Chem 1310: Introduction to physical chemistry

Part 3: Equilibria

Using ICE for equilibria

Using ICE for equilibria

• Write out the chemical equation• Setup the ICE table:

one column for each reactant and productone row each for Initial/Change/Equilibrium

• Define one change as x, express the other changes in it, fill the table

• Use stoichiometry or the rate constant expression to get an equation in x, solve it, calculate all fields

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I

C

E

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I .0175 .0175 0

C

E

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I .0175 .0175 0

C -x -x +2x

E

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I .0175 .0175 0

C -x -x +2x

E .0175-x .0175-x 2x

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I .0175 .0175 0

C -x -x +2x

E .0175-x .0175-x 2x = .0276

H2 + I2 2 HI⇋stoichiometry question

Given initial [H2] = [I2] = 0.0175 M, final [HI] = 0.0270 M, calculate K.

[H2] [I2] [HI]

I .0175 .0175 0

C -x -x +2x

E .0175-x= .0037

.0175-x= .0037

2x = .0276( x = .0138)

H2 + I2 2 HI⇋stoichiometry question

K = [HI]2/([H2][I2]) = (0.0276/0.0037)2 = 56

[H2] [I2] [HI]

I .0175 .0175 0

C -x -x +2x

E .0175-x= .0037

.0175-x= .0037

2x = .0276( x = .0138)

H2 + I2 2 HI⇋initial composition question

Given initial [H2] = [I2] = 0.01 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.01 0

C

E

H2 + I2 2 HI⇋initial composition question

Given initial [H2] = [I2] = 0.01 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.01 0

C -x -x +2x

E

H2 + I2 2 HI⇋initial composition question

Given initial [H2] = [I2] = 0.01 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.01 0

C -x -x +2x

E 0.01-x 0.01-x 2x

H2 + I2 2 HI⇋initial composition question

Use K = [HI]2/([H2][I2]) = (2x)2/(0.01-x)2 = 56 ("trick")

(2x)/(0.01-x) = 7.48 2x = 0.0748-7.48x x = 0.00789

[H2] [I2] [HI]

I 0.01 0.01 0

C -x -x +2x

E 0.01-x 0.01-x 2x

H2 + I2 2 HI⇋initial composition question

x = 0.00789

[H2] [I2] [HI]

I 0.01 0.01 0

C -x -x +2x

E 0.01-x = 0.0021

0.01-x = 0.0021

2x = 0.0158

H2 + I2 2 HI⇋initial composition question 2

Given initial [H2] = 0.01 M, [I2] = 0.02 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.02 0

C

E

H2 + I2 2 HI⇋initial composition question 2

Given initial [H2] = 0.01 M, [I2] = 0.02 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.02 0

C -x -x +2x

E

H2 + I2 2 HI⇋initial composition question 2

Given initial [H2] = 0.01 M, [I2] = 0.02 M, K = 56, what is the final composition?

[H2] [I2] [HI]

I 0.01 0.02 0

C -x -x +2x

E 0.01-x 0.02-x 2x

H2 + I2 2 HI⇋initial composition question 2

K = (2x)2/[(0.01-x)(0.02-x)] = 56

4x2 = 56(0.01-x)(0.02-x) = 56x2-1.68x+0.0112

52x2-1.68x+0.0112 = 0 x = 0.0094

[H2] [I2] [HI]

I 0.01 0.02 0

C -x -x +2x

E 0.01-x 0.02-x 2x

H2 + I2 2 HI⇋initial composition question 2

x = 0.0094

[H2] [I2] [HI]

I 0.01 0.02 0

C -x -x +2x

E 0.01-x = 0.0006

0.02-x = 0.0106

2x = 0.0188

H2 + I2 2 HI⇋mixed composition question

Given initial [H2] = 0.01 M, K = 56, which initial [I2] should we choose to arrive at a final [HI] = 0.015?

[H2] [I2] [HI]

I 0.01 y 0

C

E

H2 + I2 2 HI⇋mixed composition question

Given initial [H2] = 0.01 M, K = 56, which initial [I2] should we choose to arrive at a final [HI] = 0.015?

[H2] [I2] [HI]

I 0.01 y 0

C -x -x +2x

E

H2 + I2 2 HI⇋mixed composition question

Given initial [H2] = 0.01 M, K = 56, which initial [I2] should we choose to arrive at a final [HI] = 0.015?

[H2] [I2] [HI]

I 0.01 y 0

C -x -x +2x

E 0.01-x y-x 2x = 0.015( x = 0.0075)

H2 + I2 2 HI⇋mixed composition question

K = (0.015)2/(0.0025)(y-0.0075) = 56

y = 0.0091 final [I2] = 0.0016

Check by re-calculating K !!!

[H2] [I2] [HI]

I 0.01 y 0

C -x -x +2x

E 0.01-x = 0.0025

y-x =y-0.0075

2x = 0.015( x = 0.0075)

Ammonia formation

Given a mixture of 1 mol N2 and 3 mol H2 in 10L, we add a catalyst to form ammonia at room temperature. What percentage of the starting material will react? KC = 3.5*108.

N2 + 3 H2 2 NH⇋ 3

Ammonia formation

[N2] [H2] [NH3]

I 0.1 0.3 0

C -x -3x +2x

E 0.1-x 0.3-3x 2x

1 mol N2 and 3 mol H2 in 10L

Ammonia formation

KC = [NH3]2/([N2][H2]3) = (2x)2/[(.1-x)(.3-3x)3] = 4x2/[27(.1-x)4] = 3.5*108 2x/[5.196(.1-x)2] = 18708 (.1-x)2 = 0.0000206x x2-0.20002x+0.01 = 0 x = 0.0986

[N2] [H2] [NH3]

I 0.1 0.3 0

C -x -3x +2x

E 0.1-x 0.3-3x 2x

Ammonia formation

x = 0.0986. Conversion percentage: (x/0.1)*100 = 98.6%

Try this for yourself: what happens if the volume is increased to 100 L?

[N2] [H2] [NH3]

I 0.1 0.3 0

C -x -3x +2x

E 0.1-x = .0014

0.3-3x = .0042

2x = .0028

Formation of water -using approximations

O2 + 2 H2 2 H⇋ 2O

We start with 1 mol of O2, 2 mol of H2 in 10 L. How many molecules of reactants remain?

81

2(g)2

2(g)

2(g)2 10*3.3

][O][H

]O[HCK

Formation of water -using approximations

KC = [H2O]2/([O2][H2]2) = (2x)2/[(.1-x)(.2-2x)2] = (2x)2/[4(.1-x)3] = 3.3*1081.

Not easy to solve. Problem!

[O2] [H2] [H2O]

I 0.1 0.2 0

C -x -2x +2x

E 0.1-x 0.2-2x 2x

Formation of water -using approximations

KC = (2x)2/[4(.1-x)3] = 3.3*1081

x must be very close to 0.1.

There is a standard "trick" to solve problems like this. You want to neglect a small variable relative to a larger known quantity.

That works best if we are trying to solve for a small variable. Since the reaction is virtually complete, it is better to solve for the fraction (y) that has not reacted.

KC = [H2O]2/([O2][H2]2) = (0.2-2y)2/[y(2y)2] = (0.2-2y)2/[4y3] (0.2)2/[4y3] = 3.3*1081

y3 = (0.2)2/(4*3.3*1081) = 3.03*10-84 y = 1.45*10-28 mol/Lor a total of 0.00087 molecules. "Product-favoured"

Formation of water -using approximations

[O2] [H2] [H2O]

I 0 0 0.2

C +y +2y -2y

E y 2y 0.2-2y