Consider the reaction at 300 K:C_6H_{6(l)} + | vedclass

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(B) The reaction is $C_6H_{6(l)} + 7.5 O_{2(g)} \longrightarrow 6 CO_{2(g)} + 3 H_2O_{(l)}$.First,calculate the change in the number of moles of gas,$

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$-4900.88$

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(B) The reaction is $C_6H_{6(l)} + 7.5 O_{2(g)} \longrightarrow 6 CO_{2(g)} + 3 H_2O_{(l)}$.First,calculate the change in the number of moles of gas,$\Delta n_g = n_{g,products} - n_{g,reactants} = 6 - 7.5 = -1.5 \ mol$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.For $1 \ mol$ of benzene,$\Delta U = \Delta H - \Delta n_g RT = -3271 \ kJ - (-1.5 \ mol 	imes 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 300 \ K)$.$\Delta U = -3271 + 3.7413 = -3267.2587 \ kJ \ mol^{-1}$.For $1.5 \ mol$ of benzene,$\Delta U_{total} = 1.5 	imes (-3267.2587) = -4900.888 \ kJ$.

Consider the reaction at $300 \ K$:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \longrightarrow 6CO_{2(g)} + 3H_2O_{(l)} ; \Delta H = -3271 \ kJ$
What is $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^\circ C$? .....$kJ$

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Consider the reaction at $300 \ K$:$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \longrightarrow 6CO_{2(g)} + 3H_2O_{(l)} ; \Delta H = -3271 \ kJ$What is $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^\circ C$? .....$kJ$