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Course homeModule II-09 · Recommended before: Before equilibrium concentrations
Equilibrium expressions and ICE tables
An ICE table tracks initial values, stoichiometric changes and final equilibrium values. Concentration ratios are evaluated at equilibrium, not necessarily initially.
Learn the method
K_c = products/reactants, each raised to stoichiometric powers
- Write the balanced reaction and the correct K expression.
- List initial concentrations.
- Represent change using x with coefficients.
- Substitute equilibrium expressions into K.
- Solve and check nonnegative concentrations.
Common mistake: Do not use initial concentrations to evaluate an equilibrium constant unless they are equilibrium values.
Worked example
Problem: For A ⇌ B, [A]eq = 0.20 M and [B]eq = 0.60 M. Find Kc.
Step-by-step solution: Kc = [B]/[A] = 0.60/0.20 = 3.0.
Check the units and whether the answer is reasonable before continuing.
More applications of this mathematics
Try each extension before expanding its solution.
One-step ICE example
For A ⇌ B starting with 0.100 M A and K=3.00, x/(0.100−x)=3 gives x=0.075 M B and 0.025 M A.
Equilibrium coefficient powers
For 2A ⇌ B, Kc = [B]/[A]²; the coefficient of A becomes an exponent of its activity or approximate concentration ratio.
Reaction quotient and Kp
The expression for Q uses the same form as K but current concentrations or pressures. At equilibrium Q=K. For gas equilibria under ideal conditions, Kp=Kc(RT)^Δn using a consistent standard-state convention.
Independent practice — five questions
Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.
Supporting reference
OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.
Applied practice: Predicting direction and using an ICE table
These additional, ungraded problems cover skills beyond the scored five-question check above. Try each before expanding its solution. Your existing score and progress records are unchanged.
Challenge 1. For A ⇌ B with K = 3.0, the initial mixture has [A] = 0.500 M and [B] = 0.100 M. Which direction will the reaction proceed?
Show worked solution for challenge 1
Q = [B]/[A] = 0.100/0.500 = 0.200. Because Q is less than K (3.0), the reaction proceeds forward, producing more B.
Challenge 2. For A ⇌ B with K = 3.0, start with [A] = 0.400 M and [B] = 0 M. Find both equilibrium concentrations.
Show worked solution for challenge 2
Let x be the amount of B formed. K = x/(0.400 − x) = 3.0, so x = 0.300 M. Therefore [B]eq = 0.300 M and [A]eq = 0.100 M.
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