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 ΔH∘:
   • 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 ΔH∘ 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.