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(A) The combustion reaction for liquid benzene $(C_6H_6)$ is: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$.The change in the numbe

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$-3271.74 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The combustion reaction for liquid benzene $(C_6H_6)$ is: $C_6H_6(l) + \frac{15}{2}O_2(g) ightarrow 6CO_2(g) + 3H_2O(l)$.The change in the number of moles of gas is $\Delta n_g = n_p(g) - n_r(g) = 6 - 7.5 = -1.5$.The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is $\Delta H = \Delta U + \Delta n_g RT$.Given $\Delta U = -3268 \ kJ \ mol^{-1} = -3268000 \ J \ mol^{-1}$,$T = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$\Delta H = -3268000 + (-1.5 	imes 8.314 	imes 300)$.$\Delta H = -3268000 - 3741.3 \ J \ mol^{-1} = -3271741.3 \ J \ mol^{-1}$.Therefore,$\Delta H = -3271.74 \ kJ \ mol^{-1}$.

The heat associated with the combustion of liquid benzene at constant volume is $-3268 \ kJ \ mol^{-1}$. Calculate the change in enthalpy when this reaction occurs at $300 \ K$ $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

Similar Questions

Consider the reaction at $300 \ K$:
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