JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
At \(25^{\circ} C\) and \(1\, atm\) pressure, the enthalpy of combustion of benzene\(_{(l)}\) and acetylene\(_{(g)}\) are \(-3268 \,kJ\, mol ^{-1}\) and \(-1300\, kJ\, mol ^{-1}\), respectively. The change in enthalpy for the reaction \(3 C _{2} {H _{2}}_{(g)} \rightarrow C _{6} {H _{6}}_{(l)}\), is \(.....\,kJ \,mol ^{-1}\)
- A \(+\,324\)
- B \(+\,632\)
- C \(-\,632\)
- D \(-\,324\)
Answer & Solution
Correct Answer
(C) \(-\,632\)
Step-by-step Solution
Detailed explanation
\(\Delta H =\sum \Delta H _{\text {Combustion }}\) (Reactant) \(-\sum \Delta H _{\text {Combustion }}\) (Product) \(=3 \times(-1300)\,-\,[-3268]\) \(=-632\, kJ\, mol ^{-1}\)
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