Question

At $700K$ equilibrium constant for the reaction;
${H_{2(g)}} + {I_{2(g)}}\underset {} \leftrightarrows 2H{I_{(g)}}$ Is $54.8.$ If $0.5\,mol $$Litr{e^{ - 1}}$ of $H{I_{(g)}}$ is present at equilibrium at $700K,$ what are the concentrations of ${H_{2(g)}}$ and ${I_{(g)}}$ assuming that we initially started with $H{I_{(g)}}$ and allowed it to reach equilibrium at $700K.$

Answer 1

Hint: In a reversible process when concentration of reactant and product becomes constant and the rate of forward reaction becomes equal to the rate of backward reaction, the reaction said to be in a chemical equilibrium.
Chemical equilibrium is a dynamic process and not static.

Formula:
For a given reaction
${H_{(g)}}^ + {I_{(g)}}\underset {} \leftrightarrows 2H{I^ - }$
By applying law of mass action
$K = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}.$

Complete Step by Step Solution:
Given that
${H_{(g)}}^ + {I_{(g)}}\underset {} \leftrightarrows 2H{I^ - }$
Molar concentration of
$[HI] = 0.5\,mol/l$ at equilibrium
Equilibrium constant $K = 54.8$
Absolute temperature $T = 700\, K.$
$\therefore $ Equilibrium constant $K = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}$ . . . (1)
Where $[{H_2}] = $molar concentration of ${H_2}$
$[{I_2}] = $molar concentration of ${I_2}$
Since, We have to find molar concentration of ${H_2}$ and ${I_2}.$
Let their molar concentration be $x\, mol/l.$
Since reaction starts from
$2HI\underset {} \leftrightarrows {H_{2(g)}} + {I_{2(g)}}$
Equilibrium constant $K = $The reaction occurs in a diverse direction.
$\therefore $reversing equation (1)
We get,
$\dfrac{1}{K} = \dfrac{{[{H_2}][{I_2}]}}{{[H{I_2}]}}$________2)
By substituting the values in equation (2), we can write
$\dfrac{1}{{54.8}} = \dfrac{{x \times x}}{{{{(0.5)}^2}}}$
$ \Rightarrow \dfrac{1}{{54.8}} = \dfrac{{{x^2}}}{{0.25}}$
By cross multiplication and rearranging it, we can write
${x^2} = \dfrac{{0.25}}{{54.8}}$
$ \Rightarrow {x^2} = 0.00456$
By taking square root to both the sides, we get
$x = \sqrt {0.00456} $
$ \Rightarrow x = 0.068$
$ \Rightarrow x = 0.068\,mole/l.$

$\therefore $ Concentration of ${H_2}$ and ${I_2}$ at equilibrium is $ 0.068\,mole/l.$

Additional Information:
(1) Equilibrium constant depends on temperature.
(2) It is independent of the concentration of the reactant with which we start a reaction or the direction from which equilibrium is established.
(3) If reaction is reversed. The equilibrium constant is reversed i.e., ${K^1} = \dfrac{1}{K}.$
(4) If equation is divided by \[2\] the new ${K^1}$ is $\sqrt K .$
(5) If equation is multiply by $2$ the new ${K^1}$ is ${K^2}.$
(6) If equation is written in two steps ${K^1} = {K_1} \times {K_2}.$
(7) The value of equilibrium constant \[k\] is not affected by addition of catalyst to the reaction.

Note:
In the given question equal concentration of ${H_2}$ and ${I_2}$ are given therefore the same concentration i.e., $x\,mol/l$ is taken. The reaction starts from the \[H{I^ - }\] side therefore $K = \dfrac{1}{K}$ should be taken.