JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
Combustion of \(1\) mole of benzene is expressed at \(\mathrm{C}_6 \mathrm{H}_6(1)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(1) \text {. }\) The standard enthalpy of combustion of \(2 \mathrm{~mol}\) of benzene is - ' \(x\) ' \(k J\). \(\mathrm{x}=\) _______ \((1)\) standard Enthalpy of formation of \(1 \mathrm{~mol}\) of \(\mathrm{C}_6 \mathrm{H}_6(1)\), for the reaction \(6 \mathrm{C}\) (graphite) \(+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_n(1)\) is \(48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\). \((2)\) Standard Enthalpy of formation of \(1 \mathrm{~mol}\) of \(\mathrm{CO}_2(\mathrm{~g})\), for the reaction \(\mathrm{C}\) (graphite) \(+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_2(\mathrm{~g})\) is \(-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\). \((3)\) Standard and Enthalpy of formation of \(1 \mathrm{~mol}\) of \(\mathrm{H}_2 \mathrm{O}(1)\), for the reaction \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(1)\) is \(-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
- A \(6535\)
- B \(6540\)
- C \(6545\)
- D \(6550\)
Answer & Solution
Correct Answer
(A) \(6535\)
Step-by-step Solution
Detailed explanation
\(6 \mathrm{C} \text { (graphite) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\ell) ; \Delta \mathrm{H}=48.5 \mathrm{~kJ} / \mathrm{mol}\)…
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