The value of enthalpy change $(Δ H)$ for the reaction $C_{2} H_{5} OH (l) + 3 O_{2} (g) ⟶ 2 {CO}_{2} (g) + 3 H_{2} O (l)$ at 27 °C is $1366.5 kJ {mol}^{- 1}$ . The value of internal energy change for the above reaction at this temperature will be

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$- 1369.0 kJmol - 1$

$- 1361.5 kJ {mol}^{- 1}$

$- 1371.5 kJ {mol}^{- 1}$

$- 1364.0 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

Since $Δ_{r} H = Δ_{r} U + (Δ v_{g}) R T,$ we have $Δ_{r} U = Δ_{r} H - (Δ v_{g}) R T$

For the given reaction $(Δ v_{g}) = - 1$ . Hence

$\begin{aligned} Δ_{r} U = - 1366.5 kJ {mol}^{- 1} - (- 1) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1}) (300 K) \\ = (- 1366.5 + 2.494) kJ {mol}^{- 1} ≃ 1364.0 kJ {mol}^{- 1} \end{aligned}$

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