ThermochemistryHard
Question
At 300 K, the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are −408, −393 and −286 kJ/mol, respectively. The enthalpy of combustion of benzoic acid(s) at 300 K is
Options
A.−3201 kJ/mol, at constant pressure.
B.−3199.75 kJ/mol, at constant pressure.
C.−3201 kJ/mol, at constant volume.
D.−3199.75 kJ/mol, at constant volume.
Solution
$C_{6}H_{5}COOH(s) + \frac{15}{2}O_{2}(g) \rightarrow 7CO_{2}(g) + 3H_{2}O(l)$
$\Delta H = \left\lbrack 7( - 393) + 3( - 286) \right\rbrack - \lbrack - 408\rbrack = - 320\text{ kJ/mol} $$${\text{and }\Delta U = \Delta H - \Delta n_{g}RT }{ = ( - 3201) - \left( 7 - \frac{15}{2} \right) \times \frac{8.314}{1000} \times 300 }{ = - 3199.75\text{ kJ/mol}}$$
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