Calculating Equilibrium Concentrations (ICE Tables)计算平衡浓度（ICE 表）

Big Idea · Transformations (TRA)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 7~50 min

Learning goals

  • Given K and initial concentrations, use an ICE table to set up an equation in x and find the equilibrium concentrations
  • Solve by taking a square root, by the “small x” approximation, or with the quadratic formula, and use the 5% rule to check the approximation
  • Choose the physically meaningful root of the equation

Before you start

  • Calculating the equilibrium constant (U7.4)
  • Magnitude of the equilibrium constant (U7.5)

Using the ICE table the other way

U7.4 found K from equilibrium data; this lesson goes the other way, using K and the initial concentrations to find the equilibrium concentrations:

  1. Set up an ICE table, writing each change in terms of x (coefficient × x);
  2. Substitute the E row into the K expression to get an equation in x only;
  3. Solve for x, then calculate each equilibrium concentration;
  4. Substitute back and check that Q equals K.

Three ways to solve

Square root. When both sides are perfect squares (for example, Δn = 0 and the two reactants start at the same concentration), take the square root; no approximation is needed.

The “small x” approximation. When K is small, hardly any reactant is used up, so c₀ − x ≈ c₀ and the equation becomes much simpler. Once you have x, check it: the approximation is acceptable only if x is no more than 5% of c₀.

The quadratic formula. When the approximation fails, rearrange to ax² + bx + c = 0 and use the quadratic formula. Only one of the two roots makes sense: the one that keeps every equilibrium concentration from being negative (usually 0 < x < c₀).

Choose a reaction, drag the initial concentration, and compare the approximate and exact solutions:

Try it: find equilibrium concentrations with an ICE table

Example 1: Taking a square root

At 800 °C, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has K = 0.640. Initially [CO] = [H₂O] = 0.100 M, with no products. Find the equilibrium concentrations.

Step-by-step solution

  1. Decide the direction. There are no products at the start, so Q = 0 < K and the reaction runs forward: CO and H₂O decrease, CO₂ and H₂ increase.

  2. Set up an ICE table. All four coefficients are 1, so every change is x (concentrations in M):

    COH₂OCO₂H₂
    I0.1000.10000
    C−x−x+x+x
    E0.100 − x0.100 − xxx

  3. Substitute the equilibrium row into K.

    K=[COX2][HX2][CO][HX2O]=x2(0.100−x)2=0.640\begin{aligned} K &= \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} \\ &= \frac{x^2}{(0.100 - x)^2} = 0.640 \end{aligned}

  4. Take the square root of both sides. Both sides are perfect squares, so the square root turns this into a linear equation, and no quadratic is needed:

    x0.100−x=0.640=0.800\frac{x}{0.100 - x} = \sqrt{0.640} = 0.800

  5. Solve for x.

    x=0.800 (0.100−x)x=0.0800−0.800x1.800x=0.0800x=0.0444\begin{aligned} x &= 0.800\,(0.100 - x) \\ x &= 0.0800 - 0.800x \\ 1.800x &= 0.0800 \\ x &= 0.0444 \end{aligned}

  6. Find the equilibrium concentrations. [H₂] equals [CO₂], and [H₂O] equals [CO]:

    [COX2]=x=0.0444 M[CO]=0.100−x=0.0556 M\begin{aligned} [\ce{CO2}] &= x = \boxed{0.0444\ \text{M}} \\ [\ce{CO}] &= 0.100 - x \\ &= \boxed{0.0556\ \text{M}} \end{aligned}

  7. Check by substituting back. Put the equilibrium concentrations back into the K expression:

    (0.0444)2(0.0556)2=1.97×10−33.09×10−3=0.638≈0.640 ✓\begin{aligned} \frac{(0.0444)^2}{(0.0556)^2} &= \frac{1.97 \times 10^{-3}}{3.09 \times 10^{-3}} \\ &= 0.638 \approx 0.640 \ \checkmark \end{aligned}

    The small difference comes from rounding, so the answer is right.

Example 2: When the approximation fails

At 25 °C, N₂O₄(g) ⇌ 2NO₂(g) has K = 6.3 × 10⁻³. Initially [N₂O₄] = 0.0200 M. Find [NO₂] at equilibrium.

