Unit 7 · Topic 7.7

7.7 Calculating Equilibrium Concentrations

To find equilibrium amounts, first compare the reaction quotient Q with K to see which way the reaction must go. Then set up an ICE table with the starting amounts and K, and solve for the change, using a small-x approximation when K is small.

Key terms

ICE table
Q < K
Q > K
small-x approximation
equilibrium concentration

Step 1: compare Q with K

Calculate Q from the starting amounts and compare it with K. Think of Q as where you are and K as where you're headed.

Comparison	What it means	Net direction
Q < K	Too few products compared with equilibrium	Forward: reactants are used up, products form
Q > K	Too many products compared with equilibrium	Reverse: products are used up, reactants form
Q = K	Already at equilibrium	No net change; forward and reverse rates are equal

Step 2: set up the ICE table

Use the direction to set signs in the Change row. If the reaction goes forward, reactants get −x (times their coefficients) and products get +x. If it goes backward, it's the reverse.

If you start with zero of any product, the reaction has to go forward, since Q = 0 is less than any K.

Then write K in terms of x using the Equilibrium row and solve.

Going backward and using pressures

If Q > K, the reaction runs in reverse, so in the Change row the products get −x and the reactants get +x. For example, if a flask starts with only HI, the change row for H₂ + I₂ ⇌ 2HI is +x, +x, −2x.

Everything works the same with Kp: use partial pressures in atm in the ICE table instead of molarities. Don't mix the two kinds of units in one table.

Step 3: solve for x

Perfect square: if both sides of the K expression are squares, take the square root of both sides. This happens with H₂ + I₂ ⇌ 2HI when [H₂] and [I₂] start equal.
Small-x approximation: if K is very small compared with the starting concentrations, x will be tiny, so 0.10 − x ≈ 0.10. This turns a hard equation into an easy one. Check afterward that x is less than about 5% of the number you subtracted it from.
If x isn't small, the equation has to be solved exactly (for example with the quadratic formula or a calculator's solver). Exam questions are usually designed so you won't need to.

Step 4: check

Plug your equilibrium values back into the K expression. You should get K back (to rounding). Every concentration must be positive; a negative value means x was set up in the wrong direction or a root was picked that doesn't make physical sense.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Which way will it go?
For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50.0 at a certain temperature. A flask contains [H₂] = 0.10 M, [I₂] = 0.20 M and [HI] = 0.80 M. Is the system at equilibrium? If not, which way does it shift?
Show the solutionHide the solution
1. Step 1: Qc = [HI]² / ([H₂][I₂]) = (0.80)² ÷ (0.10 × 0.20) = 0.64 ÷ 0.020 = 32.
2. Step 2: Q = 32 is less than K = 50.0, so there are too few products.
3. Step 3: The reaction proceeds forward: H₂ and I₂ are consumed and more HI forms until Q = K.
Answer: Not at equilibrium (Q = 32 < K = 50.0); net reaction goes forward, toward HI.
Example 2Calculator allowed
Perfect-square ICE problem
For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50.0. A flask starts with 0.100 M H₂, 0.100 M I₂ and no HI. Find all equilibrium concentrations.
Show the solutionHide the solution
1. Step 1: No HI at the start, so Q = 0 < K and the reaction goes forward.
2. Step 2: ICE: E row is [H₂] = 0.100 − x, [I₂] = 0.100 − x, [HI] = 2x.
3. Step 3: 50.0 = (2x)² / (0.100 − x)². Both sides are perfect squares, so take square roots: √50.0 = 7.07 = 2x / (0.100 − x).
4. Step 4: 7.07(0.100 − x) = 2x → 0.707 = 9.07x → x = 0.0780.
5. Step 5: [HI] = 2x = 0.156 M; [H₂] = [I₂] = 0.100 − 0.0780 = 0.022 M.
6. Step 6: Check: (0.156)² ÷ (0.022)² ≈ 50. It works.
Answer: [HI] = 0.156 M; [H₂] = [I₂] = 0.022 M
Example 3Calculator allowed
Small-x approximation (and checking it)
For COCl₂(g) ⇌ CO(g) + Cl₂(g), Kc = 2.2 × 10⁻¹⁰ at a certain temperature. A flask starts with 0.10 M COCl₂. Find [CO] at equilibrium.
Show the solutionHide the solution
1. Step 1: ICE: [COCl₂] = 0.10 − x, [CO] = x, [Cl₂] = x.
2. Step 2: Kc = x² / (0.10 − x) = 2.2 × 10⁻¹⁰.
3. Step 3: K is tiny, so x will be tiny compared with 0.10. Approximate 0.10 − x ≈ 0.10: x² = (2.2 × 10⁻¹⁰)(0.10) = 2.2 × 10⁻¹¹.
4. Step 4: x = √(2.2 × 10⁻¹¹) = 4.7 × 10⁻⁶ M.
5. Step 5: Check the approximation: 4.7 × 10⁻⁶ ÷ 0.10 × 100% ≈ 0.005%, far below 5%, so it's valid. Skipping this check is a common way to lose a point when an approximation isn't justified.
Answer: [CO] = [Cl₂] ≈ 4.7 × 10⁻⁶ M; [COCl₂] ≈ 0.10 M

Common mistakes

Comparing Q and K backward. Q < K means more products are needed, so the reaction goes forward.
Calculating Q with initial values but forgetting that zero product means Q = 0.
Using the small-x approximation when K isn't small relative to the starting concentration, and never checking it.
Forgetting that a coefficient of 2 means 2x in the Change row, and then (2x)² in the expression.

On the exam

A common question gives a mixture and K and asks you to predict the direction of the net reaction. Calculate Q, compare it with K, and state the conclusion in words.
When you use an approximation, say so and show briefly that x is small compared with the starting value.

Connected topics

Videos

Equilibrium Practice Problems - The ICE Box Method - AP Chem Unit 7, Topic 7a
Jeremy Krug (krugslist)Watch on YouTube (opens in a new tab)
Using the reaction quotient | Equilibrium | AP Chemistry | Khan Academy
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Worked example: Calculating equilibrium concentrations from initial concentrations | Khan Academy
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Equilibrium Equations: Crash Course Chemistry #29
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Practice Problem: Calculating Equilibrium Concentrations
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Check yourself

4 questions on 7.7 Calculating Equilibrium Concentrations. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50 at a certain temperature. A flask at this temperature contains [H₂] = 0.50 M, [I₂] = 0.50 M and [HI] = 2.0 M. Which of the following is true?

Question 2 of 4Calculator allowed

For the reaction A(aq) ⇌ B(aq), Kc = 3.0. A solution is prepared with [A] = 0.80 M and no B. What is [B] once equilibrium is reached?

Question 3 of 4Calculator allowed

For the reaction X(aq) ⇌ Y(aq) + Z(aq), Kc = 1.0 × 10⁻⁶. A solution is made with [X] = 0.010 M and no Y or Z. Which of the following is closest to [Y] at equilibrium?

Question 4 of 4Calculator allowed

For A(g) ⇌ 2B(g), Kc = 1.6 × 10⁻⁵ at a certain temperature. A flask starts with [A] = 0.10 M and no B. Which of the following is closest to [B] at equilibrium?

0 of 4 answered