For the combustion of 1 mol of liquid benzen | vedclass

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Question

(A) The combustion reaction of benzene is: $C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6 CO_{2(g)} + 3 H_2O_{(l)}$ The change in the number of ga

Vedclass · Accepted Answer

$-3264.2 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The combustion reaction of benzene is: $C_6H_{6(l)} + \frac{15}{2} O_{2(g)} ightarrow 6 CO_{2(g)} + 3 H_2O_{(l)}$ The change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 6 - 7.5 = -1.5$. Given: $\Delta H = -3268 \ kJ \ mol^{-1}$,$T = 298 \ K$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$. The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$. Rearranging for internal energy change (heat of combustion at constant volume): $\Delta U = \Delta H - \Delta n_g RT$. Substituting the values: $\Delta U = -3268 - (-1.5 	imes 8.314 	imes 10^{-3} 	imes 298)$. $\Delta U = -3268 + 3.716 = -3264.284 \ kJ \ mol^{-1}$. Thus,the heat of combustion at constant volume is approximately $-3264.2 \ kJ \ mol^{-1}$.

For the combustion of $1 \ mol$ of liquid benzene at $298 \ K$,the heat of reaction at constant pressure is $-3268 \ kJ \ mol^{-1}$. What is the heat of combustion at constant volume? $(R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1})$

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The value of $\Delta H$ for cooling $2 \ mol$ of an ideal monoatomic gas from $225^{\circ} C$ to $125^{\circ} C$ at constant pressure will be [given $C_{p} = \frac{5}{2} R$]. (in $R$)

Which of the following relations are correct?
$(A)$ $\Delta U = q + p \Delta V$
$(B)$ $\Delta G = \Delta H - T \Delta S$
$(C)$ $\Delta S = \frac{q_{rev}}{T}$
$(D)$ $\Delta H = \Delta U - \Delta nRT$
Choose the most appropriate answer from the options given below:

The heat of reaction for $C_6H_{12}O_{6(s)} + 6O_{2(g)} \to 6CO_{2(g)} + 6H_2O_{(l)}$ at constant pressure is $-651 \, kcal$ at $17 \, ^oC$. Calculate the heat of reaction at constant volume at $17 \, ^oC$ in $kcal$.

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