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 0.766 g H2 and 96.8 g I2. At equilibrium, the flask contains 90.2 g HI.

Part A Calculate the equilibrium constant (Kc) for the reaction at this temperature. Express your answer using two significant figures. Kc= nothing SubmitPrevious AnswersRequest Answer

Answer 1

Calculate initial concentration of H2:

Molar mass of H2 = 2.016 g/mol

mass(H2)= 0.766 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(0.766 g)/(2.016 g/mol)

= 0.38 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.38/3.74

= 0.1016 M

Calculate initial concentration of I2:

Molar mass of I2 = 253.8 g/mol

mass(I2)= 96.8 g

use:

number of mol of I2,

n = mass of I2/molar mass of I2

=(96.8 g)/(2.538*10^2 g/mol)

= 0.3814 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.3814/3.74

= 0.102 M

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)= 90.2 g

use:

number of mol of HI,

n = mass of HI/molar mass of HI

=(90.2 g)/(1.279*10^2 g/mol)

= 0.7052 mol

volume , V = 3.74 L

use:

Molarity,

M = number of mol / volume in L

= 0.7052/3.74

= 0.1886 M

Now calculate Kc:

ICE Table:

[H2] [I2] [HI]

initial 0.1016 0.102 0

change -1x -1x +2x

equilibrium 0.1016-1x 0.102-1x +2x

Given at equilibrium,

[HI] = 0.1886

+2x = 0.1886

x = 0.0943

Equilibrium constant expression is

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

Kc = (+2x)^2/(0.1016-1x)(0.102-1x)

Kc = (+2*0.0943)^2/(0.1016-1*0.0943)(0.102-1*0.0943)

Kc = 633

Answer: 6.3*10^2