Use the following data to calculate the molar enthalpy of combustion of ethane, $C_{2} H_{6}$
$2 C_{2} H_{2} (g) + 5 O_{2} (g) ⟶ 4 {CO}_{2} (g) + 2 H_{2} O (l) Δ H = - 2511 kJ / mol$
$\begin{array}{ll} C_{2} H_{2} (g) + 2 H_{2} (g) ⟶ C_{2} H_{6} (g); Δ H = - 311 kJ / mol \\ 2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l); Δ H = - 484 kJ / mol \end{array}$

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$- 1428 kJ / mol$

$- 2684 kJ / mol$

$- 2856 kJ / mol$

$- 3306 kJ / mol$

answer is A.

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Detailed Solution

$\begin{array}{l} - \frac{2511}{2} - 484 - Δ H_{C} (C_{2} H_{6}) = - 311 \\ \Rightarrow Δ H_{C} (C_{2} H_{6}) = - 1428.5 \end{array}$

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