At a certain temperature 2HI - H2-I2 only 50

The correct option is A 0.25 2HI ⇔H_2 + I_2 #160; #160; #160; #160; 2 #160; #160; #160; 0 #160; #160; #160; #160;0 #160; ...

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Byju's Answer

Standard XII

Chemistry

Law of Mass Action

At a certain ...

Question

At a certain temperature $2 H I \Leftrightarrow H_{2} + I_{2}$ only 50% HI is dissociated at equilibrium. The equilibrium constant is :

0.25

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0.3

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0.5

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Solution

The correct option is A 0.25
$2 H I \Leftrightarrow H_{2} + I_{2}$
2 0 0 Initial concentration
2-2x x x At equilibrium
Since only 50% of HI is dissociated, thus
$2 x = 1 => x = 0.5$
Now, from expression for equilibrium constant for the reaction we get,
$K_{c} = \frac{x^{2}}{2 - 2 x} = 0.25$

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At a certain temperature 2HI⇔H2+I2 only 50% HI is dissociated at equilibrium. The equilibrium constant is :

The correct option is A 0.252HI⇔H2+I2 2 0 0 Initial concentration 2-2x x x At equilibriumSince only 50% of HI is dissociated, thus2x=1=>x=0.5Now, from expression for equilibrium constant for the reaction we get,Kc=x22−2x=0.25

At a certain temperature $2 H I \Leftrightarrow H_{2} + I_{2}$ only 50% HI is dissociated at equilibrium. The equilibrium constant is :

The correct option is A 0.25
$2 H I \Leftrightarrow H_{2} + I_{2}$
2 0 0 Initial concentration
2-2x x x At equilibrium
Since only 50% of HI is dissociated, thus
$2 x = 1 => x = 0.5$
Now, from expression for equilibrium constant for the reaction we get,
$K_{c} = \frac{x^{2}}{2 - 2 x} = 0.25$