For the reaction, $C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$ at constant temperature, $Δ H - Δ E$ is:

+ 3 R T

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- R T

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+ R T

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- 3 R T

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Solution

The correct option is D $- 3 R T$
$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$

$Δ n$ is the difference in the number of moles of gaseous products and gaseous reactants.
$Δ n = (3) - (1 + 5) = - 3$
$Δ H = Δ E + Δ n R T$
$Δ H - Δ E = Δ n R T$
$Δ H - Δ E = - 3 R T$
Hence.option D is correct.

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