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Calculating Equilibrium Concentrations

It is often necessary for chemists to determine the concentrations of all the species in a chemical reaction after equilibrium is achieved. The following problems will show you how to do this.

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3 of 35

Calculating Equilibrium Concentrations

Example 1:

Graphite (carbon) and carbon dioxide are kept at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

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4 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I
C
E

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5 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C
E

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6 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x
E

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7 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E

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8 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

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9 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

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10 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

Solve for x using the

quadratic formula or

calculator.

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11 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

0=4x²+ 0.021x - 2.52x10^-4

Solve for x using the

quadratic formula or

calculator.

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12 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

x = 0.0057 M

0=4x²+ 0.021x - 2.52x10^-4

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13 of 35

Calculating Equilibrium Concentrations

Graphite (carbon) and carbon dioxide are kept at constant value at 1000K until the reaction

C(graphite) + CO₂(g) ⇌ 2 CO(g)

has come to equilibrium. At this temperature, K_c = 0.021. The initial concentration of CO₂ is 0.012 M. Calculate the equilibrium concentration of all gaseous species.

R	C (s) +	CO₂(g) ⇌	2 CO (g)
I	---------	0.012 M	0 M
C	---------	- x	+2x
E	---------	0.012 M - x	2x

x = 0.0057 M

[CO]_eq=0M + 2(0.00574 M)

= 0.011 M

[CO₂]_eq=0.012M – 0.00574M

= 0.0063 M

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14 of 35

Calculating Equilibrium Concentrations

Example 2:

The reaction

N₂(g) + O₂(g) ⇋ 2 NO(g)

contributes to air pollution whenever a fuel is burned in air at a high temperature, as in a gasoline engine.

At 1500 K, K_c = 1.0x10^-5 (is this product or reactant favored?).

A sample of air is heated in a closed container to 1500 K. Initially, [N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M. Calculate the equilibrium concentrations of all species.

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15 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

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16 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I
C
E

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17 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C
E

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18 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E

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19 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x

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20 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x

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21 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x

Math trick:

If the concentrations are over 100 times greater than the K value, you can ignore the “-x” terms in the K expression. x will be inconsequentially small compared to the concentrations. Now no quadratic formula!

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22 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x

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23 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x
E

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24 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x
E	~0.80 M

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25 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x
E	~0.80 M	~0.20 M

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26 of 35

Calculating Equilibrium Concentrations

N₂(g) + O₂(g) ⇋ 2 NO(g)

[N₂]₀ = 0.80 M and [O₂]₀ = 0.20 M

K_c = 1.0x10^-5

R	N₂ (s) +	O₂(g) ⇌	2 NO (g)
I	0.80 M	0.20 M	0 M
C	- x	- x	+ 2x
E	0.80 M – x	0.20 M – x	2x
E	~0.80 M	~0.20 M	0.0013 M

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27 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g)

Hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where K_P = 1.00x10².

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

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28 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	Which way will equilibrium shift?
E

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29 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	Which way will equilibrium shift?
E

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30 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	+x	+x	-2x
E

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31 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	+x	+x	-2x
E	0.0100 + x	0.00500 + x	0.500 – 2x

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32 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	+x	+x	-2x
E	0.0100 + x	0.00500 + x	0.500 – 2x

☹ Cannot use the 100x rule.

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33 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	+x	+x	-2x
E	0.0100 + x	0.00500 + x	0.500 – 2x

0 = 96x² + 3.5x – 0.245

x = 0.0355 M

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34 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

R	H₂(g) +	I₂(g) ⇌	2 HI(g)
I	0.0100 atm	0.00500 atm	0.500 atm
C	+x	+x	-2x
E	0.0100 + 0.0355	0.00500 + 0.0355	0.500 – 2(0.0355)

0 = 96x² + 3.5x – 0.245

x = 0.0355 M

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35 of 35

Calculating Equilibrium Concentrations When Q ≠ K

H₂(g) + I₂(g) ⇌ 2 HI(g) K_P = 1.00x10²

A 5.00-L flask contains each of the following species:

HI at 5.00x10^-1 atm, H₂ at 1.00x10^-2 atm, I₂ at 5.00x10^-3 atm

Calculate the equilibrium pressures of all species.

Answers: P_H₂ = 0.0455 atm; P_I₂ = 0.0405 atm; P_HI = 0.429 atm

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