Question

\(2\) \(mol\) of \(Hg(g)\) is combusted in a fixed volume bomb calorimeter with excess of \(O_2\) at 298K and 1 atm into \(HgO(s)\) During the reaction, temperature increases from \(2980\) \(K\) to \(3128\) \(K\). If heat capacity of the bomb calorimeter and enthalpy of formation of \(Hg(g)\) are \(2000\) \(kJ \;K^{-1}\) and \(6132\) \(kJ \;mol^{-1}\) 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 \(R\) =\(83\) \(J\) \(K^{-1}\) \(mol^{-1}\)] 

Answer 1

Correct Answer: 90.39

The value of \(|X|\) is \(\underline{90.39}\).

Explanation: 
Evaluating the given values we get,

\(\triangle H^{\degree}_{combustion}\) = \(\triangle H^{\degree}_{f}(HgO)-\triangle H^{\degree}_{Hg(l)}+\frac{1}{2} \triangle H^{\degree}_{f}O_2\)

\(\triangle H^{\degree}_{f}(HgO) = -151.710+61.32\)

\(\triangle H^{\degree}_{f}=90.39\)