Heat of combustion of ethanol is -671.08Kcalmol- 1 at 27° C. What is ΔE at 27° C for this reaction.

Question

Heat of combustion of ethanol is -671.08Kcalmol- 1 at 27° C. What is ΔE at 27° C for this reaction. (A) -335.24Kcalmol- 1 (B) -670.48Kcalmol- 1 (C) +335.24Kcalmol- 1 (D) +670.48Kcalmol- 1

Tardigrade Admin · Accepted Answer

ÎH=-671.08Kcalmol- 1 C2H5OH(.ℓ .)+3O2(.g.) arrow 2(CO)2(.g.)+3H2O(.ℓ .) ÎH=ÎE+ÎnRT -671.08=ÎE+((. - 1 .) × 2 × 300/1000) ÎE=-671.08+0.6 =-670.48Kcalmol- 1

Q. Heat of combustion of ethanol is $- 671.08 Kc a l m o l^{- 1}$ at $2 7^{\circ} C .$ What is $Δ E$ at $2 7^{\circ} C$ for this reaction.

$- 335.24 Kc a l m o l^{- 1}$

$- 670.48 Kc a l m o l^{- 1}$

$+ 335.24 Kc a l m o l^{- 1}$

$+ 670.48 Kc a l m o l^{- 1}$

$Δ H = - 671.08 Kc a l m o l^{- 1}$
$C_{2} H_{5} O H (ℓ) + 3 O_{2} (g) \to 2 (CO)_{2} (g) + 3 H_{2} O (ℓ)$
$Δ H = Δ E + Δ n RT$
$- 671.08 = Δ E + \frac{( - 1 ) \times 2 \times 300}{1000}$
$Δ E = - 671.08 + 0.6$
$= - 670.48 Kc a l m o l^{- 1}$

Q. Heat of combustion of ethanol is −671.08Kcalmol−1 at 27∘C. What is ΔE at 27∘C for this reaction.

−335.24Kcalmol−1

−670.48Kcalmol−1

+335.24Kcalmol−1

+670.48Kcalmol−1