Question

Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.74 L flask at a certain temperature initially contains 0.767 g H2 and 96.9 g I2. At equilibrium, the flask contains 90.6 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature. Express your answer using two significant figures.

Answer 1

Step 1: calculate initial concentration of H2

Molar mass of H2 = 2.016 g/mol

mass(H2)= 0.767 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(0.767 g)/(2.016 g/mol)

= 0.3805 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.3805/3.74

= 0.1017 M

Step 2: calculate initial concentration of I2

Molar mass of I2 = 253.8 g/mol

mass(I2)= 96.9 g

use:

number of mol of I2,

n = mass of I2/molar mass of I2

=(96.9 g)/(2.538*10^2 g/mol)

= 0.3818 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.3818/3.74

= 0.1021 M

Step 3: calculate final concentration of HI

Molar mass of HI,

MM = 1*MM(H) + 1*MM(I)

= 1*1.008 + 1*126.9

= 127.908 g/mol

mass(HI)= 90.6 g

use:

number of mol of HI,

n = mass of HI/molar mass of HI

=(90.6 g)/(1.279*10^2 g/mol)

= 0.7083 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.7083/3.74

= 0.1894 M

Step 4: calculate Kc

ICE Table:

Given at equilibrium,

[HI] = 0.1894

+2x = 0.1894

x = 0.0947

Equilibrium constant expression is

Kc = [HI]^2/[H2][I2]

Kc = (+2x)^2/(0.1017-1x)(0.1021-1x)

Kc = (+2*0.0947)^2/(0.1017-1*0.0947)(0.1021-1*0.0947)

Kc = 693

Answer: 690 (rounding till 2 significant figures)