 kJ K^-1·mol^-1

ΔG° = −166.7 − 0.149 = −166.85 kJK^-1·mol^-1 (≈ −167 kJ) 

Conclusion: ΔG° < 0, so the reaction is spontaneous under standard conditions. (Spontaneity is driven by the strongly exothermic ΔH°, while ΔS° is approximately zero.) 

In a sealed vessel, ammonium chloride forms an equilibrium with ammonia and hydrogen chloride: 

NH₄Cl (s) ⇌ NH₃ (g) + HCl (g) 

ΔG^o = +91 kJ mol⁻¹ 

ΔG^o = −RTlnK 

A. The equilibrium mixture will contain mainly reactants and the value of K ≪ 1.

B. The equilibrium mixture will contain mainly reactants and the value of K ≫ 1.

C. The equilibrium mixture will contain mainly products and the value of K ≪ 1.

D. The equilibrium mixture will contain mainly products and the value of K ≫ 1.

Answer: A. The equilibrium mixture will contain mainly reactants and the value of K ≪ 1. 

A. Correct: 

The equilibrium mixture will contain mainly reactants and the value of K ≪ 1. 

   • ΔG^o = +91 kJmol⁻¹ ⇒ reaction not spontaneous in the forward direction.

   • K = `"e^frac{-ΔG°}{RT}=e^(-36.7)` ≈ 1 × 10⁻¹⁶, which is much less than 1.

Therefore, mostly NH₄Cl(s) and very little NH₃ and HCl gases are present. → Matches both the equilibrium composition and the magnitude of K.

B. Incorrect: 

The equilibrium mixture will contain mainly reactants and the value of K ≫ 1.

   • Although it is true that the mixture contains mainly reactants, the value of  K cannot be ≫ 1. 

   • ΔG^o positive means K < 1. 

   • K ≫ 1would only happen if ΔG^o were negative (spontaneous). 

→ Contradicts thermodynamic relationship. 

C. Incorrect: 

The equilibrium mixture will contain mainly products and the value of K ≪ 1.

   • K ≪ 1implies that the equilibrium lies to the left, not the right.

   • So if K is small, there cannot be mainly products. 

→ Internal contradiction (small K→ few products). 

D. Incorrect: 

The equilibrium mixture will contain mainly products and the value of K ≫ 1.

   • K ≫ 1would require ΔG^o < 0, but here ΔG^o = +91 kJ mol⁻¹. 

   • Reaction is non-spontaneous, so equilibrium cannot favor products.

→ Both direction and sign inconsistent with data.