13.7 Chemical Equilibrium Calculations

          Chemical Equilibrium Calculations

The chemical equilibrium calculations follow the chart of ICE table where I stands for an initial and C for a change and

E for an equilibrium.

The videos below illustrate the chemical equilibrium in some details:

https://www.youtube.com/watch?v=aJ0KNQ5-KaI

https://www.youtube.com/watch?v=54n1XppP-lA

Calculation of Equilibrium Constant Using Equilibrium Concentrations:

1. Calculating the equilibrium constant K C for the following reaction at 25 ^oC if [NO] at equilibrium = 0.106 M, and O2 at equilibrium = 0.122 M and NO2 at equilibrium = 0.129

2 NO (g) +  O2 (g)     2 NO2 (g)

K C = [ NO2 ] ²  /  { [ NO ] ² x [ O2 ] } = [ 0.129 ] ²  /  { [ 0.106 ] ² x [ 0.122 ] } = 12.1

• Given the balanced equation and the value for K C from the question 1, calculate the new K C.

1/2 NO (g)  + 1/4 O2 (g)    1/2 NO2 (g)

K C = [ NO2 ] ^1/2  / { [ NO ] ¹/² x [ O2 ] ^1/4 } = [ 0.129 ] ¹/ ² /  { [ 0.106 ] ¹/² x [ 0.122 ] ^1/4 } = 1.87

Calculation of Equilibrium Constant Using Initial Concentrations:

• Find the equilibrium constant K C for the following equilibrium. The initial concentrations of AB and A2D are

0.30 M before they are mixed and when equilibrium is reached, the equilibrium concentration of A2D is 0.20 M. Use ICE table for your calculations.

2 AB (g) + C2D (s)           A2D (g) +                   2 CB (s) K C = [ A2D ]  /            [ AB ] ²  = [ 0.20 ] / [ 0.5 ] ² = 0.8

Using ICE table:

Calculations of the Change of the Concentration Using Initial Concentration and Equilibrium Constant

• If 0.50 mol of NO2 is placed in a 2.0 L flask to create NO and O2 calculate the change in concentration if K C =

1.2 x 10 ^{– 5}

2 NO2 (g)      2 NO (g)    + O2 (g)

ICE table:

K C = [ NO2 ] ² / { [ NO ] ² * [ O2 ] }

1.2 x 10 ^-5 = { [ 2 x ] ² * [x ] / [ 0.250 – 2 x ] ²

K C = 1.2 x 10 ^-5 <<<< 0.250 , then 2x is discarded because it is insignificant:

1.2 x 10 ^-5 = { [ 2 x ] ² * [x ] / [ 0.250 – 2 x ] ² = 4 x ³ / (0.250) ²

4 x ³ = (0.250) ² * ( 1.2 x 10 ^-5 )

x  = [ (0.250) ² * ( 1.2 x 10 ^-5 ) ] 1/3 = 5.72 x 10 ^-3

[ NO2 ]  = 0.250 – 2 (5.72 x 10 ^-3) = 0.29 M

[ NO ] = 2 x = 2 (5.72 x 10 ^-3 ) = 0.011 M

[ O2 ] = 5.72 x 10 ^-3 = 0.00057 M

Calculation of the Equilibrium Concentrations Using Initial Concentrations and Equilibrium Constant

• For the system, if the start with 0.100 mol/L of CO2 and H2, what are the concentrations of the reactants and products at equilibrium given that K C = 0.64 at 900 K.

CO2 (g)  +  H2 (g)    ßà CO (g) + H2O (g)

K C = { [ CO ] * [ H2O ] } / { [ CO2 ] * [ H2 ] }

K C = 0.64 = { [ x ] * [ x ] } / { [ 0.100 – x ] * [ 0.100 – x ] } = [ x ² ] / [ 0.100 – x ] ²

√ 0.64 = √ [ x ² ] / [ 0.100 – x ] ²

0.80 = x / (0.100 – x)

0.080 – 0.80 x = x

x + 0.80 x = 0.080

1.80 x = 0.080

x = 0.0444

[ CO ] = [ H2O ] = 0.0444 M

[ CO2 ] = [ H2 ] = 0.100 – x = 0.100 – 0.0444 = 0.0556 M

• For the system, if we start with 0.010 mol/L of H2 and I2 and 0.096 mol/ L of HI, what are their concentrations at equilibrium given that K C = 0.016?

