MHT CET · Chemistry · Thermodynamics (C)
What is the difference between \(\Delta \mathrm{H}\) and \(\Delta \mathrm{U}\) for reaction given below at \(298 \mathrm{~K}\) ? \(\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) \(2 \mathrm{C}_6 \mathrm{H}_{6(\ell)}+150_{2(\mathrm{~g})} \rightarrow 12 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(\ell)}\)
- A \(-2.72 \mathrm{~kJ}\)
- B \(-7.43 \mathrm{~kJ}\)
- C \(-7.8 \mathrm{~kJ}\)
- D \(-3.72 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(B) \(-7.43 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{H}-\Delta \mathrm{U}=\Delta \mathrm{ngRT} \)
\( \Delta \mathrm{ng}=12-15=-3 \)
\( \Delta \mathrm{H}-\Delta \mathrm{U}=-3 \times 8.314 \times 298 \)
\( =-7432.7 \mathrm{~J}=-7.432 \mathrm{~kJ}\)
\( \Delta \mathrm{ng}=12-15=-3 \)
\( \Delta \mathrm{H}-\Delta \mathrm{U}=-3 \times 8.314 \times 298 \)
\( =-7432.7 \mathrm{~J}=-7.432 \mathrm{~kJ}\)
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