Question

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

2 ICl(g) ⇌ I2(g) + Cl2(g)

What is the equilibrium concentration of ICl if 0.75 mol of I2 and 0.75 mol of Cl2 are initially mixed in a 2.0-L flask?

options:
A) 0.45 M
B) 1.45 M
C) 0.14 M
D) 0.90 M

Answer 1

D) 0.90 M   is the answer.

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Answer 2

At a certain temperature the equilibrium constant , $K_c$ = 0.11 for the reaction :

2ICl (g)  ⇌ I₂ (g) + Cl₂ (g)

We have to find the equilbrium concentration of ICl if 0.75 mol of I₂ and 0.75 mol of Cl₂ are initially mixed in a 2.0 L flask.

What is equilibrium of chemical reaction ?

there are so many ways to define this term. but the simplest way is " A reaction is said to be in equilibrium when the rate of forward reaction equals to the rate of backward reaction." the rate constant at the equilibrium is known asn equilibrium constant.

it is given as $K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$ when chemical reaction is aA + bB ⇌ cC + dD

the reaction is ,

                  2ICl (g)  ⇌   I₂ (g) + Cl₂ (g) , $K_c=0.11$

reverse the reaction,

                   I₂ (g)   +   Cl₂ (g)   ⇌   2ICl (g) , $K'_c=\frac{1}{K_c}=\frac{1}{0.11}=9.09$

at t = 0       0.75      0.75               0

at eq       0.75 - x    0.75 - x         2x

∴$K'_c=\frac{[ICl]^2}{[I_2][Cl_2]}$

⇒ 9.09 = $\frac{(2x)^2}{(0.75-x)^2}$

⇒ 3 ≈ $\frac{2x}{0.75-x}$

⇒ 2.25 - 3x = 2x

⇒ x = 0.45

now the concentration in the 2 L flask, [ICl] = moles/volume = 2x/2 = x = 0.45 M

Therefore the equilibrium concentration of ICl is 0.45 mol.