The value of enthalpy change (Δ H) for the reaction C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(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

Question

The value of enthalpy change (Δ H) for the reaction C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(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 (A) -1371.5 kJ (B) -1369.0 kJ (C) -1364.0 kJ (D) -1361.5 kJ

Tardigrade Admin · Accepted Answer

Relation between Δ H (enthalpy change) and Δ E (internal energy change) is Δ H=Δ E+ Δ ngRT where Δ ng = (moles of gaseous products) - (moles of gaseous reactants) For the given reaction, Δ ng = 2 - 3 = -1 ⇒ -1366.5=Δ E-1×8.314×10-3×300 ∴ Δ E=-1364.0 kJ mol -1

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

$- 1371.5 k J$

$- 1369.0 k J$

$- 1364.0 k J$

$- 1361.5 k J$

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

−1371.5kJ

−1369.0kJ

−1364.0kJ

−1361.5kJ

Relation between ΔH (enthalpy change) and ΔE (internal energy change) is ΔH=ΔE+Δng​RT where Δng​= (moles of gaseous products) - (moles of gaseous reactants) For the given reaction, Δng​=2−3=−1 ⇒−1366.5=ΔE−1×8.314×10−3×300 ∴ΔE=−1364.0kJmol−1

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

$- 1371.5 k J$

$- 1369.0 k J$

$- 1364.0 k J$

$- 1361.5 k J$