Question

Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentration of 0.0546 M for each of SbCl3 and Cl2 and a concentration of 0.0 M for SbCl5, what is the equilibrium concentration (in molarity) of Cl2? SbCl5(g)↽−−⇀SbCl3(g)+Cl2(g)

Answer 1

Answer : The concentration of
`Cl_2` at equilibrium will be, 0.0086 M

Explanation : Given,

Equilibrium constant =
`1.7* 10^(-3)`

The balanced equilibrium reaction is,

`SbCl_3(g)+Cl_2(g)\rightleftharpoons SbCl_5(g)`

Initial conc. 0.0546 0.0546 0

At eqm. (0.0.546-x) (0.0.546-x) x

The expression of equilibrium constant for the reaction will be:

`K_c=([SbCl_5])/([SbCl_3][Cl_2])`

For the given the value of
`K_c` will be,
`(1)/(1.7* 10^(-3))`

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

`(1)/(1.7* 10^(-3))=((x))/((0.0546-x)* (0.0546-x))`

By solving the term 'x', we get:

`x=0.065M\text{ and }0.046M`

From the values of 'x' we conclude that, x = 0.065 can not be more than initial concentration. So, the value of 'x' which is equal to 0.065 is not consider.

So, x = 0.045 M

Thus, the concentration of
`Cl_2` at equilibrium =
`(0.0546-x)M=[0.0546-2(0.045)]M=0.0086M`

Answer 2

[Cl2] equilibrium = 0.0089 M

Step-by-step explanation:

Given:

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

To determine:

The equilibrium concentration of Cl2

Calculation:

Set-up an ICE table for the given reaction:

`SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)`

I 0 0.0546 0.0546

C +x -x -x

E x (0.0546-x) (0.0546-x)

`Kc = ([SbCl3][Cl2])/([SbCl5])\\\\1.7*10^(-3) =((0.0546-x)^(2) )/(x) \\\\x = 0.0457 M`

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M