Application on Hess’s Law medium

Question 1: Given the following reactions and their respective enthalpy changes:

1. C(s) + O2(g) → CO2(g) ΔH1 = -393.5 kJ/mol

2. H2(g) + 1/2O2(g) → H2O(l) ΔH2 = -286.0 kJ/mol

3. C(s) + H2(g) → CH4(g) ΔH3 = -74.8 kJ/mol

Calculate the enthalpy change for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Answer 1: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes.

Multiplying reaction 1 by 2 gives: 2C(s) + 2O2(g) → 2CO2(g) 2ΔH1 = 2(-393.5 kJ/mol) = -787.0 kJ/mol

Multiplying reaction 2 by 2 gives: 2H2(g) + O2(g) → 2H2O(l) 2ΔH2 = 2(-286.0 kJ/mol) = -572.0 kJ/mol

Adding reactions 3, 2, and 1 gives: C(s) + H2(g) + 2H2(g) + O2(g) + 2O2(g) → CH4(g) + 2H2O(l) + 2CO2(g) ΔH3 + 2ΔH2 + 2ΔH1 = -74.8 kJ/mol + (-572.0 kJ/mol) + (-787.0 kJ/mol) = -1433.8 kJ/mol

Since the given reaction is the reverse of the calculated reaction, the enthalpy change for the given reaction is the negative of the calculated value. ΔH = -(-1433.8 kJ/mol) = 1433.8 kJ/mol

Question 2: Given the following reactions and their respective enthalpy changes:

1. 2SO2(g) + O2(g) → 2SO3(g) ΔH1 = -198.2 kJ/mol

2. S(s) + O2(g) → SO2(g) ΔH2 = -296.8 kJ/mol

3. 2S(s) + 3O2(g) → 2SO3(g) ΔH3 = -792.0 kJ/mol

Calculate the enthalpy change for the reaction: 2SO2(g) + O2(g) → 2SO3(g) + 198.2 kJ

Answer 2: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes.

Multiplying reaction 2 by 2 gives: 2S(s) + 2O2(g) → 2SO2(g) 2ΔH2 = 2(-296.8 kJ/mol) = -593.6 kJ/mol

Adding reactions 1 and 2 gives: 2SO2(g) + O2(g) + 2S(s) + 2O2(g) → 2SO3(g) + 2