Question

Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [HCl]eq = 0.13 M, [HI]eq = 5.6 × 10-16 M, [Cl2]eq = 0.0019 M. 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s)

Answer 1

Answer: The value of $K_c$ is $2.836\times 10^{31}$

Explanation:

Concentration equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by $K_{c}$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

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

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

For the given chemical reaction:

$2HI(g)+Cl_2(g)\rightleftharpoons 2HCl(g)+I_2(s)$

The expression for $K_c$ is written as:

$K_c=\frac{[I_2][HCl]^2}{[HI]^2[Cl_2]}$

The concentration for solid substances are taken to be 1.

We are given:

$[HI]_{eq}=5.6\times 10^{-16}M$

$[HCl]_{eq}=0.13M$

$[Cl_2]_{eq}=0.0019M$

Putting values in above equation, we get:

$K_c=\frac{(0.13)^2}{(5.6\times 10^{-16})^2\times (0.0019)}\\\\K_c=2.836\times 10^{31}$

Hence, the value of $K_c$ is $2.836\times 10^{31}$