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 : The concentration of at equilibrium will be, 0.0086 M

Explanation : Given,

Equilibrium constant =

The balanced equilibrium reaction is,

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:

For the given the value of will be,

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

By solving the term 'x', we get:

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 at equilibrium =

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