The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C6H6(.l.)+15O2(.g.) arrow 12CO2(.g.)+6H2O(.l.) at 25° C in kJ is

Question

The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C6H6(.l.)+15O2(.g.) arrow 12CO2(.g.)+6H2O(.l.) at 25° C in kJ is (A) -7.43 (B) +3.72 (C) -3.72 (D) +7.43

Tardigrade Admin · Accepted Answer

Δ H-Δ E=Δ ngRT Δ ng=12-15=-3 So Δ H -Δ E =-3 × 8.314 J mol -1 K -1 × 298 K =-7.43 kJ mol -1

Q. The difference between heats of reaction at constant pressure and constant volume for the reaction,
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l)$ at $2 5^{\circ} C$ in kJ is

-7.43

+3.72

-3.72

+7.43

$Δ H - Δ E = Δ n_{g} RT$
$Δ n_{g} = 12 - 15 = - 3$
So $Δ H - Δ E = - 3 \times 8.314 J m o l^{- 1} K^{- 1} \times 298 K$ $= - 7.43 k J m o l^{- 1}$

Q. The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C6​H6​(l)+15O2​(g)→12CO2​(g)+6H2​O(l) at 25∘C in kJ is