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The standard heat of combustion of propane is $- 2220.1 kJ {mol}^{- 1} .$ The standard heat of vaporisation of liquid water is $44.0 kJ {mol}^{- 1} .$ What is $∆ H^{0} of$
$C_{3} H_{8} (g) + 5 O_{2} (g) \overset{}{\to} 3 {CO}_{2} (g) + 4 H_{2} O (g)$

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$- 2220.1 kJ$

$- 2044.1 kJ$

$- 2396.1 kJ$

$- 2176.1 kJ$

answer is B.

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Detailed Solution

$∆ H_{(vap)}^{0} of H_{2} O = 44.0 kJ {mol}^{- 1} ∴ ∆ H_{(vap)}^{0} of 4 mol H_{2} O = 176.0 kJ ∴ ∆ H_{(Rxn)}^{0} = ∆ H_{comb}^{0} (C_{3} H_{8}) + ∆ H_{(vap)}^{0} (4 H_{2} O) = - 2220.1 + 176.0 = - 2044.1 kJ$

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