Chapter 17.4, Problem 36SP

Interpretation Introduction

Interpretation: The standard heat of reaction for the formation of ${\text{NO}}_2\text{(g)}$ is to be calculated.

Concept Introduction: The standard heat of the reaction is equal to the sum of all the standard heat of the formation of the products minus the sum of all the standard heat of the formation of the reactants.

The standard enthalpy of formation of a pure substance in its elemental form or pure state is zero.

Answer to Problem 36SP

The standard heat of the reaction for the formation of ${\text{NO}}_2\text{(g)}$ is $\text{-114}\text{.14 kJ}$

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

  ${\text{2NO(g)+O}}_{\text{2}}\text{(g)}\to 2{\text{NO}}_2\text{(g)}$

To calculate the standard heat of reaction, we need to know the heat of the formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [H}}^{\text{0}}{}_{\text{f}}{\text{(products)-H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [2×H}}^{\text{0}}{}_{\text{f}}{\text{(NO}}_{\text{2}}{\text{)]-[2×H}}^{\text{0}}{}_{\text{f}}{\text{(NO)+1×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(NO}}_{\text{2}}\text{)= 33}\text{.18 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(NO)= 90}\text{.25 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [2 mol×(33}\text{.18kJ/mol)]-[2mol×(90}\text{.25kJ/mol)+1mol×(0kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [66}\text{.36kJ]-[180}\text{.5kJ+0kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [66}\text{.36kJ-180}\text{.5kJ]=-114}\text{.14 kJ}\end{array}$