Observe the following reaction $2 \mathrm{~A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g})…
Observe the following reaction
$2 \mathrm{~A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \xrightarrow{\mathrm{T}(\mathrm{K})} 2 \mathrm{~A}_2 \mathrm{~B}(\mathrm{~g})+600 \mathrm{~kJ}$
The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\prime}\right)$ of $\mathrm{A}_2 \mathrm{~B}(\mathrm{~g})$ is
$600 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Missing
Solution
Given reaction $-2 \mathrm{~A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{~A}_2 \mathrm{~B}+600 \mathrm{~kJ}$
By this equation indicates that the formation of 2 moles of $\mathrm{A}_2 \mathrm{~B}$ releases 600 kJ energy.
$\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{A}_2 \mathrm{~B}\right)=\frac{\text { Total enthalpy change }}{\text { Number of moles of } \mathrm{A}_2 \mathrm{~B}}$
$\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{A}_2 \mathrm{~B}\right)=\frac{600 \mathrm{~kJ}}{2}=300 \mathrm{~kJ} / \mathrm{mol}$
Since reaction ${ }^2$ releases energy (Exothermic). The enthalpy of formation is negative.
$\Delta H_f^{\circ}=-300 \mathrm{~kJ} / \mathrm{mol}$