Step-by-step solution

  1. Set up an ICE table. Only N₂O₄ is present at the start, so the reaction runs forward. N₂O₄ decreases by x, and NO₂ increases by 2x, following the coefficients (concentrations in M):

    N₂O₄2NO₂
    I0.02000
    C−x+2x
    E0.0200 − x2x

  2. Substitute the equilibrium row into K. [NO₂] is 2x, so the 2 is squared as well:

    K=[NOX2]2[NX2OX4]=(2x)20.0200−x=6.3×10−3\begin{aligned} K &= \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} \\ &= \frac{(2x)^2}{0.0200 - x} = 6.3 \times 10^{-3} \end{aligned}

  3. Try the approximation first. Assume x ≪ 0.0200, so the denominator is 0.0200:

    4x20.0200=6.3×10−3x2=3.15×10−5x=5.6×10−3\begin{aligned} \frac{4x^2}{0.0200} &= 6.3 \times 10^{-3} \\ x^2 &= 3.15 \times 10^{-5} \\ x &= 5.6 \times 10^{-3} \end{aligned}

  4. Test the approximation. Express x as a percentage of the starting concentration:

    5.6×10−30.0200×100%=28%>5%\frac{5.6 \times 10^{-3}}{0.0200} \times 100\% = 28\% > 5\%

    That is far more than 5%, so the approximation fails and the full equation must be solved.

  5. Rearrange into a quadratic. Multiply both sides by (0.0200 − x), then move every term to one side (keep the letter K for now and substitute its value at the end):

    4x2=K (0.0200−x)4x2+Kx−0.0200K=0\begin{gathered} 4x^2 = K\,(0.0200 - x) \\ 4x^2 + Kx - 0.0200K = 0 \end{gathered}

  6. Use the quadratic formula. Matching ax² + bx + c = 0: a = 4, b = K = 6.3 × 10⁻³ and c = −0.0200K = −1.26 × 10⁻⁴.

    b2=3.97×10−5−4ac=2.016×10−3b2−4ac=2.056×10−3x=−b+b2−4ac2a=−6.3×10−3+0.045348=4.9×10−3\begin{aligned} b^2 &= 3.97 \times 10^{-5} \\ -4ac &= 2.016 \times 10^{-3} \\ b^2 - 4ac &= 2.056 \times 10^{-3} \\ x &= \frac{-b + \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-6.3 \times 10^{-3} + 0.04534}{8} \\ &= 4.9 \times 10^{-3} \end{aligned}

    The other root (with the “−” sign) is negative; it would make [NO₂] negative, so it is rejected.

  7. Find the equilibrium concentrations.

    [NOX2]=2x=9.8×10−3 M[NX2OX4]=0.0200−0.0049=0.0151 M\begin{aligned} [\ce{NO2}] &= 2x \\ &= \boxed{9.8 \times 10^{-3}\ \text{M}} \\ [\ce{N2O4}] &= 0.0200 - 0.0049 \\ &= 0.0151\ \text{M} \end{aligned}

  8. Check by substituting back. Use the unrounded x = 4.88 × 10⁻³:

    (9.76×10−3)20.01512=6.30×10−3 ✓\frac{(9.76 \times 10^{-3})^2}{0.01512} = 6.30 \times 10^{-3} \ \checkmark
Practice: solving for x with the approximation c₀ − x ≈ c₀, does the result pass the 5% check?
  1. HA ⇌ H⁺ + A⁻, Ka = 1.8 × 10⁻⁵, c₀ = 0.10 M
  2. HA ⇌ H⁺ + A⁻, Ka = 6.4 × 10⁻⁴, c₀ = 0.010 M
  3. HA ⇌ H⁺ + A⁻, Ka = 6.4 × 10⁻⁴, c₀ = 1.0 M
  4. HA ⇌ H⁺ + A⁻, Ka = 4.9 × 10⁻¹⁰, c₀ = 0.0010 M
  5. N₂ + O₂ ⇌ 2NO, K = 4.1 × 10⁻⁴, with N₂ and O₂ both starting at 0.50 M
  6. N₂O₄ ⇌ 2NO₂, K = 6.3 × 10⁻³, c₀ = 0.010 M
  7. N₂O₄ ⇌ 2NO₂, K = 6.3 × 10⁻³, c₀ = 1.0 M
  8. HA ⇌ H⁺ + A⁻, Ka = 1.8 × 10⁻⁴, c₀ = 0.0050 M

Common mistakes

Mistake 1: “Whenever K is less than 1, the approximation works.” Whether the approximation works depends on the ratio of x to c₀, which you have to calculate and check. When K isn’t very small and c₀ is small, the approximation can be far off.

