The enthalpy of combustion of propane, graphite and dihydrogen at $298 K are - 2220.0 {kJ mol}^{- 1}, - 393.5 {kJ mol}^{- 1} and - 285.8 {kJ mol}^{- 1}$ respectively.
The magnitude of enthalpy of formation of propane
$(C_{3} H_{8})$ is ~\text{ kJ mol}^{–1}\). (Nearest integer)

104.00

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104

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104.0

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Solution

Enthalpy of combustion of propane, graphite and $H_{2} at 298 K$ are
$C_{3} H_{6} (g) 5 O_{2} (g) \to 3 C O_{2} (g) 4 H_{2} O (1), Δ H_{1} = - 2220 {kJ mol}^{- 1}$
$C (graphite + O_{2} (g) \to C O_{2} (g), Δ H_{2} = - 393.5 {kJ mol}^{- 1}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), δ H_{3} = - 285.8 {kJ mol}^{- 1}$
The desired reaction is
$3 C (graphite) + 4 H 2 (g) \to C_{3} H_{8} (g)$
$Δ H_{f} = 3 Δ H_{2} + 4 Δ H_{3} - Δ H_{1}$
$= 3 (- 393.5) + 4 (- 285.8) - (- 2220)$
$= - 103.7 {kJ mol}^{- 1}$
$| Δ H_{f} | = 104 {kJ mol}^{- 1}$

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Similar questions

Q. The enthalpy of combustion of methane graphite and dihydrogen at

298 K

are

- 890.3

m o l^{- 1}

and

- 285.8

m o l^{- 1}

respectively. Enthalpy of formation of

C H_{4} (g)

will be:

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$