U7.7 · Calculating Equilibrium Concentrations (ICE Tables)Practice · 计算平衡浓度（ICE 表）
Lessons covered:U7.7 Calculating Equilibrium Concentrations (ICE Tables)
- 1Numeric
At some temperature, K = 36 for A(g) + B(g) ⇌ 2C(g). Initially [A] = [B] = 0.150 M and there is no C. What is [C] at equilibrium?
MExplanationICE: A and B are each 0.150 − x, and C is 2x. (2x)² ÷ (0.150 − x)² = 36; take the square root of both sides: 2x ÷ (0.150 − x) = 6, so 2x = 0.90 − 6x and x = 0.1125. [C] = 2x = 0.225 M, with 0.0375 M each of A and B left.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)
- 2Numeric
At a high temperature, K = 1.0 × 10−6 for 2H₂S(g) ⇌ 2H₂(g) + S₂(g). Initially there is only H₂S, at 0.100 M. What is [S₂] at equilibrium?
× 10MExplanationICE: H₂S 0.100 − 2x, H₂ 2x, S₂ x. K = (2x)²·x ÷ (0.100 − 2x)². K is small, so take 0.100 − 2x ≈ 0.100: 4x³ = 1.0 × 10−6 × 0.100² = 1.0 × 10−8, x³ = 2.5 × 10−9, x ≈ 1.36 × 10−3. Check: 2x is only 2.7% of 0.100, so the shortcut holds; the exact solution is x = 1.33 × 10−3 M.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)
- 3Multiple choice
At some temperature, K = 4 for A(g) + B(g) ⇌ C(g) + D(g), starting from [A] = [B] = 0.30 M. A student writes x² ÷ (0.30 − x)² = 4, expands it into a quadratic, and gets x = 0.20 and x = 0.60. Which root should be used?
Explanationx can’t exceed the starting concentration: x = 0.60 would give [A] = 0.30 − 0.60 = −0.30 M, which is meaningless. Use x = 0.20: C and D are 0.20 each, A and B 0.10 each, and 0.20² ÷ 0.10² = 4. Taking the square root of both sides, x ÷ (0.30 − x) = 2, is quicker here.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)
- 4Numeric
At some temperature, K = 36 for A(g) + B(g) ⇌ 2C(g). This time the container starts with only C, at 0.300 M. What is [A] at equilibrium?
MExplanationWith no reactants present, the reaction runs in reverse. Let A and B each form x while C falls by 2x: (0.300 − 2x)² ÷ x² = 36. Taking the square root, (0.300 − 2x) ÷ x = 6, so 0.300 = 8x and x = 0.0375 M. [A] = 0.0375 M and [C] = 0.225 M.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)
- 5Numeric
At 400 °C, K = 50 for H₂(g) + I₂(g) ⇌ 2HI(g). Initially [H₂] = 0.200 M and [I₂] = 0.100 M, with no HI. What is [HI] at equilibrium?
MExplanationICE: H₂ 0.200 − x, I₂ 0.100 − x, HI 2x. (2x)² ÷ [(0.200 − x)(0.100 − x)] = 50. The starting concentrations differ, so no square root, and K isn’t small, so no shortcut. Expanding: 4x² = 50(0.02 − 0.3x + x²), i.e. 46x² − 15x + 1 = 0, so x = 0.0934 or 0.2326; the second exceeds the starting [I₂] and is rejected. [HI] = 2x = 0.187 M.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)
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