2 mol of Hg_{(g)} is combusted in a fixed vo | vedclass

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(A) The heat released in the bomb calorimeter is $Q = C \Delta T = 20.00 \ kJ \ K^{-1} 	imes (312.8 - 298.0) \ K = 20.00 	imes 14.8 = 296 \ kJ$.Sinc

Vedclass · Accepted Answer

$90.39$

Vedclass · Answer

(A) The heat released in the bomb calorimeter is $Q = C \Delta T = 20.00 \ kJ \ K^{-1} 	imes (312.8 - 298.0) \ K = 20.00 	imes 14.8 = 296 \ kJ$.Since $2 \ mol$ of $Hg_{(g)}$ are reacted,the internal energy change for the reaction $Hg_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow HgO_{(s)}$ is $\Delta U = -\frac{296 \ kJ}{2 \ mol} = -148 \ kJ \ mol^{-1}$.Using $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta n_g = 0 - (1 + 0.5) = -1.5 \ mol$:$\Delta H = -148 + (-1.5 	imes 8.3 	imes 10^{-3} 	imes 298) = -148 - 3.7101 = -151.7101 \ kJ \ mol^{-1}$.This is the enthalpy change for the reaction $Hg_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow HgO_{(s)}$.Given $\Delta H_f(Hg_{(g)}) = 61.32 \ kJ \ mol^{-1}$,we use the relation $\Delta H_{reaction} = \Delta H_f(HgO_{(s)}) - [\Delta H_f(Hg_{(g)}) + \frac{1}{2} \Delta H_f(O_{2(g)})]$.$-151.7101 = \Delta H_f(HgO_{(s)}) - 61.32$.$\Delta H_f(HgO_{(s)}) = -151.7101 + 61.32 = -90.39 \ kJ \ mol^{-1}$.Thus,$|X| = 90.39$.

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