Question

The standard enthalpy of vaporization of Freon-11, CFCl3, is 25.21 kJ/mol at its normal boiling point of 17°C. What is the change of entropy for 1 mol of liquid Freon-11 when it vaporizes at its normal boiling point?

For the process Cl2(g) → 2Cl(g), Question 18 options: ΔH is + and ΔS is + for the reaction. ΔH is + and ΔS = 0 for the reaction. ΔH is – and ΔS is – for the reaction. ΔH is – and ΔS is + for the reaction. ΔH is + and ΔS is – for the reaction Arrange the following reactions in order of increasing ΔS° value.

1. H2(g) + F2(g) → 2HF(g) 2. NH4NO3(s) → N2O(g) + 2H2O(l) 3. (NH4)2Cr2O7(s) → Cr2O3(s) + 4H2O(l) + N2(g)

From these two reactions at 298 K,

    V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K

    V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234.2 kJ; ΔS° = 0.2 J/K

calculate ΔG° for the following at 298 K:

    2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Answer 1

Q1. Freon-11

Standard enthalpy of vaporization = 25.21 kJ/mol = 25.21 x 10³ J/mol

normal boiling point = 17 ^oC = 290 K

change in entropy = (moles Freon-11) * (Standard enthalpy of vaporization) / (normal boiling point)

change in entropy = (1 mol) * (25.21 x 10³ J/mol) / (290 K)

change in entropy = 86.93 J/K

Q2. Cl₂ (g) 2 Cl (g)

Correct option is : H is + and S is +

Q3. Highest S^o : (NH4)2Cr2O7(s) → Cr2O3(s) + 4H2O(l) + N2(g)

middle S^o : NH4NO3(s) → N2O(g) + 2H2O(l)

Lowest S^o : H2(g) + F2(g) → 2HF(g)

Q4. For reaction 1, V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g)

G^o = 267.3 kJ

If we reverse this reaction, then G^o = -267.3 kJ

For reaction 2, V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g)

G^o = -234.3 kJ

If we reverse this reaction, then G^o = +234.3 kJ

If we add both reverse reactions, then we get the desired reaction, 2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

G^o = (-267.3 kJ) + (234.3 kJ)

G^o = -33.0 kJ