Chapter 10, Problem 10.95QP

You are given the following data.

$\begin{array}{l}{H}_2(\text{g})\to 2H(\text{g})\Delta \text{H}°=436.4kJ/mol\\ B{r}_2(\text{g})\to 2Br(\text{g})\Delta \text{H}°=192.5kJ/mol\\ H_2(\text{g})\text{ + }B{r}_2(\text{g})\to 2Br(\text{g})\Delta \text{H}°=-72.4kJ/mol\end{array}$

Calculate ΔH° for the reaction

$H_2(\text{g})\text{ + }Br(\text{g})\to HBr(\text{g})$

Interpretation Introduction

Interpretation:

The standard enthalpy of reaction has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

To calculate: The enthalpy of the reaction

Answer to Problem 10.95QP

The enthalpy of the reaction is $\text{-350}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

Record the given info

$\begin{array}{l}{\text{H}}_{\left(\text{g}\right)}\to 2{\text{H}}_{\text{2(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=(-436}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{Br}}_{\left(\text{g}\right)}\to 2{\text{Br}}_{\text{2(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=(-192}\text{.5kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{H}}_{\text{2(g)}}{\text{+Br}}_{\text{2(g)}}\to 2{\text{HBr}}_{\left(\text{g}\right)}{\text{ΔH°}}_{\text{reaction}}\text{=(-72}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\end{array}$

To obtain the desired equation, the given equations are arranged.

$\begin{array}{l}{\text{H}}_{\left(\text{g}\right)}\to \frac{\text{1}}{\text{2}}{\text{H}}_{\text{2(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=}\frac{\text{1}}{\text{2}}\text{(-436}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\text{)=-218}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{Br}}_{\left(\text{g}\right)}\to \frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=}\frac{\text{1}}{\text{2}}\text{(-192}\text{.5kJ}{\text{mol}}^{\text{-1}}\text{)=-96}\text{.25}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ \frac{\text{1}}{\text{2}}{\text{H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2(g)}}\to {\text{HBr}}_{\left(\text{g}\right)}{\text{ΔH°}}_{\text{reaction}}\text{=}\frac{\text{1}}{\text{2}}\text{(-72}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\text{)=-36}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

${\text{H}}_{\left(\text{g}\right)}{\text{+Br}}_{\left(\text{g}\right)}\to {\text{HBr}}_{\left(\text{g}\right)}\text{ΔH°=-350}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$

Conclusion

The enthalpy of the reaction was calculated using the standard enthalpies of reactants and the products.  The enthalpy of reaction was found to be $\text{-350}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$ .