Question

In reaction 2 ICl (g) <=> I2 (g) + Cl2 (g) in a 625 ml container. Once equilibrium is reached, 0.0383g of I2 is found in the mixture. What is the value of Kc for the reaction?

Answer 1

2 ICl (g) <=====> I2 (g) + Cl2 (g)

Inicial: 0.682 g 0 0

(first we have only ICl, it doesn't react yet)

When reaction takes place: -2.x +x +x

(x is the amount that reacts,

e.g. -2 . x is the amount of ICl

that is consumed, for I2 and Cl2

we call +x because they are products)

Then in equilibrium:

First, we calculate molarity M

ICL = g/molecular mass . V(L) = 0.682 g/(162.35 g/mol . 0.625 L)=6.72x10^-3M

I2 = 0.0383 g / (253.80 g/mol . 0.625 L) = 2.41x10^-4 M

Remember everything is in the same volume

2 ICl (g) <=====> I2 (g) + Cl2 (g)

Inicial: 6.72x10^-3M 0 0

Reaction takes place: -2.x x x

Equilibrium: 6.72x10^-3M-2.x x x=2.41x10^-4M

`Kc\text{ = }(\lbrack Cl2\rbrack.\lbrack I2\rbrack)/(\lbrack ICl\rbrack^2)=\text{ }\frac{x.x}{(6.72x10^(-3)M-2.x)2^{}}=\frac{2.41x10^(-4)M\text{.}2.41x10^(-4)M}{(6.72x10^(-3)M-2.2.41x10^(-4)M)^2}=8.64x10^(-6)`

Kc = 8.64x10^-6