Question

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
2 ICl(g) ? I2(g) + Cl2(g).
What is the equilibrium concentration of ICl if 0.45 mol of I2 and 0.45 mol of Cl2 are initially mixed in a 2.0-L flask?

a.) 0.17 M

b.) 0.27 M

c.) 0.34 M

d.) 0.14 M

Answer 1

Answer: The equilibrium concentration of ICl is 0.27 M

Explanation:

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

Initial concentration of iodine gas = $\frac{0.45}{2}=0.225M$

Initial concentration of chlorine gas = $\frac{0.45}{2}=0.225M$

For the given chemical equation:

$2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11$

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      $I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091$

Initial:         0.225      0.225

At eqllm:   0.225-x    0.225-x     2x

The expression of $K_c$ for above equation follows:

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

Putting values in above equation, we get:

$9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668$

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M