For the reaction CaCO_{3(s)} rightarrow CaO_{ | vedclass

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(B) The chemical equation is $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$.The change in the number of moles of gaseous species is $\Delta n_g = n_{

Vedclass · Accepted Answer

$163.6$

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(B) The chemical equation is $CaCO_{3(s)} ightarrow CaO_{(s)} + CO_{2(g)}$.The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0 = 1$.The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.Given $T = 977 + 273 = 1250 \, K$ and $R = 8.314 	imes 10^{-3} \, kJ \cdot K^{-1} \cdot mol^{-1}$.Substituting the values: $174 = \Delta U + (1 	imes 8.314 	imes 10^{-3} 	imes 1250)$.$174 = \Delta U + 10.3925$.$\Delta U = 174 - 10.3925 = 163.6075 \, kJ \approx 163.6 \, kJ$.

For the reaction $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$ at $977 \, ^\circ C$,$\Delta H = 174 \, kJ/mol$. Calculate $\Delta U$ in $kJ$.

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