Enthalpy of formation of methane is $-75 \mathrm{~kJ} / \mathrm{mol}$. What is the enthalpy change for…
Enthalpy of formation of methane is $-75 \mathrm{~kJ} / \mathrm{mol}$. What is the enthalpy change for formation of $24 \mathrm{~g}$ of methane?
$-112.5 \mathrm{~kJ}$
$-75 \mathrm{~kJ}$
$-150 \mathrm{~kJ}$
$-130 \mathrm{~kJ}$
Solution
Enthalpy of formation for $1 \mathrm{~mol}$ methane $\left(\mathrm{CH}_4\right)$ is $-75 \mathrm{~kJ} / \mathrm{mol}$ moles of $\mathrm{CH}_4=\frac{24}{16}=1.5 \mathrm{moles}$
Change in enthalpy of formation $=-75 \times 1.5$ $=-112.5 \mathrm{~kJ}$