Calculate the work done in the following reaction at constant pressure and at $423 K$ .
(Given : $R = 8.314 J K^{- 1} m o l^{- 1})$ $4 H C I_{(g)} + O_{2 (g)} \to 2 C I_{2 (g)} + 2 H_{2} O_{(g)}$

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Solution

$4 {H C l}_{(g)} + {O_{2}}_{(g)} ⟶ 2 {C l}_{2}_{(g)} + 2 {H_{2} O}_{(g)}$
For the above reaction-
$Δ n_{g} = (2 + 2) - (4 + 1) = - 1$
As we know that-
$W = P Δ V = Δ n_{g} R T (∵ P V = n R T)$
At $423 K$ -
$W = Δ n_{g} R T$
$W = (- 1) \times 8.314 \times 423$
$\Rightarrow W = - 3516.822 J$
Hence the work done for the given reaction is $- 3516.822 J$ .

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Calculate the work done in the following reaction at constant pressure and at 423 K.(Given : R=8.314 JK−1 mol−1) 4HCI(g)+O2(g)→2 CI2(g)+2H2O(g)

4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g)For the above reaction-Δng=(2+2)−(4+1)=−1As we know that-W=PΔV=ΔngRT(∵PV=nRT)At 423K-W=ΔngRTW=(−1)×8.314×423⇒W=−3516.822JHence the work done for the given reaction is −3516.822J.

Calculate the work done in the following reaction at constant pressure and at $423 K$ .
(Given : $R = 8.314 J K^{- 1} m o l^{- 1})$ $4 H C I_{(g)} + O_{2 (g)} \to 2 C I_{2 (g)} + 2 H_{2} O_{(g)}$