Question

Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask is found to contain CH2Cl2 (3.42×10-2 M), CH4 (3.69×10-2 M), and CCl4 (4.12×10-2 M). What is the value of the equilibrium constant for this reaction at 517 K?

Answer 1

Answer: The value of equilibrium constant for the given reaction at 517 K is 1.30

Step-by-step explanation:

The chemical equation for the dissociation of
`CH_2Cl_2` follows:

`2CH_2Cl_2(g)\rightleftharpoons CH_4(g)+CCl_4(g)`

The expression of
`K_(eq)` for above equation follows:

`K_(eq)=([CH_4][CCl_4])/([CH_2Cl_2]^2)`

We are given:

`[CH_4]_(eq)=3.69* 10^(-2)M`

`[CCl_4]_(eq)=4.12* 10^(-2)M`

`[CH_2Cl_2]_(eq)=3.42* 10^(-2)M`

Putting values in above expression, we get:

`K_(eq)=((3.69* 10^(-2))* (4.12* 10^(-2)))/((3.42* 10^(-2))^2)\\\\K_(eq)=1.30`

Hence, the value of equilibrium constant for the given reaction at 517 K is 1.30