Question

Equilibrium concentration of hi, i2 and h2 is 0.7,0.1 and 0.1 moles/litre. Calculate the equilibrium constant for the reaction?

Answer 1

If the reaction is:-

H2+I2 = 2HI

Then,

Kc= [HI]^2 / [H2] +[I2]

= (0.7)^2 / (0.1) x (0.1)

=49

But,

If the reaction is:-

2HI = H2+I2

Then,

Kc= 1/49

Answer 2

Answer: The equilibrium constant for the reaction is 49.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Assuming that the reaction is the formation of hydrogen iodide, the chemical equation follows:

$H_2+I_2\rightarrow 2HI$

The expression of $K_c$ for above equation follows:

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

We are given:

[HI] = 0.7 mol/L

$[H_2]=0.1mol/L$

$[I_2]=0.1mol/L$

Putting values in above equation follows:

$K_c=\frac{(0.7)^2}{0.1\times 0.1}\\\\K_c=49$

Hence, the equilibrium constant for the reaction is 49.