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Application on Hess’s Law

Grade 10 SABIS

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

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

  2. CO(g) + 1/2O2(g) → CO2(g) ΔH2 = -283.0 kJ/mol

  3. C(graphite) + 1/2O2(g) → CO(g) ΔH3 = -110.5 kJ/mol

Calculate the enthalpy change for the reaction: C(graphite) + 1/2O2(g) → CO2(g)

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.

Adding reactions 2 and 3 gives: 2CO(g) + O2(g) → 2CO2(g) ΔH2 + ΔH3 = -283.0 kJ/mol + (-110.5 kJ/mol) = -393.5 kJ/mol

Since this reaction is the reverse of reaction 1, the enthalpy change for the given reaction is the negative of ΔH1. ΔH = -(-393.5 kJ/mol) = 393.5 kJ/mol

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

  1. N2(g) + O2(g) → 2NO(g) ΔH1 = 180.6 kJ/mol

  2. 1/2N2(g) + O2(g) → NO2(g) ΔH2 = 33.2 kJ/mol

Calculate the enthalpy change for the reaction: NO(g) + NO2(g) → N2O3(g)

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: N2(g) + 2O2(g) → 2NO2(g) 2ΔH2 = 2(33.2 kJ/mol) = 66.4 kJ/mol

Adding reactions 1 and 2 gives: 2N2(g) + 2O2(g) → 4NO(g) 2ΔH1 + 2ΔH2 = 2(180.6 kJ/mol) + 66.4 kJ/mol = 427.6 kJ/mol

Since this reaction is the reverse of the desired reaction, the enthalpy change for the given reaction is the negative of the calculated value. ΔH = -427.6 kJ/mol

Question 3: Given the following reactions and their respective enthalpy changes:

  1. 2H2(g) + O2(g) → 2H2O(l) ΔH1 = -572 kJ/mol

  2. 2H2O(l) → 2H2(g) + O2(g) ΔH2 = 572 kJ/mol

Calculate the enthalpy change for the reaction: H2(g) + 1/2O2(g) → H2O(l)

Answer 3: 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

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