Question

$2$ $mol$ of $Hg(g)$ is combusted in a fixed volume bomb calorimeter with excess of $O_2$ at 298K and 1 atm into $HgO(s)$ During the reaction, temperature increases from $2980$ $K$ to $3128$ $K$. If heat capacity of the bomb calorimeter and enthalpy of formation of $Hg(g)$ are $2000$ $kJ \;K^{-1}$ and $6132$ $kJ \;mol^{-1}$ at $298$ $K$, respectively, the calculated standard molar enthalpy of formation of $HgO(s)$ at $298$ $K$ is $X$ $kJ \;mol^{-1}$ The value of | $X$ | is [Given : Gas constant $R$ =$83$ $J$ $K^{-1}$ $mol^{-1}$]

Answer 1

The value of $|X|$ is $\underline{90.39}$.

**Explanation: **
Evaluating the given values we get,

$\triangle H^{\degree}_{combustion}$ = $\triangle H^{\degree}_{f}(HgO)-\triangle H^{\degree}_{Hg(l)}+\frac{1}{2} \triangle H^{\degree}_{f}O_2$

$\triangle H^{\degree}_{f}(HgO) = -151.710+61.32$

$\triangle H^{\degree}_{f}=90.39$