Solving a Simple Kc Problem

Introduction

Writing an ICE table is the setup; solving its equilibrium equation yields actual composition. A simple one-to-one reaction can often be solved by rearranging a linear ratio, while other stoichiometries lead to quadratic equations. Physical bounds and back-substitution decide whether a mathematical root is chemically valid.

Core explanation

For A ⇌ B with initial [A] = C and [B] = 0, a forward change x gives equilibrium [A] = C − x and [B] = x. Then Kc = x/(C − x). Rearranging gives x = KC/(1 + K). This formula shows that x remains between zero and C for positive K, as it must. It also shows the limiting behavior: large K gives x near C, small K gives x near zero.

If B is present initially, use [B] = B₀ + x and [A] = A₀ − x. The equation K = (B₀ + x)/(A₀ − x) still solves linearly. The allowed x range is limited by positive concentrations. If initial Q > K, the solution may yield negative x under a forward-positive convention, indicating net reverse change; do not discard it solely because it is negative.

For A ⇌ 2B with B initially zero, Kc = (2x)²/(C − x) = 4x²/(C − x). This produces a quadratic equation. Mathematical solutions may include a root greater than C or a negative root, which would imply impossible concentrations under the chosen forward setup. Select the root satisfying 0 ≤ x ≤ C and check it in the original expression.

When using calculator output, carry adequate digits in x until all concentrations and K are checked. Near-complete reactions may leave a small residual reactant; premature rounding to zero makes a ratio undefined. If an approximation is used to simplify an equation, verify afterward that the neglected change is small relative to the baseline it replaced.

Every result is tied to the stated temperature, reaction equation and initial composition. A new initial mixture can yield different final concentrations with the same K. A numerical equilibrium calculation therefore combines the thermodynamic constant with a mass balance rather than treating K as a direct concentration by itself.

Step-by-step reasoning

1. Write the balanced equation and ICE table. 2. Substitute equilibrium expressions into Kc. 3. Solve algebraically and impose nonnegative-concentration bounds. 4. Back-substitute into Kc and the material balance.

Visual explanation

Draw a solution flow from initial A₀ to x, then to [A]eq and [B]eq, and finally back into the K ratio as a verification loop.

Real-world analogy

A ledger equation may yield a negative account balance mathematically, but that solution is impossible if no borrowing is allowed. Physical concentration bounds similarly reject invalid algebraic roots.

Real-world example

An analyst measuring a reaction's Kc can combine that value with the initial reactor charge to forecast the product fraction at equilibrium before an experiment is run.

Why?

Why must a root be checked physically? Algebra manipulates symbols without enforcing that species amounts remain nonnegative and that a reaction cannot consume more material than supplied.

Common misconception

“Every quadratic root is a possible equilibrium.” Only roots satisfying the original unsquared equation and material constraints describe a physical composition.

Worked example

For A ⇌ B, start with 0.60 M A and no B, with Kc = 2.0. Let B = x and A = 0.60 − x. Then x/(0.60 − x) = 2.0, so x = 1.20 − 2x and 3x = 1.20, giving x = 0.40 M. Final A = 0.20 M, B = 0.40 M, and B/A = 2.0.

Quick check

1. If A ⇌ B begins with C of A and K = 1, what fraction becomes B? Answer: One half, since x = KC/(1 + K) = C/2.

Exam focus

Show the ICE row and solve from the original K expression. Reject negative species concentrations and verify the answer by back-substitution.

Advanced insight

For several simultaneous reactions, numerical solvers may be needed, but the same structure remains: stoichiometric extent variables, conservation constraints, activity models and equilibrium constants at fixed conditions.

Summary

An ICE table turns initial amounts into equilibrium expressions in x. Solve the K equation, apply physical bounds and check the resulting concentrations in the original quotient.

Practice questions

1. For A ⇌ B, C = 1.0 M and K = 4, find B at equilibrium. Answer: x = 4(1.0)/(1 + 4) = 0.80 M. 2. Why is x = 1.2 M impossible when initial A is 1.0 M and B begins at zero? Answer: It would make equilibrium A negative under A ⇌ B. 3. Does changing initial C change K at fixed temperature? Answer: No. It changes final concentrations but not the temperature-specific constant.