At 300 , K,the standard enthalpies of formati | vedclass

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(D) The combustion reaction of benzoic acid is: $C_6H_5COOH_{(s)} + \frac{15}{2} O_{2_{(g)}} \rightarrow 7CO_{2_{(g)}} + 3H_2O_{(l)}$The enthalpy of c

Vedclass · Accepted Answer

$-3199$

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(D) The combustion reaction of benzoic acid is: $C_6H_5COOH_{(s)} + \frac{15}{2} O_{2_{(g)}} ightarrow 7CO_{2_{(g)}} + 3H_2O_{(l)}$The enthalpy of combustion $\Delta H_c^o$ is given by: $\Delta H_c^o = [7 	imes \Delta H_f^o(CO_2) + 3 	imes \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_6H_5COOH)]$$\Delta H_c^o = [7 	imes (-393) + 3 	imes (-286)] - [-408]$$\Delta H_c^o = [-2751 - 858] + 408 = -3609 + 408 = -3201 \, kJ \, mol^{-1}$For the reaction,$\Delta n_g = n_{products(g)} - n_{reactants(g)} = 7 - 7.5 = -0.5 \, mol$Using the relation $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta U$ is the heat of combustion at constant volume:$-3201 = \Delta U + (-0.5) 	imes (8.314 	imes 10^{-3} \, kJ \, K^{-1} \, mol^{-1}) 	imes 300 \, K$$-3201 = \Delta U - 1.2471$$\Delta U = -3201 + 1.2471 = -3199.7529 \, kJ \, mol^{-1}$

At $300 \, K$,the standard enthalpies of formation of $C_6H_5COOH_{(s)}$,$CO_{2_{(g)}}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. What is the heat of combustion of benzoic acid at constant volume in $kJ$ (in $.75$)? $(R = 8.31 \, J \, mol^{-1} \, K^{-1})$

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