The enthalpy change for the reaction $C_{3} H_{8} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g) + {C H}_{4} (g)$ at 25C is -55.7 KJ/mol. Calculate the enthalpy of combustion of $C_{2} H_{6} (g)$ . The enthalpy of combustion of $H_{2}$ and ${C H}_{4}$ are -285.8 and -890.0 KJ/mol respectively. Enthalpy of combustion of propane is -2220 $K J {m o l}^{- 1}$ .

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Solution

$C_{3} H_{8} (g) + H_{2} (g) \to C_{2} H_{6} (g) + C H_{4} (g) . . . (1) △ H = - 55.7 K j / m o l e$
$H_{2} + \frac{1}{2} O_{2} \to C O_{2} + H_{2} O . . . (i i) △ H = - 285 .8 K J / m o l e$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O . . . (i i i) △ H = - 890 K J / m o l e$
$C H_{3} H_{8} + S O_{2} \to 3 C O_{2} + 4 H_{2} O . . . (i v) H = - 2220 K J$
$C_{2} H_{6} \frac{+ 70}{2} 2 \to 2 C O_{2} + 3 H_{2} O △ H = ?$
Reverse the eq (1)
$C_{2} H_{6} + C H_{4} \to C_{3} H_{8} + H_{2} . . . (v) △ = 55.7$
(v) + (iv)
$C_{2} H_{6} + C H_{4} + C_{3} H_{8} + S O_{2} \to 3 C O_{2} + 4 H_{2} O + C_{3} H_{8} + H_{2}$
$(55.7 - 2220) K J / m$
add eq (II) & $e q^{n} (I I I)$ reserve
$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O (55.7 - 2220 - 205.8 + 390)$ KJ/mol
$△ H = - 1560.1 K J / m o l e$

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