MDCAT Chemistry chapter 6: Chemistry · Chemical equilibrium MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
49 exam-style questions on Chemistry · Chemical equilibrium, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Equilibrium and Le Chatelier's principle29 questions
Ionic equilibria20 questions
Ten Chemistry · Chemical equilibrium questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. The reaction A ⇌ B is started with pure A. By time t, [A] has fallen from 1.0 to 0.30 mol dm−3 and [B] has risen to 0.70 mol dm−3; after t, both stay constant. Which statement describes the system after time t?
A
no more A turns into B, as the reaction has stopped
B
the forward rate exceeds the reverse rate, since [B] > [A]
C
the system is not at equilibrium, since [A] ≠ [B]
D
A turns into B and B turns into A at equal rates
Show answer
Answer: D. Constant concentrations with both A and B present mean dynamic equilibrium: the forward and reverse reactions continue at equal rates. Equilibrium does not need equal concentrations; here Kc = 0.70 ÷ 0.30 ≈ 2.3.
2. For N2(g) + 3H2(g) ⇌ 2NH3(g) at 500 K, how does the numerical value of Kp (pressures in atm) compare with that of Kc (concentrations in mol dm−3)? (R = 0.0821 dm3 atm K−1 mol−1)
A
Kp is larger, because Δn = +2
B
they are equal, because all the species are gases
C
Kp is smaller, because Δn = −2
D
Kp is larger, because the reaction is exothermic
Show answer
Answer: C. Δn = 2 − 4 = −2, so Kp = Kc(RT)−2 = Kc ÷ (RT)2; with RT = 0.0821 × 500 ≈ 41, Kp is much smaller than Kc. Taking Δn as reactant moles minus product moles gives the wrong sign.
3. 2.0 mol of HI is placed in a 1.0 dm3 flask and heated: 2HI(g) ⇌ H2(g) + I2(g). At equilibrium 0.40 mol of H2 is present. What is Kc?
A
0.0625
B
9.0
C
0.11
D
0.13
Show answer
Answer: C. Forming 0.40 mol H2 also forms 0.40 mol I2 and uses 0.80 mol HI, leaving 1.2 mol. Kc = (0.40 × 0.40) ÷ 1.22 = 0.16 ÷ 1.44 = 0.11. Using 1.6 mol HI (a 1 : 1 ratio) gives 0.0625, and not squaring [HI] gives 0.13.
4. A sparingly soluble salt MX has Ksp = 1.6 × 10−9 mol2 dm−6. What is its solubility in pure water?
A
8.0 × 10−10 mol dm−3
B
4.0 × 10−5 mol dm−3
C
1.6 × 10−9 mol dm−3
D
2.6 × 10−18 mol dm−3
Show answer
Answer: B. For MX, Ksp = s2, so s = √(16 × 10−10) = 4.0 × 10−5 mol dm−3. Halving Ksp gives 8.0 × 10−10, and squaring it gives 2.6 × 10−18.
5. Two weak acids have pKa(HX) = 4.0 and pKa(HY) = 9.0. Which statement is correct?
A
HX is the stronger acid and Y− the stronger base
B
HY is the stronger acid and X− the stronger base
C
HX is the stronger acid and X− the stronger base
D
HY is the stronger acid and Y− the stronger base
Show answer
Answer: A. pKa = −log Ka, so the smaller pKa goes with the larger Ka: HX (Ka = 10−4) is stronger than HY (10−9). The weaker the acid, the stronger its conjugate base, so Y− is the stronger base.
6. A buffer of pH 5.74 is to be made from ethanoic acid (pKa = 4.74) and sodium ethanoate. What concentration ratio [CH3COO−] : [CH3COOH] is needed?
A
1 : 10
B
1 : 1
C
10 : 1
D
100 : 1
Show answer
Answer: C. pH = pKa + log([salt]/[acid]), so 5.74 = 4.74 + log(ratio), log(ratio) = 1 and the ratio is 10 : 1. A 1 : 1 mixture would give pH = pKa = 4.74, and 1 : 10 would give 3.74.
7. In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1, which single change would increase both the rate of reaction and the equilibrium yield of ammonia?
A
increasing the temperature
B
using a more active catalyst
C
increasing the pressure
D
lowering the temperature
Show answer
Answer: C. Higher pressure crowds the molecules together, so they collide more often, and it also shifts the equilibrium to the side with fewer gas moles (4 → 2). A higher temperature speeds the reaction but lowers the yield of this exothermic reaction; a lower temperature does the opposite, and a catalyst raises only the rate.
8. The solubility of magnesium hydroxide, Mg(OH)2, in water is s mol dm−3. In terms of s, its Ksp is
A
s2
B
2s3
C
4s2
D
4s3
Show answer
Answer: D. Mg(OH)2 ⇌ Mg2+ + 2OH−, so [Mg2+] = s and [OH−] = 2s. Ksp = s × (2s)2 = 4s3. 2s3 forgets to square the 2 along with s.
9. What is the solubility product expression for silver chromate, Ag2CrO4?
A
[Ag+]2[CrO42−]
B
[Ag+][CrO42−]
C
2[Ag+][CrO42−]
D
[Ag+]2[CrO42−] / [Ag2CrO4]
Show answer
Answer: A. Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42−(aq), so Ksp = [Ag+]2[CrO42−]. The coefficient 2 becomes a power, and the solid does not appear in the expression.
10. For PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium concentrations at a certain temperature are [PCl5] = 0.20, [PCl3] = 0.10 and [Cl2] = 0.10 mol dm−3. What is the value of Kc in mol dm−3?
A
0.050
B
0.010
C
0.50
D
20
Show answer
Answer: A. Kc = [PCl3][Cl2]/[PCl5] = (0.10 × 0.10) ÷ 0.20 = 0.050 mol dm−3. 0.010 forgets to divide by [PCl5], and 20 is the expression upside down.
Chemistry · Chemical equilibrium: questions students ask
How many Chemistry · Chemical equilibrium MCQs are there?
49, each with the answer and a short explanation, across 2 topics: Equilibrium and Le Chatelier's principle, Ionic equilibria. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Chemistry chapter 6 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
