Chapter 17, Problem 83A

Interpretation Introduction

Interpretation: The enthalpy change for the formation of nitrogen monoxide is to be calculated.

Concept Introduction: Hess’s Law states that the overall change in enthalpy for the solution is the sum of all changes independent of the different phases or steps of a reaction.

Answer to Problem 83A

The heat of the reaction ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to 2\text{NO(g)}$ is

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

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

The thermochemical equations given are

  $\begin{array}{l}{\text{4NH}}_3{\text{(g)+3O}}_{\text{2}}\text{(g)}\to 2{\text{N}}_{\text{2}}{\text{(g)+6H}}_{\text{2}}\text{O(l) }\Delta \text{H= -1}\text{.53}\times {\text{10}}^3\text{ kJ (2)}\\ {\text{4NH}}_3{\text{(g)+5O}}_{\text{2}}\text{(g)}\to 4{\text{NO(g)+6H}}_{\text{2}}\text{O(l) }\Delta \text{H= -1}\text{.17}\times {\text{10}}^3\text{ kJ (3)}\end{array}$

To get the required chemical reaction, first, we must reverse the reaction $\text{(2)}$ which will change the sign of enthalpy.

  $2{\text{N}}_{\text{2}}{\text{(g)+6H}}_{\text{2}}\text{O(l)}\to {\text{4NH}}_3{\text{(g)+3O}}_{\text{2}}\text{(g) }\Delta \text{H= 1}\text{.53}\times {\text{10}}^3\text{ kJ (4)}$

Now, adding the three reactions $(3)$ and $\text{ (4)}$ , we get $\text{(5) }$ and the enthalpies also get added

  $\begin{array}{l}{\text{4NH}}_3{\text{(g)+5O}}_{\text{2}}\text{(g)}\to 4{\text{NO(g)+6H}}_{\text{2}}\text{O(l) }\Delta \text{H= -1}\text{.17}\times {\text{10}}^3\text{ kJ (3) }\\ 2{\text{N}}_{\text{2}}{\text{(g)+6H}}_{\text{2}}\text{O(l)}\to {\text{4NH}}_3{\text{(g)+3O}}_{\text{2}}\text{(g) }\Delta \text{H= 1}\text{.53}\times {\text{10}}^3\text{ kJ (4)}\\ \overline{\underset{¯}{2{\text{N}}_{\text{2}}{\text{(g)+2O}}_{\text{2}}\text{(g)}\to \text{4NO(g) }\Delta \text{H= 0}\text{.36}}}\times {\text{10}}^3\text{ kJ (5) }\end{array}$

Now, multiply $\text{(5)}$ by half to get reaction $\text{(1)}$ and the enthalpy will also be halved.

  ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to 2\text{NO(g) }\Delta \text{H= }\frac{1}{2}\times 0.36\times {\text{10}}^3\text{ kJ=0}\text{.18}\times {\text{10}}^3\text{ kJ}$

The heat of the reaction ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to 2\text{NO(g)}$ is $\text{0}\text{.18}\times {\text{10}}^3\text{ kJ}$