Heat of formation of H2O(g) at 1 atm and 25Â°C is -243 kJ. Δ E for the reaction H2(g) + 30 JK-1 mol-1(1/2) O(g)→ H2O(g) at 25Â°C is

Question

Heat of formation of H2O(g) at 1 atm and 25Â°C is -243 kJ. Δ E for the reaction H2(g) + 30 JK-1 mol-1(1/2) O(g)→ H2O(g) at 25Â°C is (A) 241.8 kJ (B) -241.8 kJ (C) -243 kJ (D) 243 kJ

Tardigrade Admin · Accepted Answer

Δ H = Δ E + Δ nRT Δ n = - 1 /2 ∴ -43= Δ E+(-1/2)×8.314×298×10-3 ∴ Δ E=-241.76

Q. Heat of formation of $H_{2} O (g)$ at $1$ atm and 25°C is
$- 243 k J .Δ E$ for the reaction $H_{2} (g) + 30 J K^{- 1} m o l^{- 1} \frac{1}{2} O (g) \to H_{2} O (g)$ at 25°C is

$241.8 k J$

$- 241.8 k J$

$- 243 k J$

$243 k J$

$Δ H = Δ E + Δ n RT$
$Δ n = - 1/2$
$∴ - 43 = Δ E + (- 1/2) \times 8.314 \times 298 \times 1 0^{- 3}$
$∴ Δ E = - 241.76$

Q. Heat of formation of H2​O(g) at 1 atm and 25°C is −243kJ.ΔE for the reaction H2​(g)+30JK−1mol−121​O(g)→H2​O(g) at 25°C is

241.8kJ

−241.8kJ

−243kJ

243kJ