Question

At certain temperature, a 10 litre vessel contains 0.4 mole i2 & 0.1 mole HI at equilibrium. Then Kp for H2 + I2 =2HI

Answer 1

Answer: The value of the equilibrium constant is 0.0625

Explanation:

Moles of  $I_2$ at equilibrium= 0.4 mole

Volume of container = 10 L

equilibrium concentration of $I_2=\frac{moles}{volume}=\frac{0.4moles}{10L}=0.04M$

Moles of  $HI$ at equilibrium= 0.1 mole

Volume of container = 10 L

equilibrium concentration of $HI=\frac{moles}{volume}=\frac{0.1moles}{10L}=0.01M$

The given balanced equilibrium reaction is,

                       $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initial conc           x             x         0

At eqm. conc.    (x-y)  M   (x-y) M   (2y) M

The expression for equilibrium constant for this reaction will be,

$K_p=\frac{[HI]^2}{[H_2]\times [I_2]}$

$K_p=\frac{(2y)^2}{(x-y)(x-y)}$

we are given : (x-y) = 0.04  

2y = 0.01

Now put all the given values in this expression, we get :

$K_p=\frac{(0.01)^2}{0.04\times 0.04}$

$K_p=0.0625$

Thus the value of the equilibrium constant is 0.0625

Answer 2

Explanation:

so this is the correct answer