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Question: At \(700K\) equilibrium constant for the reaction; \({H_{2(g)}} + {I_{2(g)}}\underset {} \leftrigh...

At 700K700K equilibrium constant for the reaction;
H2(g)+I2(g)⇆2HI(g){H_{2(g)}} + {I_{2(g)}}\underset {} \leftrightarrows 2H{I_{(g)}} Is 54.8.54.8. If 0.5 mol0.5\,mol Litre−1Litr{e^{ - 1}} of HI(g)H{I_{(g)}} is present at equilibrium at 700K,700K, what are the concentrations of H2(g){H_{2(g)}} and I(g){I_{(g)}} assuming that we initially started with HI(g)H{I_{(g)}} and allowed it to reach equilibrium at 700K.700K.

Explanation

Solution

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)⇆2HI−{H_{(g)}}^ + {I_{(g)}}\underset {} \leftrightarrows 2H{I^ - }
By applying law of mass action
K=[HI]2[H2][I2].K = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}.

Complete Step by Step Solution:
Given that
H(g)+I(g)⇆2HI−{H_{(g)}}^ + {I_{(g)}}\underset {} \leftrightarrows 2H{I^ - }
Molar concentration of
[HI]=0.5 mol/l[HI] = 0.5\,mol/l at equilibrium
Equilibrium constant K=54.8K = 54.8
Absolute temperature T=700 K.T = 700\, K.
∴\therefore Equilibrium constant K=[HI]2[H2][I2]K = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}} . . . (1)
Where [H2]=[{H_2}] = molar concentration of H2{H_2}
[I2]=[{I_2}] = molar concentration of I2{I_2}
Since, We have to find molar concentration of H2{H_2} and I2.{I_2}.
Let their molar concentration be x mol/l.x\, mol/l.
Since reaction starts from
2HI⇆H2(g)+I2(g)2HI\underset {} \leftrightarrows {H_{2(g)}} + {I_{2(g)}}
Equilibrium constant K=K = The reaction occurs in a diverse direction.
∴\therefore reversing equation (1)
We get,
1K=[H2][I2][HI2]\dfrac{1}{K} = \dfrac{{[{H_2}][{I_2}]}}{{[H{I_2}]}}________2)
By substituting the values in equation (2), we can write
154.8=x×x(0.5)2\dfrac{1}{{54.8}} = \dfrac{{x \times x}}{{{{(0.5)}^2}}}
⇒154.8=x20.25\Rightarrow \dfrac{1}{{54.8}} = \dfrac{{{x^2}}}{{0.25}}
By cross multiplication and rearranging it, we can write
x2=0.2554.8{x^2} = \dfrac{{0.25}}{{54.8}}
⇒x2=0.00456\Rightarrow {x^2} = 0.00456
By taking square root to both the sides, we get
x=0.00456x = \sqrt {0.00456}
⇒x=0.068\Rightarrow x = 0.068
⇒x=0.068 mole/l.\Rightarrow x = 0.068\,mole/l.

**∴\therefore Concentration of H2{H_2} and I2{I_2} at equilibrium is 0.068 mole/l. 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., K1=1K.{K^1} = \dfrac{1}{K}.
(4) If equation is divided by 22 the new K1{K^1} is K.\sqrt K .
(5) If equation is multiply by 22 the new K1{K^1} is K2.{K^2}.
(6) If equation is written in two steps K1=K1×K2.{K^1} = {K_1} \times {K_2}.
(7) The value of equilibrium constant kk is not affected by addition of catalyst to the reaction.

Note:
In the given question equal concentration of H2{H_2} and I2{I_2} are given therefore the same concentration i.e., x mol/lx\,mol/l is taken. The reaction starts from the HI−H{I^ - } side therefore K=1KK = \dfrac{1}{K} should be taken.