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  • The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are <img alt="-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-285.8 \mathrm{~kJ} \mathr

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are -2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. The magnitude of enthalpy of formation of propane \left(\mathrm{C}_{3} \mathrm{H}_{8}\right) is ___________\mathrm{kJ} \; \mathrm{mol}^{-1}. (Nearest integer)

Option: 1

103.7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given the combustion enthalpy -

\text{(1)}\; \mathrm{C}_3 \mathrm{H_8(g)}+5\; \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}_1=-2220.0 \mathrm{~kJ} /\mathrm{~mol}

\text{(2)}\; \mathrm{C_{(s)}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 \quad \mathrm{\Delta }_c \mathrm{H}_2=-393.5 \mathrm{~kJ} / \mathrm{mol}}

\text{(3)}\; \mathrm{\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \quad \Delta _{\mathrm{c}} \mathrm{H}_3=-285.8 \mathrm{~kJ} / \mathrm{mol}}

After doing  3 \times (2) + 4 \times (3) -----(1)

\text{(4)}\; \mathrm{3C_{(s)} + 4H_{2(g)}\rightarrow C_{3}H_{8(g)}}\; \; \; \; \; \; \mathrm{\Delta H = 103.7\; kJ/mol}

(4) is the enthalpy of formation propane \mathrm{(C_{3}H_{6})}

Answer = 103.7

Posted by

Rishabh

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