Question

Consider the following reaction:

SO2Cl2(g)⇌SO2(g)+Cl2(g)
A reaction mixture is made containing an initial [SO2Cl2] of 2.4×10−2 M . At equilibrium, [Cl2]= 1.3×10−2 M .

Calculate the value of the equilibrium constant (Kc).

Answer 1

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

`SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)`

Initial conc.
`2.4* 10^(-2)` 0 0

At eqm.
`(2.4* 10^(-2)-x)` x x

As we are given:

Concentration of
`Cl_2` at equilibrium =
`1.3* 10^(-2)M`

That means,

`x=1.3* 10^(-2)M`

The expression for equilibrium constant is:

`K_c=([SO_2][Cl_2])/([SO_2Cl_2])`

Now put all the given values in this expression, we get:

`K_c=((x)* (x))/(2.4* 10^(-2)-x)`

`K_c=((1.3* 10^(-2))* (1.3* 10^(-2)))/(2.4* 10^(-2)-1.3* 10^(-2))`

`K_c=0.0154`

Thus, the value of equilibrium constant (Kc) is, 0.0154