Consider the following reaction:

N₂ (g) + O₂ (g) ⇄ 2 NO (g)

The initial concentration of N₂ is 0.75 M and O₂ is 0.15 M. K_c is equal to 1.0 x 10^-5 at 1227 ^oC. Calculate the equilibrium concentrations of all species.

\(\displaystyle K_c\;=\;\frac{[NO]^2}{[N_2][O_2]}\;=\;1.0 \times 10^{-5}\)

Set up the ICE table.

Put the equilibrium concentrations into the equilibrium constant expression.

\(\displaystyle \frac{(2x)^2}{(0.75\;+\;x)(0.15\;+\;x)}\;=\;1.0 \times 10^{-5}\)

From the equation it looks like we can only solve for x with the quadratic formula. But, the value of k is very small — 1.0 x 10^-5 — when comparing to the initial concentrations. In other words, the amount of x that is subtracted from the initial concentrations, is not going to affect the initial concentrations. This means we can assume x is negligible, and we can neglect it. Our equilibrium constant expression becomes:

\(\displaystyle \frac{(2x)^2}{(0.75)(0.15)}\;=\;1.0 \times 10^{-5}\)

We will make the assumption that 0.75 – x ≈ 0.75 M and 0.15 – x ≈ 0.15 M. We will check the assumption using the 5% rule after we solve for x.

\(\displaystyle x^2\;=\;\frac{1.0 \times 10^{-5}\;(0.75)(0.15)}{4}\)

\(\displaystyle x\;=\;\sqrt{\frac{1.0 \times 10^{-5}\;(0.75)(0.15)}{4}}\)

\(\displaystyle x\;=\;5.3 \times 10^{-4}\)

Next, we need to check our assumption using the 5% rule. We divide x by the initial concentration and multiply by 100. If the answer is less than 5%, the assumption is valid. If it is 5% or greater, the assumption is not valid and the quadratic formula must be used to calculate x.

\(\displaystyle \frac{x}{[\text{initial conc.}]}\;\times 100\)

Check, using the 5% rule for nitrogen and oxygen.

\(\displaystyle \frac{5.3 \times 10^{-4}}{[0.75\;M]}\;\times 100\;=\;0.071\%\)

0.071% < 5% so our assumption was valid. Now we check for oxygen.

\(\displaystyle \frac{5.3 \times 10^{-4}}{[0.15\;M]}\;\times 100\;=\;0.35\%\)

0.35% < 5% and our assumption was valid. The assumptions introduced an error of less than 5%. We can now calculate our equilibrium concentrations.

[N₂]_eq = 0.75 M – 5.3 x 10^-4 M = 0.75 M
[O₂]_eq = 0.15 M – 5.3 x 10^-4 M = 0.15 M
[NO]_eq = 2 x (5.3 x 10^-4 M) = 0.0011 M

In summary, if K is relatively small and [A]_o (the initial concentration) is relatively large, use the 5% rule to calculate x. Always check your assumption. To determine if the 5% rule can be used, you can divide the initial concentration by K, and if it is greater than 400, less than 5% error will be introduced.

\(\displaystyle \text {If}\;\frac {[A]_o}{K_c}\;>\;400\;\;\text{assumption is justified, neglecting x will result in less than 5% error}\)
 
If x is neglected, always check your assumption using the 5% rule. See the figure below for steps on how to solve equilibrium problems.

Watch the Following Videos

Calculation of the Equilibrium Constant
Solving Equilibrium Problems Part 1
Solving Equilibrium Problems Part 2
Solving Equilibrium Problems Part 3
Solving Equilibrium Problems Part 4
Solving Equilibrium Problems Part 5
Solving Equilibrium Problems Part 6

Exercises

Exercise 1. The following reaction has K_c = 4.60 x 10^-3 at a certain temperature.

N₂O₄ (g) ⇄ 2 NO₂ (g)

The initial concentration of N₂O₄ was 3.60 M. What are the concentrations of N₂O₄ and NO₂ at equilibrium?

Check Answer/Solution to Exercise 1

Exercise 2. The following reaction has K_p = 0.0503.

2 Fe (l) + 3 CO₂ (g) ⇄ Fe₂O₃ (s) + 3 CO (g)

Calculate the pressures of all gas species at equilibrium if the initial pressure of CO₂ was 28.25 atm.

Check Answer/Solution to Exercise 2

Exercise 3. What are the equilibrium concentrations of all species in a 0.45 M NH₃ solution?

NH₃ (aq) + H₂O (l) ⇄ NH₄⁺ (aq) + OH^– (aq)    Kc = 1.8 x 10^-5

Check Answer/Solution to Exercise 3

Exercise 4. Calculate the equilibrium concentrations of all species if the initial concentration of HCN is 0.25 M.

HCN (aq) ⇄ H⁺ (aq) + CN^– (aq)     K_c = 4.9 x 10^-10

Check Answer/Solution to Exercise 4

Exercise 5. The following reaction has K_c = 4.60 x 10^-3 at a certain temperature.

N₂O₄ (g) ⇄ 2 NO₂ (g)

The initial concentration of N₂O₄ was 0.500 M. What are the concentrations of N₂O₄ and NO₂ at equilibrium?

Check Answer/Solution to Exercise 5

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