The heat of reaction for C_6H_{12}O_{6(s)} + | vedclass

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(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change ($\Delta U$ or $\Delta E$) is given by: $\Delta H = \Delta U + \D

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$-654.48$

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(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change ($\Delta U$ or $\Delta E$) is given by: $\Delta H = \Delta U + \Delta n_g RT$.For the reaction: $C_6H_{12}O_{6(s)} + 6O_{2(g)} 	o 6CO_{2(g)} + 6H_2O_{(l)}$,the change in the number of gaseous moles is $\Delta n_g = n_{g(products)} - n_{g(reactants)} = 6 - 6 = 0$.Given $\Delta H = -651 \, kcal$,$R = 1.987 	imes 10^{-3} \, kcal \, K^{-1} \, mol^{-1}$,and $T = 17 + 273 = 290 \, K$.Since $\Delta n_g = 0$,$\Delta H = \Delta U$.Therefore,$\Delta U = -651 \, kcal$.

The heat of reaction for $C_6H_{12}O_{6(s)} + 6O_{2(g)} \to 6CO_{2(g)} + 6H_2O_{(l)}$ at constant pressure is $-651 \, kcal$ at $17 \, ^oC$. Calculate the heat of reaction at constant volume at $17 \, ^oC$ in $kcal$.

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