Chapter 10.3, Problem 5RQ

Interpretation Introduction

Interpretation: The $\Delta H$ for the following reaction needs to be calculated.

  $2{\text{NO}}_{\text{2}}(g)\to {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}(g)$

Concept introduction: Hess's law states that the enthalpy change in moving from a set of reactants to a set of products does not depend on the number of steps in the reaction. So, enthalpy change for the net reaction can be calculated if the enthalpy change of all the steps is known.

Answer to Problem 5RQ

  $\Delta H$ for the reaction is $-58\text{kJ}$ .

Explanation of Solution

The chemical equation for which $\Delta H$ needs to be calculated is,

  $2{\text{NO}}_{\text{2}}(g)\to {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}(g)$  ...... (1)

The given reactions are,

  ${\text{N}}_{\text{2}}(g)+2{\text{O}}_2(g)\to 2{\text{NO}}_{\text{2}}(g)\Delta H=67.7\text{kJ}$   ...... (2)

  ${\text{N}}_{\text{2}}(g)+2{\text{O}}_2(g)\to {\text{N}}_2{\text{O}}_4(g)\Delta H=9.7\text{kJ}$   ...... (3)

Observe the equation (1). Here, ${\text{NO}}_{\text{2}}$ is reactant. So, reverse equation (2).This changes the sign of enthalpy of equation (1).

  $2{\text{NO}}_{\text{2}}(g)\to {\text{N}}_{\text{2}}(g)+2{\text{O}}_2(g)\Delta H=-67.7\text{kJ}$   ...... (4)

  ${\text{N}}_{\text{2}}(g)+2{\text{O}}_2(g)\to {\text{N}}_2{\text{O}}_4(g)\Delta H=9.7\text{kJ}$   ...... (5)

Now add equation (4) and (5).

  $\begin{array}{l}2{\text{NO}}_{\text{2}}(g)+\cancel{{\text{N}}_{\text{2}}(g)}+2{\text{O}}_2(g)\to \cancel{{\text{N}}_{\text{2}}(g)}+\cancel{2{\text{O}}_2(g)}+{\text{N}}_2{\text{O}}_4(g)\Delta H=-67.7\text{kJ+}9.7\text{kJ}\\ 2{\text{NO}}_{\text{2}}(g)\to {\text{N}}_2{\text{O}}_4(g)\Delta H=-58\text{kJ}\end{array}$

Conclusion

By Hess’s law, enthalpy change for a reaction can be calculated.