Question

At 700 K, equilibrium constant for the reaction
H2(g) + I2(g) ⟷ 2HI(g)
is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

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Answer 1

Answer :

• Equilibrium Concentration of H₂ & I₂ = 0.068 M.

Step-by-step explanation :

Given that,

• The equilibrium concentration of HI = 0.5 M.

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A.T.Q,

The equilibrium constant, $\rm{K_c}$ for the reaction given below = 54.8

$\rm{H_2(g)+I_2(g)\leftrightarrow{}2HI(g)}$

Then, the equilibrium constant for the reverse reaction, $\rm{K'_c}$ (given below) = 1/54.8

$\rm{2HI(g)\leftrightarrow{}H_2(g)+I_2(g)}$

$\rule{350}{3}$

Now, let x M be the concentrations of hydrogen and iodine at equilibrium.

By applying the law of chemical equilibrium, we get,

$\implies\rm{K'_c=\dfrac{[H_2][I_2]}{[HI]}}$

Putting the values, we get,

$\implies\rm{\dfrac{1}{54}=\dfrac{x\times{}x}{(0.5)^2}}$

$\implies\rm{\dfrac{1}{54.8}=\dfrac{x^2}{0.25}}$

$\implies\rm{x^2=\dfrac{0.25}{54.8}=0.00456}$

$\implies\rm{x=[H_2]=[I_2]=}\:\bold{0.068\:M.}$