$C_{2} H_{6} (g) + 3 .5 O_{2} (g) ⟶ 2 {CO}_{2} (g) + 3 H_{2} O (g)$
$∆ S_{vap} (H_{2} O, l) = x_{1} cal K^{- 1}$ (boiling point $= T_{1}$ )
$\begin{array}{ll} {ΔH}_{f} (H_{2} O, l) = x_{2} \\ {ΔH}_{f} ({CO}_{2}) = x_{3} \\ {ΔH}_{f} (C_{2} H_{6}) = x_{4} \end{array}$
Hence, $∆ H$ for the reaction is :

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$\begin{array}{r} 2 x_{3} + 3 x_{2} - x_{4} + 3 x_{1} T_{1} \end{array}$

$\begin{aligned} 2 x_{3} + 3 x_{2} - x_{4} \end{aligned}$

$\begin{array}{r} 2 x_{3} + 3 x_{2} - x_{4} - 3 x_{1} T_{1} \end{array}$

$x_{1} T_{1} + x_{2} + x_{3} - x_{4}$

answer is B.

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Detailed Solution

${ΔH}_{vap} (H_{2} O, l) = x_{1} T_{1}$

${ΔH}_{f} (reaction) = 2 x_{3} + 3 (x_{2} + x_{1} T_{1}) - x_{4}$ .

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