Question

A mixture of 0.307 M Cl 2 , 0.465 M F 2 , and 0.706 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g ) − ⇀ ↽ − 2 ClF ( g ) K c = 20.0 at 2500 K Calculate the equilibrium concentration of each gas at 2500 K .

Answer 1

Answer: The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

Explanation:

We are given:

Initial concentration of chlorine gas = 0.307 M

Initial concentration of fluorine gas = 0.465 M

Initial concentration of ClF gas = 0.706 M

The given chemical equation follows:

                            $Cl_2(g)+F_2(g)\rightleftharpoons 2ClF(g)$

Initial:                  0.307       0.465       0.706

At eqllm:           0.307-x    0.465-x       0.706+2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[ClF]^2}{[Cl_2][F_2]}$

We are given:

$K_c=20.0$

Putting values in above equation, we get:

$20.0=\frac{(0.706+2x)^2}{(0.307-x)(0.465-x)}\\\\x=0.148,0.993$

Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible

So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M

Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M

Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M

Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.