ThermochemistryMedium

Question

The enthalpy of formation of methane(g) at constant pressure is −18,500 cal/mol at 27°C. The enthalpy of formation at constant volume would be

Options

A.−19,700 cal
B.−17,300 cal
C.−18,498.8 cal
D.−18,500 cal

Solution

$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l);\Delta n_{g} = 1 - 3 = - 2$

Now,

$\Delta E = \Delta H - \Delta n_{g}RT = ( - 18500) - ( - 2) \times 2 \times 300 = - 17300\text{ cal/mol}$

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