Chapter 10, Problem 34A

Interpretation Introduction

Interpretation: The $\Delta \text{H}$ for the following reaction needs to be calculated:

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

Concept Introduction: For the overall processes, the enthalpy change can be determined by adding the enthalpy change of all the steps involved in the process is known as Hess’s Law. The equation to show Hess’s law is:

  ${\text{ΔH}}_{\text{total}}{\text{= ΔH}}_{\text{1}}{\text{+ΔH}}_{\text{2}}{\text{+ΔH}}_{\text{3}}\text{+}\mathrm{...}{\text{+ΔH}}_{\text{n}}$

Answer to Problem 34A

The $\Delta \text{H}$ for the given reaction is $-233\text{ kJ}$ .

Explanation of Solution

Given:

  ${\text{2O}}_{\text{3}}\left(g\right)\to {\text{3O}}_{\text{2}}\left(g\right)\Delta \text{H=}-427\text{ kJ}$

  ${\text{O}}_2\left(g\right)\to 2\text{O}\left(g\right)\Delta \text{H = }495\text{ kJ}$

  $\text{NO}\left(g\right)+{\text{O}}_{\text{3}}\to {\text{NO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\Delta \text{H=}-199\text{ kJ}$

Rules related to enthalpies of the reaction are:

• When a reaction is inverted, then the sign of enthalpy is also inverted.
• When a reaction is multiplied by ‘n’ coefficient, then the value of enthalpy is multiplied by ‘n’ value.

The given equations are as follows:

  ${\text{2O}}_{\text{3}}\left(g\right)\to {\text{3O}}_{\text{2}}\left(g\right)\Delta \text{H=}-427\text{ kJ}$- (1)

  ${\text{O}}_2\left(g\right)\to 2\text{O}\left(g\right)\Delta \text{H = }495\text{ kJ}$- (2)

  $\text{NO}\left(g\right)+{\text{O}}_{\text{3}}\to {\text{NO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\Delta \text{H=}-199\text{ kJ}$ - (3)

The required equation is:

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

In order to obtain the required equation, the reaction in equation (1) and (2) are reversed as:

  ${\text{3O}}_{\text{2}}\left(g\right)\to {\text{2O}}_{\text{3}}\left(g\right)\Delta \text{H = +}427\text{ kJ}$- (4)

  $2\text{O}\left(g\right)\to {\text{ O}}_2\left(g\right)\Delta \text{H = -}495\text{ kJ}$- (5)

The reaction in equation (3) is multiplied by 2 as:

  $\text{2NO}\left(g\right)+2{\text{O}}_{\text{3}}\to 2{\text{NO}}_{\text{2}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\Delta \text{H=}-2\left(199\text{ kJ}\right)\text{ = -398 kJ}$ - (6)

Now, adding equations (4), (5) and (6) as:

  $\begin{array}{l}{\text{3O}}_{\text{2}}\left(g\right)\to {\text{2O}}_{\text{3}}\left(g\right)\Delta \text{H = +}427\text{ kJ}\\ 2\text{O}\left(g\right)\to {\text{ O}}_2\left(g\right)\Delta \text{H = -}495\text{ kJ}\\ \underset{\_}{\text{2NO}\left(g\right)+2{\text{O}}_{\text{3}}\to 2{\text{NO}}_{\text{2}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\Delta \text{H= -398 kJ}}\\ \text{2NO}\left(g\right)+2\text{O}\left(g\right)\to 2{\text{NO}}_{\text{2}}\left(g\right)\Delta \text{H = }\left(\text{+427-495-398}\right)\text{ kJ}\\ \Delta \text{H = -466 kJ }\end{array}$

The obtained reaction is multiplied by $\frac{1}{2}$ as:

  $\begin{array}{l}\frac{1}{2}\left(\text{2NO}\left(g\right)+2\text{O}\left(g\right)\to 2{\text{NO}}_{\text{2}}\left(g\right)\right)\Delta \text{H = }\frac{1}{2}\left(\text{-466 kJ}\right)\\ \text{NO}\left(g\right)+\text{O}\left(g\right)\to {\text{NO}}_{\text{2}}\left(g\right)\Delta \text{H = }-233\text{ kJ}\end{array}$

Hence, the $\Delta \text{H}$ for the given reaction is $-233\text{ kJ}$ .

Conclusion

  $-233\text{ kJ}$ is the $\Delta \text{H}$ for the given reaction.