For the reaction FeCO_{3(s)} to FeO_{(s)} + C | vedclass

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Question

(B) For the given reaction $FeCO_{3(s)} 	o FeO_{(s)} + CO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_{P(g)} - n_{R(g)} = 1

Vedclass · Accepted Answer

$80.32$

Vedclass · Answer

(B) For the given reaction $FeCO_{3(s)} 	o FeO_{(s)} + CO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_{P(g)} - n_{R(g)} = 1 - 0 = 1$.Given $\Delta H = 82.8 \ kJ$ and $T = 25 \ ^{\circ}C = 298 \ K$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$,we have $\Delta U = \Delta H - \Delta n_g RT$.Substituting the values: $\Delta U = 82.8 \ kJ - (1 	imes 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 298 \ K)$.$\Delta U = 82.8 - 2.4776 \approx 82.8 - 2.48 = 80.32 \ kJ$.Thus,the value of $\Delta U$ is $80.32 \ kJ$.

For the reaction $FeCO_{3(s)} \to FeO_{(s)} + CO_{2(g)}$,$\Delta H = 82.8 \ kJ$ at $25 \ ^{\circ}C$. What is $\Delta U$ at $25 \ ^{\circ}C$ in $kJ$?

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