If $\mathrm{\Delta }{H}_1=900\mathrm{kJ}$

$\mathrm{\Delta }{H}_2=650\mathrm{kJ}$

Then find $\mathrm{\Delta }{H}_{eg}$ of $X$.

Solution:

Given,

$\mathrm{\Delta }{H}_1=900\mathrm{kJ}$

$\mathrm{\Delta }{H}_2=650\mathrm{kJ}$

The change in the enthalpy is

$\mathrm{\Delta }H=\mathrm{\Delta }{H}_2-\mathrm{\Delta }{H}_1$

.$\mathrm{\Delta }H=900-650$

$\mathrm{\Delta }H=250 \mathrm{kJ}$

Hence, $\mathrm{\Delta }{H}_{eg}$ of $X$is 250 kJ.