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 = $1.7\times 10^{-3}$

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:

$K_c=\frac{[SbCl_5]}{[SbCl_3][Cl_2]}$

For the given the value of $K_c$ will be, $\frac{1}{1.7\times 10^{-3}}$

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

$\frac{1}{1.7\times 10^{-3}}=\frac{(x)}{(0.0546-x)\times (0.0546-x)}$

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

Answer:

[Cl2] equilibrium = 0.0089 M

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)

$Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M$

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M