In the reaction at 300 K:C_6H_6(ell) + frac{ | vedclass

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(B) For the combustion of $1 \ mol$ of benzene:$\Delta n_g = \sum n_g(	ext{products}) - \sum n_g(	ext{reactants}) = 6 - 7.5 = -1.5 \ mol$.The relati

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(B) For the combustion of $1 \ mol$ of benzene:$\Delta n_g = \sum n_g(	ext{products}) - \sum n_g(	ext{reactants}) = 6 - 7.5 = -1.5 \ mol$.The relation between $\Delta H$ and $\Delta U$ is given by $\Delta H = \Delta U + \Delta n_g RT$.Given $\Delta H = -3271 \ kJ$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$:$-3271 = \Delta U + (-1.5) 	imes (8.314 	imes 10^{-3}) 	imes 300$.$-3271 = \Delta U - 3.7413$.$\Delta U = -3271 + 3.7413 = -3267.2587 \ kJ \ mol^{-1}$.For $1.5 \ mol$ of benzene,the total change in internal energy is:$\Delta U_{total} = 1.5 	imes (-3267.2587) \approx -4900.89 \ kJ$.

In the reaction at $300 \ K$:
$C_6H_6(\ell) + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O(\ell) \quad \Delta H = -3271 \ kJ$
What is the value of $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^oC$? $..... \ kJ$

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In the reaction at $300 \ K$:$C_6H_6(\ell) + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O(\ell) \quad \Delta H = -3271 \ kJ$What is the value of $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^oC$? $..... \ kJ$