The difference between heat of reaction at constant pressure and constant volume for the reaction given below at $25^{\circ} C$ in $k J$ is :
$A (g) + 3 B (g) \to C (g)$

- 7.42 k J

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+ 3.72 k J

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- 3.72 k J

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7.42 k J

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Solution

The correct option is A $- 7.42 k J$
We know,
Heat of a reaction at constant pressure = $Δ H$
Heat of reaction at constant volume = $Δ U$

Again,
$Δ H = Δ U + P Δ V$
$\Rightarrow Δ H = Δ U + Δ n_{g} R T$
$\Rightarrow Δ H - Δ U = Δ n_{g} R T$
$T = 25^{\circ} C$ or $298 K$

For the reactipon :
$A (g) + 3 B (g) \to C (g)$
$Δ n_{g} = n_{P} - n_{R}$
$= 1 - 4$
$= - 3$
On substituting the values;
$Δ H - Δ U = (- 3) (8.3) \times 298$
$= - 7420.2 J$
$= 7.42 k J$

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The difference between heat of reaction at constant pressure and constant volume for the reaction given below at 25 ∘C in kJ is : A(g)+3B(g)→C(g)

The difference between heat of reaction at constant pressure and constant volume for the reaction given below at $25^{\circ} C$ in $k J$ is :
$A (g) + 3 B (g) \to C (g)$