The value of enthalpy change (ΔH) for the reaction C2H5OH(I) + 3O2(g) → 2CO2(g) + 3H2O(I) at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be :

Question

The value of enthalpy change (ΔH) for the reaction C2H5OH(I) + 3O2(g) → 2CO2(g) + 3H2O(I) at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be : (A) -1369.0 kJ (B) -1364.0 kJ (C) -1361.5 kJ (D) -1371.5 kJ

Tardigrade Admin · Accepted Answer

C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ) Δ ng = 2-3 = -1 Δ U = Δ H - Δ ng RT = -1366.5-(-1)×(8.314/103)×300 = - 1366.5 + 0.8314 Ã 3 = - 1364 KJ

Q. The value of enthalpy change ( $Δ$ H) for the reaction
$C_{2} H_{5} O H_{(I)} + 3 O_{2 (g)} \to 2 C O_{2 (g)} + 3 H_{2} O_{(I)}$
at $2 7^{\circ} C$ is $- 1366.5 k J m o l^{- 1} .$ The value of internal energy change for the above reaction at this temperature will be :

$- 1369.0 k J$

$- 1364.0 k J$

$- 1361.5 k J$

$- 1371.5 k J$

$C_{2} H_{5} O H (ℓ) + 3 O_{2 (g)} \to 2 C O_{2 (g)} + 3 H_{2} O (ℓ)$
$Δ n_{g} = 2 - 3 = - 1$
$Δ U = Δ H - Δ n_{g} RT$
$= - 1366.5 - (- 1) \times \frac{8.314}{1 0 ^{3}} \times 300$
$= - 1366.5 + 0.8314 \times 3 = - 1364 K J$

Q. The value of enthalpy change (ΔH) for the reaction C2​H5​OH(I)​+3O2(g)​→2CO2(g)​+3H2​O(I)​ at 27∘C is −1366.5kJmol−1. The value of internal energy change for the above reaction at this temperature will be :

−1369.0kJ

−1364.0kJ

−1361.5kJ

−1371.5kJ