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 4…

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asked Feb 8, 2021 137k views

1 Answer

Grant Zukowski · Answer 1 · 2021-02-14T23:18:15+0000

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