Mistake 2: “(2x)² is the same as 2x².” (2x)² = 4x². A substance with coefficient 2 has concentration 2x, and the whole 2x is squared.

Mistake 3: “Either root of the quadratic will do.” Only the root that keeps every concentration from being negative makes sense. Reject a negative root or one larger than c₀.

Quick check

Solving x² ÷ (0.10 − x) = 0.50 gives two roots: x = 0.085 and x = −0.59. What is x at equilibrium?

Starting from pure HI, 2HI(g) ⇌ H₂(g) + I₂(g), with [HI]₀ = c₀ and H₂ increasing by x. Which is the equation for K?

A(g) + B(g) ⇌ C(g) + D(g) has K = 9.0. Initially [A] = [B] = 1.0 M, with no C or D. What is [C] at equilibrium?

X(g) ⇌ 2Y(g) has K = 1.0 × 10⁻⁶, and initially [X] = 0.25 M. [Y] at equilibrium is about:

The approximation gives x = 0.012 M, and the initial concentration is c₀ = 0.10 M. What should you do?

Which situation can be solved by taking a square root, with no approximation?

AP-style free response

AP-style free response

At 2000 °C, N₂(g) + O₂(g) ⇌ 2NO(g) has K = 4.1 × 10⁻⁴ (OpenStax). A reactor starts with [N₂] = 0.040 M and [O₂] = 0.010 M (close to the ratio in air) and no NO.

(a) Set up an ICE table.

(b) Write the equation for x.

(c) Use an approximation to find [NO] at equilibrium.

(d) Check whether the approximation in (c) is acceptable.

(e) This reaction has ΔH = +180.6 kJ/mol (from NBS enthalpies of formation). If the reaction were run at a lower temperature, would [NO] at equilibrium be larger or smaller? Justify your answer.

Model answer and scoring

(a) I: 0.040, 0.010, 0; C: −x, −x, +2x; E: 0.040 − x, 0.010 − x, 2x.

(b) (2x)² ÷ [(0.040 − x)(0.010 − x)] = 4.1 × 10⁻⁴.

(c) Approximating 0.040 − x ≈ 0.040 and 0.010 − x ≈ 0.010: 4x² ÷ (4.0 × 10⁻⁴) = 4.1 × 10⁻⁴, so x² = 4.1 × 10⁻⁸, x = 2.0 × 10⁻⁴, and [NO] = 2x = 4.0 × 10⁻⁴ M.

(d) x ÷ 0.010 = 2%, which is under 5%, so the approximation is acceptable (the exact solution is also 4.0 × 10⁻⁴ M).

(e) Smaller. The forward reaction is endothermic, so lowering the temperature decreases K, and less NO forms at equilibrium.

Scoring: (a) ICE table; (b) the equation; (c) solving with the approximation; (d) the 5% check against the smaller initial concentration; (e) K decreases on cooling for an endothermic reaction.

Summary

  • Given K and initial concentrations: ICE table → equation in x → solve for x → equilibrium concentrations → check by substituting back.
  • Take a square root for a perfect square; use c₀ − x ≈ c₀ when K is small, checking that x is no more than 5% of c₀; otherwise solve the quadratic.
  • Keep only the root that makes every concentration non-negative.

AP tip: Whenever you use an approximation, show the check (“x/c₀ = …% < 5%”). The AP exam rarely requires the quadratic formula; most questions can be solved with a square root or an approximation that works. When the approximation fails, say so.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.7 · Calculating Equilibrium Concentrations (ICE Tables) · 5 questions

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