Chapter 5, Problem 5.67QE

Interpretation Introduction

Interpretation:

$\Delta H$ has to be determined for the reaction, ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)\text{ + 2}{\text{H}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right).$

Concept Introduction:

According to Hess’s Laws, change in enthalpy in an overall reaction can be calculated from the change in enthalpy of other reactions.

• If an equation is obtained by the addition of more than one thermochemical equations, then the enthalpy change of that equation is the sum of change in enthalpy of all the equations added.
• If an equation is the reverse direction of a thermochemical equation, then the change in enthalpy has same numerical value but opposite sign.
• The enthalpy change depends on the mass of reacting substance. If the coefficients present in an equation is multiplied with a factor, then change in enthalpy also should be multiplied with that same factor.

Answer to Problem 5.67QE

$\Delta H$ of the given reaction is $-312\text{ kJ}\text{.}$

Explanation of Solution

Given reactions are shown below,

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)\text{ + }\frac{7}{2}{O}_2\left(g\right)\underset{}{\to }\text{ 2}{\text{CO}}_{\text{2}}\left(g\right)\text{ + 3}{\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta H=-1560kJ\left(1\right)\\ \\ \text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)\text{ + 5}{\text{O}}_{\text{2}}\left(g\right)\underset{}{\to }\text{ 4}{\text{CO}}_{\text{2}}\left(g\right)\text{ + 2}{\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta H=-2599kJ\left(2\right)\\ \\ H_2\left(g\right)\text{ + }\frac{1}{2}{\text{O}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta H=-286kJ\left(3\right)\\ \\ {\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)\text{ + 2}{\text{H}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)\Delta H=?\left(4\right)\end{array}$

From the above equations, it is clear equation (4) can be obtained by adding the reverse of equation (1), equation (2) divided with 2 and equation (3) multiplied with 2.

  $\begin{array}{c}\bcancel{\text{ 2}{\text{CO}}_{\text{2}}\left(g\right)}\text{ + }\bcancel{\text{3}{\text{H}}_{\text{2}}\text{O}\left(l\right)}\underset{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)\text{ + }\bcancel{\frac{7}{2}{O}_2\left(g\right)}\\ \\ {\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)\text{ + }\bcancel{\frac{\text{5}}{2}{\text{O}}_{\text{2}}\left(g\right)}\underset{}{\to }\bcancel{\text{2}{\text{CO}}_{\text{2}}\left(g\right)}\text{ + }\bcancel{{\text{H}}_{\text{2}}\text{O}\left(l\right)}\\ \\ \underset{\_}{\text{ 2 }{H}_2\left(g\right)\text{ + }\bcancel{{\text{O}}_{\text{2}}\left(g\right)}\underset{}{\to }\bcancel{\text{2}{\text{H}}_{\text{2}}\text{O}\left(l\right)}}\\ \\ {\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)\text{ + 2}{\text{H}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)\end{array}$

According to Hess’s Law, when the direction of reaction reverses, then the enthalpy change will get opposite sign. Also, when the number of reactants and products are multiplied or divided by a factor, then the enthalpy change must also be multiplied or divided by the same factor.

$\Delta H$ of the given reaction can be determined as follows,

  $\begin{array}{c}\Delta H_{eqn\text{ 4}}=\left(-\Delta H_{eqn\text{ 1}}\right)+\left(\frac{\Delta H_{eqn\text{ 2}}}{2}\right)+\left(2\times \Delta H_{eqn\text{ 3}}\right)\\ \\ =\left(-\left(-1560\text{ kJ}\right)\right)+\left(\frac{-2599\text{ kJ}}{2}\right)+\left(2\times -286\text{ kJ}\right)\\ \\ =-312\text{ kJ}\end{array}$

$\Delta H$ of the given reaction is $-312kJ.$