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

Step-by-step 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 =
`(0.45)/(2)=0.225M`

Initial concentration of chlorine gas =
`(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=(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=([ICl]^2)/([Cl_2][I_2])`

Putting values in above equation, we get:

`9.091=((2x)^2)/((0.225-x)* (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