Heat of combustion of ethanol at constant pressure and at temperature T K (= 298 K) is found to be -q J mol-1. Hence, heat of combustion (in J mol-1) of ethanol at the same temperature at constant volume will be:

Question

Heat of combustion of ethanol at constant pressure and at temperature T K (= 298 K) is found to be -q J mol-1. Hence, heat of combustion (in J mol-1) of ethanol at the same temperature at constant volume will be: (A) RT-q (B) -(q + RT) (C) q -RT (D) q + RT

Tardigrade Admin · Accepted Answer

C2H5OH(ℓ)+3O2(g)→2CO2(g)+3H2O(ℓ) Δ ng=2-3=-1 So Δ U=Δ H-Δ ngRT =-q+RT

Q. Heat of combustion of ethanol at constant pressure and at temperature $T K (= 298 K)$ is found to be $- q J m o l^{- 1} .$ Hence, heat of combustion $(in J m o l^{- 1})$ of ethanol at the same temperature at constant volume will be:

$RT - q$

$- (q + RT)$

$q - RT$

$q + RT$

$C_{2} H_{5} O H (ℓ) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (ℓ)$
$Δ n_{g} = 2 - 3 = - 1$
So $Δ U = Δ H - Δ n_{g} RT$
$= - q + RT$

Q. Heat of combustion of ethanol at constant pressure and at temperature TK(=298K) is found to be −qJmol−1. Hence, heat of combustion (inJmol−1) of ethanol at the same temperature at constant volume will be:

RT−q

−(q+RT)

q−RT

q+RT