Question

For combustion of one mole of magnesium in an open container at $300 \,K$ and $1$ bar pressure, $\Delta_{ C } H ^{\Theta}$ $=-601.70\, kJ \,mol ^{-1}$, the magnitude of change in internal energy for the reaction is ____$KJ$(Nearest integer) (Given $: R =8.3 \,J K ^{-1} \,mol ^{-1}$ )

Answer 1

$Mg ( s )+\frac{1}{2} O _{2}( g ) \rightarrow MgO ( s )$
$\Delta H =\Delta U +\Delta n _{ g } RT$
$-601.70 \times 10^{3}=\Delta U -\frac{1}{2} \times 8.3 \times 300$
$-601.70 \,kJ =\Delta U -1.245\, kJ$
$\Delta U =-600.455\, kJ$
Ans. $600$