Question

Calculate the standard enthalpy change and standard internal energy change for the following reaction at 300K :
$O{F}_2\left(g\right)+H_{2}O\left(g\right)\to O_2\left(g\right)+2HF\left(g\right)$
Given that the standard enthalpy of formation of $O{F}_2$, $H_{2}O$ and $HF$ are $23kjmo{l}^{-1}$,$-241.8kjmo{l}^{-1}$ and $-268.6kjmo{l}^{-1}$ respectively.

Answer 1

Solution:-

${O{F}_2}_{\left(g\right)}+{H_{2}O}_{\left(g\right)}⟶{O_2}_{\left(g\right)}+2{HF}_{\left(g\right)}$

$\Delta H_R={\sum \Delta H_f}_{\left(product\right)}-{\sum \Delta H_f}_{\left(reactant\right)}$

$\therefore \Delta H_R=(2\times (-268.6))-(23+(-241.8))$

$\Rightarrow \Delta H_R=-756kJ/mol$

Hence the standard enthalpy change will be $-756kJ/mol$

Now, as we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

Now from the given reaction,

$\Delta n_g=n_P-n_R=(2+1)-(1+1)=1$

$T=300K\left(\text{Given}\right)$

$R=8.314\times {10}^{-3}J/mol-K$

$\therefore \Delta U=(-756)-(1\times 8.314\times {10}^{-3}\times 300)$

$\Rightarrow \Delta U=-756-2.494=-753.506kJ/mol$

Hence the standard internal energy change will be $-753.506kJ/mol$.