Question

A student ran the following reaction in the laboratory at 294 K:

2CH₂Cl₂(g) ____> CH₄(g) + CCl₄(g)

When she introduced 7.04×10^-2 moles of CH₂Cl₂(g) into a 1.00 Liter container, she found the equilibrium concentration of CCl₄(g) to be 3.25×10^-2 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

Answer 1

Ans :- Equilibrium constant i.e. k_c = 3.62 x 10^-1

Explanation :-

Given,

Initial number of moles of CH₂Cl₂ = 7.04 x 10^-2 mol

Volume of the container = 1.00 L

As,

Concentration of Substance = Number of moles of solute / Volume of the solution in L

So,

Initial concentration of CH₂Cl₂ = Number of moles of CH₂Cl₂ / Volume of the solution in L

= 7.04 x 10^-2 mol / 1 L

= 7.04 x 10^-2 M

ICE table is :

...........................2 CH₂Cl₂ (g) <-----------------------------> CH₄ (g)...............+...................CCl₄ (g)

Initial (I)................7.04 x 10^-2 M........................................0.0 M.......................................0.0 M

Change (C).............-2y ......................................................+y............................................+y

Equilibrium (E)....(7.04 x 10^-2 M - 2y) M ................................y M.........................................y M

Here,

y = Amount dissociated per mole

Given,

Equilibrium concentration of CCl₄ (g) = 3.25 x 10^-2 M = y

So,

y = 3.25 x 10^-2  

Also,

Equilibrium concentration of CH₂Cl₂ (g) = (7.04 x 10^-2 M - 2y) M = (7.04 x 10^-2 - 2x3.25 x 10^-2 ) M = 5.4 x 10^-3 M

Equilibrium concentration of CH₄ (g) = Equilibrium concentration of CCl₄ (g) = y = 3.25 x 10^-2 M

Expression of Equilibrium constant i.e. K_c(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage) is.

k_c = [CH₄].[CCl₄] / [CH₂Cl₂]²

k_c = (3.25 x 10^-2 M)² / (5.4 x 10^-3 M)²

k_c = 0.362 = 3.62 x 10^-1

Hence, Equilibrium constant = kc = 3.62 x 10-1