Question

If the heats of combustion of ethane, hydrogen, and graphite are $-1560,-393.5,\mathrm{and}-286\mathrm{kJ}/\mathrm{mol}$, respectively, the standard heat of production ${}_{\mathrm{f}}{\mathrm{H}}_{298°}$ of ethane $(\mathrm{in}\mathrm{kJ}/\mathrm{mol})$ is

Answer 1

Given:

$\begin{array}{rcl}{\mathrm{\Delta H}}_{{\mathrm{C}}_2{\mathrm{H}}_6}°& =& -286\mathrm{kJ}/\mathrm{mol}\dots \left(1\right)\\ {\mathrm{\Delta H}}_{\mathrm{H}}°& =& -393.5\mathrm{kJ}/\mathrm{mol}\dots \left(2\right)\\ {\mathrm{\Delta H}}_{\mathrm{c}}°& =& -1560\mathrm{kJ}/\mathrm{mol}\dots .\left(3\right)\\ \mathrm{\Delta H}°& =& ?\end{array}$

$$\begin{gathered} \begin{array}{rcl}2{\mathrm{C}}_{\left(\mathrm{graphite}\right)}+3{\mathrm{H}}_2\left(\mathrm{g}\right)& \to & {\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)\end{array} \\ \mathrm{Graphite}\mathrm{Hydrogen}\mathrm{Ethane} \end{gathered}$$

We may obtain this by inverting (3) and multiplying (1) by 2 and (2) by 3 and adding.

$\begin{array}{rcl}{\mathrm{\Delta H}}^{°}& =& ∆{{\mathrm{H}}^{°}}_{\mathrm{Product}}-∆{{\mathrm{H}}^{°}}_{\mathrm{Reactant}}\\ & =& {\mathrm{\Delta H}}_{\mathrm{c}}°-(2\times {\mathrm{\Delta H}}_{{\mathrm{C}}_2{\mathrm{H}}_6}°+3\times {\mathrm{\Delta H}}_{\mathrm{H}}°)\\ & =& (–1560)-(2\times (–286)+3\times (–393.5\left)\right)\\ & =& 192.5\mathrm{kJ}/\mathrm{mol}\end{array}$

Final Answer: The value of $\mathrm{\Delta H}°=192.5\mathrm{kJ}/\mathrm{mol}$