Question at position 15 Which of the following is/are true for a reaction whose equilibrium constant is relatively small (K

Question

Question at position 15 Which of the following is/are true for a reaction whose equilibrium constant is relatively small (K < 1)? 1. Equilibrium is reached quickly2. The reaction contains mostly reactants at equilibrium3. The reaction contains mostly products at equilibrium4. It will take a long time to reach equilibrium only statement 3 is correct only statement 2 is correctonly statement 4 is correct only statement 1 is correct more than one of these statements is correct单项选择题

A

only statement 3 is correct

B

only statement 2 is correct

C

only statement 4 is correct

D

only statement 1 is correct

E

more than one of these statements is correct

BlackTom AI · Accepted Answer

A small equilibrium constant (K < 1) means the equilibrium position lies toward reactants; product concentrations at equilibrium are lower than reactant concentrations.

1. Relevant knowledge points and concepts
- Equilibrium constant K for a simple reaction A ⇌ B (or general aA + bB ⇌ cC + dD) is a ratio of product activities to reactant activities at equilibrium. For a simple A ⇌ B, K = [B]eq/[A]eq.
- Thermodynamics (equilibrium position) and kinetics (rate to reach equilibrium) are distinct: K is determined by thermodynamic free energies or the ratio of rate constants (K = kf/kr), while the absolute speed of reaching equilibrium depends on the magnitudes of kf and kr (rate constants), activation energies, and other kinetic factors (catalysts, temperature).
- If K < 1, products are less favored at equilibrium; reactants predominate.

2. Step-by-step reasoning
- Statement 2: If K < 1, then [Products]eq < [Reactants]eq; thus the reaction contains mostly reactants at equilibrium → true.
- Statement 3: Claims mostly products at equilibrium → contradicts K < 1 → false.
- Statements 1 and 4: Concern time to reach equilibrium. K being small only gives the ratio kf/kr, not the absolute values. The time depends on kf and kr individually, so K < 1 does not imply fast or slow approach universally → both statements false.

3. Calculation example (intermediate steps)
Take A ⇌ B with K = 0.01 and start with [A]0 = 1.0 M, [B]0 = 0.
Let x = [B] formed at equilibrium, so [B]eq = x and [A]eq = 1.0 − x.
K = x/(1.0 − x) = 0.01
Solve: x = 0.01(1.0 − x) = 0.01 − 0.01x
x + 0.01x = 0.01 → 1.01x = 0.01 → x ≈ 0.00990099 M
Thus [B]eq ≈ 0.0099 M and [A]eq ≈ 0.9901 M, confirming mostly reactants.

4. Why the correct answer is correct
Statement 2 directly follows from the definition of K: K < 1 means product concentrations are lower relative to reactants at equilibrium. The numeric example illustrates this clearly.

5. Why the other options are incorrect
- Statement 3 is the opposite of the truth for K < 1.
- Statement 1 (equilibrium is reached quickly) is incorrect because K does not determine kinetic rates; a small K can occur with large rate constants (fast equilibration) or with very small rate constants (slow equilibration).
- Statement 4 (it will take a long time to reach equilibrium) is likewise incorrect for the same reason: K alone cannot predict time to equilibrium.
- Therefore only statement 2 is correct.

6. Conclusion
Only statement 2 is correct, which matches the provided answer: only statement 2 is correct.

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