Question

Heat of combustion of ethanol is $-671.08Kcalmol^{- 1}$ at $27^\circ C.$ What is $ΔE$ at $27^\circ C$ for this reaction.

• A. $-335.24Kcalmol^{- 1}$
• B. $-670.48Kcalmol^{- 1}$
• C. $+335.24Kcalmol^{- 1}$
• D. $+670.48Kcalmol^{- 1}$

Answer 1

Answer: (B)

$ΔH=-671.08Kcalmol^{- 1}$
$C_{2}H_{5}OH\left(\right.\ell \left.\right)+3O_{2}\left(\right.g\left.\right) \rightarrow 2\left(CO\right)_{2}\left(\right.g\left.\right)+3H_{2}O\left(\right.\ell \left.\right)$
$ΔH=ΔE+ΔnRT$
$-671.08=ΔE+\frac{\left(\right. - 1 \left.\right) \times 2 \times 300}{1000}$
$ΔE=-671.08+0.6$
$=-670.48Kcalmol^{- 1}$