The difference between heats of reaction at c | vedclass

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Question

(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

Vedclass · Accepted Answer

$-7.43$

Vedclass · Answer

(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 $2C_6H_{6(l)} + 15O_{2(g)} 	o 12CO_{2(g)} + 6H_2O_{(l)}$,the change in the number of moles of gaseous species is: $\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)} = 12 - 15 = -3$. Given $T = 25\,^{\circ}C = 298\,K$ and $R = 8.314\,J\,K^{-1}\,mol^{-1}$. Substituting the values: $\Delta H - \Delta U = -3 	imes 8.314 	imes 298 = -7432.7\,J$. Converting to $kJ$: $-7432.7\,J \approx -7.43\,kJ$.

Process	Entropy Change
$(a)$ Liquid to vapor conversion	$(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature	$(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas	$(3)$ $\Delta S = (-)$

The difference between heats of reaction at constant pressure and at constant volume for the reaction $2C_6H_{6(l)} + 15O_{2(g)} \to 12CO_{2(g)} + 6H_2O_{(l)}$ at $25\,^{\circ}C$ in $kJ$ is

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