How can you predict which direction a reversible reaction will shift—and calculate exactly what remains when it reaches equilibrium? The key is to separate three ideas that students often mix together: a balanced chemical equation, the reaction quotient Q, and the equilibrium constant K. An ICE table then turns those concepts into a solvable math problem.
This college-level guide walks through a complete numerical example, explains what your answer means physically, and finishes with practice questions and detailed solutions. You should already be comfortable with molarity and balancing chemical equations.
Learning goals
- Explain why chemical equilibrium is dynamic, not a stopped reaction.
- Write a concentration-based equilibrium expression from a balanced reaction.
- Use Q versus K to predict the direction of net reaction.
- Construct and solve an ICE table without losing stoichiometric factors.
- Check your equilibrium concentrations and recognize common calculation traps.
1. Equilibrium does not mean the reaction has stopped
For a reversible reaction, reactants can form products while products simultaneously form reactants. At dynamic equilibrium, the forward and reverse reaction rates are equal. Concentrations remain constant over time because the two rates balance—not because molecules stop reacting.
Importantly, the reactant and product concentrations do not have to be equal to each other. Equilibrium tells us about the relative rates and mixture composition, not that every species reaches the same concentration.
For our example, consider nitrogen dioxide and dinitrogen tetroxide: N2O4(g) ⇌ 2 NO2(g). One N2O4 molecule produces two NO2 molecules in the forward direction. Those coefficients will be crucial.
2. How to write Qc and Kc
For the reaction above, the concentration-based reaction quotient is Qc = [NO2]2 / [N2O4]. Brackets denote molar concentrations. The product concentration is squared because the balanced reaction has a coefficient of 2 before NO2. A coefficient is an exponent in the equilibrium expression, not a multiplier in front of a concentration.
The equilibrium constant Kc has the same expression, but evaluated at equilibrium. Formally, equilibrium constants are defined using activities; the molar-concentration equations taught in introductory chemistry are useful approximations under the stated assumptions.
| Comparison | What it tells you |
|---|---|
| Qc < Kc | Net forward reaction: make more products to reach equilibrium |
| Qc = Kc | System is at equilibrium at the specified temperature |
| Qc > Kc | Net reverse reaction: make more reactants to reach equilibrium |
A quick direction-of-shift calculation
Suppose, at a particular temperature, Kc = 0.200 for this reaction. These are illustrative numerical data for a practice problem, not a claimed experimental constant at room temperature. A mixture contains [NO2] = 0.0200 M and [N2O4] = 0.0500 M.
Calculate Qc = (0.0200)2 / 0.0500 = 0.00800. Because 0.00800 < 0.200, the mixture contains too little NO2 relative to equilibrium. The reaction proceeds net forward until Q reaches K.
3. Worked example: solve an ICE table from scratch
Problem. A closed reaction mixture initially contains 0.100 M N2O4 and no NO2. At the chosen temperature, Kc = 0.200. Find the equilibrium concentrations of both substances.
Step 1: define x from the balanced equation
Let x be the amount of N2O4 consumed in molarity units. Each mole of N2O4 consumed produces two moles of NO2, so the concentration changes are −x and +2x.
| ICE | N2O4 (M) | NO2 (M) |
|---|---|---|
| Initial | 0.100 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.100 − x | 2x |
Step 2: substitute equilibrium expressions into Kc
0.200 = (2x)2 / (0.100 − x). Multiply both sides by the denominator: 0.0200 − 0.200x = 4x2, or 4x2 + 0.200x − 0.0200 = 0.
Factoring this convenient example gives (2x − 0.100)(2x + 0.200) = 0, so x = 0.0500 M or x = −0.100 M. We reject x = −0.100 M because the initial condition contains no product and that value would imply a negative NO2 concentration.
Step 3: calculate and verify
[N2O4]eq = 0.100 − 0.0500 = 0.0500 M
[NO2]eq = 2(0.0500) = 0.100 M.
Check by substituting both concentrations into the original expression: Kc = (0.100)2 / 0.0500 = 0.200. Both concentrations are nonnegative and the ratio matches the specified equilibrium constant. The solution passes both checks.
4. What Le Châtelier’s principle actually predicts
Le Châtelier’s principle helps predict a system’s response to a disturbance. Adding NO2 increases Q and can drive the net reaction toward N2O4. Removing NO2 can drive the reaction forward. For this all-gas reaction, compression at constant temperature favors the side with fewer gas molecules: the N2O4 side.
Three subtleties matter: First, the direction of the net shift is not a statement about the final magnitude of every concentration. Second, a catalyst helps a system approach equilibrium faster but does not change K or the equilibrium composition at a fixed temperature. Third, changing temperature can change K; merely changing starting concentrations does not.
5. Mistakes that cost points on college chemistry exams
- Forgetting the coefficient: when N2O4 decreases by x, NO2 increases by 2x, not x.
- Confusing Q with K: use initial or current concentrations for Q; use equilibrium concentrations for K.
- Keeping the impossible quadratic root: a physical concentration cannot be negative.
- Assuming equilibrium means 50:50: concentrations may differ significantly.
- Applying the “5% approximation” automatically: make an approximation only when justified and verify it after solving.
- Including pure solids or liquids in simple K expressions: their activities are effectively constant and are omitted from conventional equilibrium expressions.
6. Practice problems: try these before viewing the solutions
Problem A — Direction. For N2O4(g) ⇌ 2 NO2(g), Kc = 0.200. A mixture contains [NO2] = 0.0600 M and [N2O4] = 0.0400 M. Calculate Q and predict the net shift.
Problem B — ICE table. At another temperature, suppose Kc = 0.300. Starting with 0.150 M N2O4 and zero NO2, calculate the equilibrium concentrations.
Problem C — Reversing reactions. If Kc = 0.200 for N2O4 ⇌ 2 NO2 at a particular temperature, what is Kc for the reverse reaction at that same temperature?
Practice answer key with reasoning
A: Q = (0.0600)2/0.0400 = 0.0900. Since Q < K, the net shift is forward, toward NO2.
B: Set 0.300 = (2x)2/(0.150 − x). The physically valid root is x = 0.0750 M. Therefore [N2O4] = 0.0750 M and [NO2] = 0.150 M. Check: 0.1502/0.0750 = 0.300.
C: Reversing a balanced equation inverts its equilibrium constant. Kreverse = 1/0.200 = 5.00.
Exam habit: write the K expression first, make the ICE table second, solve third, and always substitute your answer back into K.
Explore the concept and keep practicing
For additional conceptual discussion, read equilibrium constants and equilibrium calculations in OpenStax Chemistry 2e. If your algebra or concentration setup needs strengthening, review the IntoChemistry stoichiometry guide. All numerical examples and practice prompts in this article were prepared specifically for this guide.