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Standard VIII

Physics

Free Body Diagram

At 700 K, t...

Question

At $700 K$ , the equilibrium constant for the reaction;
$H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}$ is $54.8$ .

If $0.5 m o l l i t r e^{- 1}$ of $H I_{(g)}$ is present at equilibrium at $700 K$ , what are the concentrations of $H_{2 (g)}$ and $I_{2 (g)}$ assuming that we initially started with $H I_{(g)}$ and allowed it to reach equilibrium at $700 K$ .

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Solution

$H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}$
$K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]} = 54.8$

$\Rightarrow [H_{2}] [I_{2}] = \frac{[H I^{2}]}{K_{c}}$

$∵ [H I] = 0.5 m o l l i t r e^{- 1}$ ;
$∴ [H_{2}] [I_{2}] = \frac{[0.5]^{2}}{54.8} = 4.56 \times 10^{- 3}$
At equilibrium, equal concentration of $H_{2}$ and $I_{2}$ will exist if dissociation of $H I$ is carried out. Thus,

$[H_{2}] = [I_{2}]$

$\Rightarrow [H_{2}]^{2} = 4.56 \times 10^{- 3}$
$\Rightarrow [H_{2}] = 0.068 m o l l i t r e^{- 1}$

$\Rightarrow [I_{2}] = 0.068 m o l l i t r e^{- 1}$

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At 700 K, the equilibrium constant for the reaction;H2(g)+I2(g)⇌2HI(g) is 54.8.If 0.5 mol litre−1 of HI(g) is present at equilibrium at 700 K, what are the concentrations of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K.