Question

HA 25

PART B: The following reaction was performed in a sealed vessel at 712 ∘C: H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.55M and [I2]=2.30M . The equilibrium concentration of I2 is 0.0300 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Express your answer numerically.

Answer 1

The given chemical reaction is H_2(g) + I_2(g) ⇌ 2HI_(g)

Now the equilibrium constant, Kc, for a reaction is the ratio between the product of concentration of the products, each raised to the power of the number of moles involved to the product of concentration of the reactants, each raised to the power of the number of moles involved, at equilibrium.

i.e for the above given reaction,

Kc = [HI]² / [H₂]*[I₂]

where all the concentration values should be the concentrations at equilibrium only. Now to find these concentrations sat equilibrium, we have to form the ICE table-

Given initial concentrations of

[HI] = 3.55M

[I2] = 2.30M

Again given at equilibrium, we have [I2] = 0.0300M

That means concentration of I2 reacted = 2.30M - 0.0300M = 2.27‬ M

Again from the chemical equation we can see, mols of H2 and I2 reacted are equimolar, so we can say, concentration of I2 reacted = 2.27‬ M

Again from the equation, we can see from one mole of H2 and I2, we get 2 moles of HI. So we can say concentration of HI reacted = 2.27 * 2 = 4.54‬‬ M

So the ICE table can be-

Reaction	H2(g) +	I2(g) ⇌	2HI(g)
Initial	3.55M	2.30M	0
Change	-2.27‬ M	-2.27‬ M	+ 2*2.27‬ M
Equilibrium	1.28‬ M	0.0300M	4.54‬‬ M

So putting these values-

Kc = [HI]² / [H₂]*[I₂]

= [4.54‬‬ M]² / [1.28‬ M]*[0.0300M ]

= 536.76