Consider the following processes $\Delta H(\mathrm{~kJ} / \mathrm{mol})$ $\begin{array}{ll}1 / 2 A…
Consider the following processes
$\Delta H(\mathrm{~kJ} / \mathrm{mol})$
$\begin{array}{ll}1 / 2 A \rightarrow B & +150 \\ 3 B \rightarrow 2 C+D & -125 \\ E+A \rightarrow 2 D & +350\end{array}$
For $B+D \rightarrow E+2 C, \Delta H$ will be
$525 \mathrm{~kJ} / \mathrm{mol}$
$-175 \mathrm{~kJ} / \mathrm{mol}$
$-325 \mathrm{~kJ} / \mathrm{mol}$
$325 \mathrm{~kJ} / \mathrm{mol}$
Solution
$\begin{aligned}
& \frac{1}{2} A \longrightarrow B ; \quad \Delta H=150 \mathrm{~kJ} / \mathrm{mol} \\
& 3 B \longrightarrow 2 C+D ; \Delta H=-125 \mathrm{~kJ} / \mathrm{mol} \\
& E+A \longrightarrow 2 D ; \Delta H=+350 \mathrm{~kJ} / \mathrm{mol} \\
& \hline \text { By }[2 \times(\mathrm{i})+(\mathrm{ii})]-(\mathrm{iii}), \text { we have } \\
& B+D \longrightarrow E+2 C \\
& \therefore \quad \Delta H=150 \times 2+(-125)-350 \\
& \quad=-175 \mathrm{~kJ} / \mathrm{mol} .
\end{aligned}$