U7.4 · Calculating the Equilibrium ConstantPractice · 计算平衡常数
Lessons covered:U7.4 Calculating the Equilibrium Constant
- 1Numeric
At some temperature, CO(g) + Cl₂(g) ⇌ COCl₂(g) is at equilibrium with [CO] = 0.0120 M, [Cl₂] = 0.0250 M, and [COCl₂] = 0.180 M. Calculate K.
Explanation= 0.180 ÷ (0.0120 × 0.0250) = 0.180 ÷ 3.0 × 10−4 = 600. K is large, so products are favored at equilibrium.
- 2Numeric
At some temperature, 0.400 mol of SO₃ is placed in a sealed 2.00 L container, and 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) takes place. At equilibrium [O₂] = 0.0300 M. Calculate K.
× 10ExplanationInitial [SO₃] = 0.400 ÷ 2.00 = 0.200 M. O₂ rises by 0.0300, so by the coefficients SO₂ rises by 2 × 0.0300 = 0.0600 and SO₃ falls by 0.0600, leaving [SO₃] = 0.1400 at equilibrium. K = 0.0600² × 0.0300 ÷ 0.1400² = 5.5 × 10−3.
- 3Multiple choice
For a given balanced equation, which change alters the value of its equilibrium constant K?
ExplanationK depends only on temperature. Changing starting concentrations or the volume changes the equilibrium concentrations, but the ratio K they form stays the same, and a catalyst speeds both directions equally, leaving K unchanged too.
- 4Numeric
At another temperature, PCl₅(g) is placed in a rigid container at an initial partial pressure of 1.00 atm, and PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) takes place. At equilibrium the total pressure is 1.40 atm. Calculate .
ExplanationLet PCl₅ fall by x, so PCl₃ and Cl₂ are each x. Total pressure = (1.00 − x) + x + x = 1.00 + x = 1.40, so x = 0.40 atm. Equilibrium pressures: PCl₅ 0.60, PCl₃ and Cl₂ 0.40 atm each. Kp = 0.40 × 0.40 ÷ 0.60 = 0.267.
- 5Multiple choice
A student calculating K for 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) in a 2.00 L container substitutes the equilibrium amounts in moles directly into the expression instead of converting them to concentrations. How does the result compare with the correct K?
Explanationc = n ÷ V. K = (n²SO₂/V²)(nO₂/V) ÷ (n²SO₃/V²) = (value from moles) ÷ V. There are three concentration factors on top and two on the bottom, so one 1/V is left over. The value from moles = V × K = 2 × K. The volume cancels only when both sides have the same total gas coefficients.
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