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disa [49]
3 years ago
13

The equilibrium constant for the reaction

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

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

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

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