JEE Advanced 2022 Paper 1 · Q01 · First Law & Internal Energy
2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O$_2$ at 298 K and 1 atm into HgO(s). During the reaction, temperature increases from 298.0 K to 312.8 K. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g) are 20.00 kJ K$^{-1}$ and 61.32 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 = 8.3 J K$^{-1}$ mol$^{-1}$]
Reveal answer + step-by-step solution
Correct answer:90.39
Solution
Step 1 — heat released in the bomb. $\Delta T = 312.8 - 298.0 = 14.8$ K. Heat absorbed by the calorimeter $q = C \Delta T = 20.00 \times 14.8 = 296$ kJ for 2 mol of Hg, so per mole $\Delta U_{\text{combustion}} = -296/2 = -148$ kJ mol$^{-1}$ (bomb calorimeter measures $\Delta U$).
Step 2 — convert $\Delta U$ to $\Delta H$ for Hg(g) + (1/2) O$_2$(g) $\rightarrow$ HgO(s): $\Delta n_g = 0 - (1 + 1/2) = -3/2$. $\Delta H = \Delta U + \Delta n_g RT = -148 + (-3/2)(8.3 \times 10^{-3})(298) = -148 - 3.71 = -151.71$ kJ mol$^{-1}$.
Step 3 — relate to $\Delta H_f^\circ$(HgO). For the formation reaction we wrote, $\Delta H = \Delta H_f^\circ(\text{HgO}) - \Delta H_f^\circ(\text{Hg, g}) - (1/2)\Delta H_f^\circ(\text{O}_2)$. With $\Delta H_f^\circ(\text{Hg, g}) = 61.32$ and $\Delta H_f^\circ(\text{O}_2) = 0$: $-151.71 = \Delta H_f^\circ(\text{HgO}) - 61.32$, so $\Delta H_f^\circ(\text{HgO}) = -90.39$ kJ mol$^{-1}$.
Therefore X = -90.39 and |X| = 90.39.
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