AP Chemistry · Topic 7.3
Reaction Quotient and Equilibrium Constant Practice
Part of Equilibrium.(TRA-7.A)
Practice questions
6
Sample questions
5 of 6 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 2/5
For aA + bB ⇌ cC + dD, K =
- A
[A]^a[B]^b / [C]^c[D]^d
- B
abK − cd
- Ccheck_circle
[C]^c[D]^d / [A]^a[B]^b
- D
([A]+[B]) / ([C]+[D])
Why
Products over reactants, each raised to its stoichiometric coefficient.
- A
Sample 2difficulty 2/5
Q is computed
- A
Only at equilibrium
- B
Only at standard state
- C
Only with K
- Dcheck_circle
At any moment using current concentrations
Why
Q is the same expression as K, but with current (not necessarily equilibrium) values.
- A
Sample 3difficulty 2/5
Pure solids and pure liquids are excluded from K expressions.
What is the correct equilibrium expression?
- Acheck_circle
K = [CO2]
- B
K = [CaO][CO2]/[CaCO3]
- C
K = [CO2]/[CaCO3]
- D
K = [CaO][CO2]
Why
Pure solids do not appear in K. Only the gaseous CO2 contributes: K = [CO2] (or Kp = P_CO2).
- A
Sample 4difficulty 3/5
For N₂ + 3H₂ ⇌ 2NH₃ (Kc = 0.50 at T), a vessel contains [N₂] = 1.0 M, [H₂] = 1.0 M, [NH₃] = 1.0 M.
What is Q, and which way does the reaction shift?
- A
Q = 0.50; system at equilibrium.
- B
Q = 2.0; shifts left.
- C
Q = 1.0; shifts right.
- Dcheck_circle
Q = 1.0; shifts left (reverse).
Why
Q = [NH₃]²/([N₂][H₂]³) = 1²/(1·1³) = 1.0. Q > Kc, so reverse.
- A
Sample 5difficulty 3/5
For A + B ⇌ C, [A]=0.10, [B]=0.10, [C]=0.05. K = 100. The reaction
- A
Is at equilibrium
- B
Shifts left
- C
Cannot tell
- Dcheck_circle
Shifts right
Why
Q = 0.05/(0.10·0.10) = 5; 5 < 100 → forward shift.
- A