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

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(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. Here,$\Delta

Vedclass · Accepted Answer

$80.32$

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(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. Here,$\Delta n_g$ is the change in the number of moles of gaseous species,which is $1 - 0 = 1$. The temperature $T = 25 + 273 = 298 \ K$. The gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$. Substituting the values: $82.8 \ kJ = \Delta U + (1 	imes 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 298 \ K)$. $82.8 \ kJ = \Delta U + 2.477 \ kJ$. $\Delta U = 82.8 - 2.477 = 80.323 \ kJ \approx 80.32 \ kJ$.

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

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