2 HI (g)   H2 (g) + I2 (g)

Q = { [ H2 ] * [ I2 ] } / [ HI ] ² = [ 0.010] ² / [0.096] = 0.0010 è Q < K C and hence the reaction shifts to the right (products’ side).

K C = [ 0.010 + x ] * [ 0.010 + x] / { [ 0.096 ] ² } = 0.016 K C = [ 0.010 + x ] ² / [ 0.096 -2x ] ² = 0.016

√[ 0.010 + x ] ² / [ 0.096 -2x ] ² = √0.016 [ 0.010 + x ] / 0.096 – 2x ] = 0.1265 (0.1265) * (0.096 – 2x) = 0.010

0.01214 – 0.253 x = 0.010

0.01214 – 0.010 = 0.253 x

x = 0.008458 = 0.00846

[ H2 ] = [ I2 ] = 0.010 + 0.00846 = 0.01846 M [ HI ] = 0.096 – 2x = 0.07908 M

• At 650 ^oC, the reaction below has a K C value of 0.771. If 2.00 mol pf both hydrogen and carbon dioxide are placed in 4.00 L container and allowed to react, what will be the equilibrium concentrations of all four gases?

H2 (g) + CO2 (g)  ßà CO (g) + H2O (g)

[ H2 ] = [ CO2 ] = [2.00 mol / 4.00 L ] = 0.500 mol / L = 0.500 M

K C = { [ CO ] * [ H2O ] } / { [ H2 ] * [ CO2 ] } = 0.771

K C = { [ x ] * [ x ] } / { [ 0.500 – x ] * [ 0.500 – x ] } = 0.771 0.771 = x² / (0.500 – x) ²

√0.771 = √ x² / (0.500 – x) ²

0.8781 = x / (0.500 – x)

0.4391 – 0.8781x = x è x = 0.239 [ CO ] = [ H2O ] = 0.239 M

[ H2 ] = [ CO2 ] = 0.500 – 0.239 = 0.261 M

In all these examples, the simplified solutions are used with the assumption that the initial concentrations are much larger than the equilibrium concentrations.

In the example below will used the more intensive method using the quadratic equation without any assumptions:

Calculations of the Change of the Concentration Using Initial Concentration and Equilibrium Constant Using the quadratic solution

• Carbonyl bromide, COBr2 can be formed by reacting CO with Br2. A mixture of 0.400 mol CO, 0.300 mol Br2 and 0.0200 mol COBr2 is sealed in a 5.00 L flask. Calculate equilibrium concentration for all gases given that the K C = 0.680

CO (g) + Br2 (g)   COBr2 (g)

[ CO ] = 0.400 mol /5.00 L = 0.0800 M [ Br2 ] = 0.300 mol /5.00 L = 0.0600 M

[ COBr2 ] = 0.0200 mol ‘ 5.00 L = 0.00400 M

Q = { [ COBr2 ] } / { [ CO ] * { Br2 } } = { [ 0.00400 ] } / { [ 0.0800 ] * { 0.0600 } } Q = 0.833

K C = 0.680

Q > K C, therefore the reaction shifts toward the reactants (to the left).

K C = 0.680 = { [ COBr2 ] } / { [ CO ] * { Br2 } }

0.680 = { [ 0.00400 – x ] } / { [ 0.0800 + x ] * { 0.0600 + x } }

0.680 ( 0.004800 + 0.140 x + x² ) = ( 0.00400 – x )

( 3.264 x 10 ³ + 0.0952 x  + 0.680 x² ) = ( 0.00400 – x )

0.680 x² + 1.0952 x – (7.36 x 10 ^-4) = 0

Applying the quadratic equation:

0.680 x² + 1.0952 x – (7.36 x 10 ^-4) = 0

Where:

a = 0.680;  b = 1.0952; c = – 7.36 x 10 ^-4

x = { [ – 1.0952 ] -/+[ √ (1.09520² – 4*(0.680)*( 7.36 x 10 ^-4) ] } / (2* 0.680) x = 6.72 x 10 ^{– 4}

[ COBr2 ] = 0.00400 – 6.72 x 10 ^{– 4} = 0.00333 M

[ CO ] = 0.0800 + 6.72 x 10 ^{– 4} = 0.0807 M

[ Br2 ] = 0.0600 + 6.72 x 10 ^{– 4} = 0.607 M