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

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(B) For $1\, mol$ of combustion of benzene:$\Delta n_g = n_{p(g)} - n_{r(g)} = 6 - 7.5 = -1.5\, mol$Using the relation $\Delta H = \Delta U + \Delta n

Vedclass · Accepted Answer

$-4900.88$

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(B) For $1\, mol$ of combustion of benzene:$\Delta n_g = n_{p(g)} - n_{r(g)} = 6 - 7.5 = -1.5\, mol$Using the relation $\Delta H = \Delta U + \Delta n_g RT$:$-3271 = \Delta U + (-1.5) 	imes (8.314 	imes 10^{-3} \, kJ \, K^{-1} \, mol^{-1}) 	imes 300 \, K$$-3271 = \Delta U - 3.7413$$\Delta U = -3271 + 3.7413 = -3267.2587 \, kJ \, mol^{-1}$For $1.5\, mol$ of benzene:$\Delta U_{total} = 1.5 	imes (-3267.2587) \approx -4900.89 \, kJ$Rounding to the nearest provided option,the result is $-4900.88\, kJ$.

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