Question

Consider the following reaction: 2CH2Cl2(g) CH4(g) CCl4(g) If 0.203 moles of CH2Cl2(g), 0.323 moles of CH4, and 0.240 moles of CCl4 are at equilibrium in a 15.2 L container at 477 K, the value of the equilibrium constant, Kc, is

Answer 1

Answer: Thus value of the equilibrium constant is 1.86

Step-by-step explanation:

Moles of
`CH_2Cl_2` at equilibrium= 0.203 mole

Moles of
`CH_4` at equilibrium = 0.323 mole

Moles of
`CCl_4` at equilibrium = 0.240mole

Volume of solution = 15.2 L

Equilibrium concentration of
`CH_2Cl_2` =
`(moles)/(Volume)=(0.203)/(15.2)=0.0134M`

Equilibrium concentration of
`CH4` =
`(moles)/(Volume)=(0.323)/(15.2)=0.0212M`

Equilibrium concentration of
`CCl_4` =
`(moles)/(Volume)=(0.240)/(15.2)=0.0158M`

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant for this reaction will be,

`K_c=([CH_4]* [CCl_4])/([CH_2Cl_2]^2)`

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

`K_c=(0.0212* 0.0158)/((0.0134)^2)`

`K_c=1.86`

Thus value of the equilibrium constant is 1.86