For the reaction, C (graphic) + $\frac{1}{2}O_{2}(g) \rightarrow CO(g)$ at 298 K and 1 atm, $\mathrm{\Delta}H = - 26.4$ kcal. What is $\mathrm{\Delta}E$, if the molar volume of graphite is 0.0053 L? $(R = 0.002\ kcal\ mol^{- 1}\ K^{- 1}$)

Question

Dr. Abubacker · Accepted Answer

(a)
C (graphite)+$\frac{1}{2}O_{2}(g) ightarrow CO(g)$
$\mathrm{\Delta}n_{g} = 1 - \frac{1}{2} = \frac{1}{2}$
$\mathrm{\Delta}H = \mathrm{\Delta}E + \mathrm{\Delta}n_{g}RT$
$\mathrm{\Delta}E = \mathrm{\Delta}E - \mathrm{\Delta}n_{g}RT$
$= - 26.4 - \frac{1}{2} 	imes 0.002 	imes 298$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - 26.7$ kcal

Q. For the reaction, C (graphic) + $\frac{1}{2}O_{2}(g) \rightarrow CO(g)$ at 298 K and 1 atm, $\mathrm{\Delta}H = - 26.4$ kcal. What is $\mathrm{\Delta}E$, if the molar volume of graphite is 0.0053 L? $(R = 0.002\ kcal\ mol^{- 1}\ K^{- 1}$)

Explanation

Reviewed by Dr. Abubacker

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