The difference between heats of reaction at constant pressure and at constant volume for the reaction $C_{2} H_{6 (l)} + 5 O_{2 (g)} \to 2 C O_{2 (g)} + 6 H_{2} O_{(g)}$ at $25 C$ at $k J$ is?

- 7.43

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+ 3.72

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- 3.72

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+ 7.43

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Solution

The correct option is C $+ 7.43$
${C_{2} H_{6}}_{(l)} + 5 {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} + 6 {H_{2} O}_{(g)}$

For the above reaction,
$Δ n_{g} = n_{P} - n_{R} = (6 + 2) - 5 = 3$

Now as we know that
$Δ H = Δ U + Δ n_{g} R T$
$\Rightarrow Δ H - Δ U = Δ n_{g} R T$

Here, $Δ H$ and $Δ U$ are the heats of reaction at constant pressure and volume respectively.

Given that:- $T = 25 ℃ = (25 + 273) K = 298 K$
$∴ Δ H - Δ U = 3 \times 8.314 \times 298 = 7432.71 J = 7.43 k J$

Hence the difference between heats of reaction at constant pressure and at constant volume for the given reaction is $+ 7.43 k J$ .

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