JEE Advance - Chemistry (2022 - Paper 1 Online - No. 1)
$2 \mathrm{~mol} \,\mathrm{of}\, \mathrm{Hg}(\mathrm{g})$ is combusted in a fixed volume bomb calorimeter with excess of $\mathrm{O}_{2}$ at $298 \mathrm{~K}$ and 1 atm into $\mathrm{HgO}(s)$. During the reaction, temperature increases from $298.0 \mathrm{~K}$ to $312.8 \mathrm{~K}$. If heat capacity of the bomb calorimeter and enthalpy of formation of $\mathrm{Hg}(g)$ are $20.00 \mathrm{~kJ} \mathrm{~K}^{-1}$ and $61.32 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ at $298 \mathrm{~K}$, respectively, the calculated standard molar enthalpy of formation of $\mathrm{HgO}(s)$ at 298 $\mathrm{K}$ is $\mathrm{X}\, \mathrm{kJ}\, \mathrm{mol}^{-1}$. The value of $|\mathrm{X}|$ is _________ .
[Given: Gas constant $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
Explanation
$$2Hg(g)+O_2(g) \longrightarrow 2HgO(s)$$
Heat capacity of calorimeter = 20 kJ K$$-1$$
Rise in temperature = 14.8 K
$$\Delta H^\circ=\Delta U^\circ+\Delta n_g RT$$
$$=-296-3\times8.3\times298\times10^{-3}$$
$$\simeq -303.42$$ kJ
$$\Delta H^\circ = \Delta H{^\circ _f}(HgO(s)) - \Delta H{^\circ _f}(Hg(g))$$
$$ - 303 - 42 = \Delta H{^\circ _f}(HgO(s)) - 2 \times 61.32$$
$$\Delta H{^\circ _f}(HgO(s)) = - 303.42 - 122.64$$
$$ = - 180.78$$ kJ
$$\left| {\Delta H{^\circ _f}(HgO(s))} \right| = 90.39$$ kJ mol$$-$$1
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