Question

2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O2 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]

Answer 1

1. Heat Evolved by the Reaction :
   • The reaction $2\text{Hg(g)} + \text{O}_2(\text{g}) \rightarrow 2\text{HgO(s)}$ releases heat, and it's found to be 296kJ. This is determined from the rise in temperature in the calorimeter, which has a heat capacity of 20 kJ K−1, and the temperature rose by 14.8K.
2. Calculation of Δ H∘ :
   • Using the relation $\Delta H^\circ = \Delta U^\circ + \Delta (n_g RT)$, we find:

$\Delta H^\circ = -296 \, \text{kJ} - (3 \times 8.3 \, \text{J K}^{-1} \text{mol}^{-1} \times 298 \, \text{K} \times 10^{-3}) \approx -303.42 \, \text{kJ}$

1. Calculation of Δ H∘ for HgO(s) :
   • By using Hess's Law, we can relate the enthalpy change for the reaction to the enthalpies of formation for the substances involved:

$\Delta H^\circ(\text{HgO(s)}) = \Delta H^\circ(\text{HgO(s)}) - \Delta H^\circ(\text{Hg(g)}) - 2 \times \Delta H^\circ(\text{Hg(s)})$

• Substituting the values gives:

$\Delta H^\circ(\text{HgO(s)}) = -303.42 + 122.64 - 180.78 = -303.42 + 90.39 \, \text{kJ mol}^{-1}$

Thus, the absolute value of the enthalpy of formation for solid mercury oxide HgO(s)) is 90.39 kJ mol−1.