U7.6 · Properties of the Equilibrium ConstantPractice · K 的性质（反转、乘系数、相加）
Lessons covered:U7.6 Properties of the Equilibrium Constant
- 1Numeric
At 25 °C, K = 160 for 2NO₂(g) ⇌ N₂O₄(g). Find K at the same temperature for NO₂(g) ⇌ ½N₂O₄(g).
ExplanationThe new coefficients are ½ of the original, so K is raised to the ½ power: K = 160½ = √160 = 12.6. Multiplying the coefficients by n raises K to the nth power; it doesn’t multiply K by ½ (that would give 80).
- 2Multiple choice
If every coefficient in an equilibrium equation is multiplied by 3, what is the equilibrium constant for the new equation?
ExplanationEvery exponent in the expression is multiplied by 3, so the whole expression is cubed: Knew = K³.
- 3Numeric
At some temperature: ① N₂(g) + O₂(g) ⇌ 2NO(g), K₁ = 2.5 × 10−3; ② 2NO(g) + O₂(g) ⇌ 2NO₂(g), K₂ = 1.6 × 102. Find K at the same temperature for 2NO₂(g) ⇌ N₂(g) + 2O₂(g).
Explanation① + ② gives N₂(g) + 2O₂(g) ⇌ 2NO₂(g) (2NO cancels), with K = K₁ × K₂ = 2.5 × 10−3 × 1.6 × 102 = 0.40. The target is the reverse, so take the reciprocal: K = 1 ÷ 0.40 = 2.5.
- 4Multiple choice
At 25 °C, formic acid HCOOH has Ka = 1.8 × 10−4 and Kw = 1.0 × 10−14. What is Kb for the formate ion (HCOO⁻ + H₂O ⇌ HCOOH + OH⁻)?
ExplanationAdding HCOOH ⇌ H⁺ + HCOO⁻ (Ka) and HCOO⁻ + H₂O ⇌ HCOOH + OH⁻ (Kb) gives H₂O ⇌ H⁺ + OH⁻ (Kw), so Ka × Kb = Kw and Kb = 1.0 × 10−14 ÷ 1.8 × 10−4 = 5.6 × 10−11.
- 5Multiple choice
Two equations are added to give a third. How are ΔH and K for the third equation obtained from the first two?
ExplanationWhen equations are added, ΔH values add (Hess’s law), while the equilibrium expressions multiply, so K values multiply. Likewise, reversing an equation flips the sign of ΔH and inverts K, and multiplying by n multiplies ΔH by n and raises K to the nth power.
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