AP Chemistry · Topic 7.7
Calculating Equilibrium Concentrations Practice
Part of Equilibrium.(TRA-7.E)
Practice questions
14
Sample questions
5 of 14 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 3/5
For A(g) <-> B(g) + C(g), Kc = 0.040 with the ICE table shown.
If x is small relative to 1.0, what is x approximately?
- A
0.040
- B
0.50
- Ccheck_circle
0.20
- D
0.0040
Why
Kc ~ x*x/1.0 = x^2 = 0.040, so x = sqrt(0.040) = 0.20 (small-x approximation).
- A
Sample 2difficulty 3/5
For N2O4(g) <-> 2 NO2(g), at equilibrium [NO2] = 0.20 M.
What is [N2O4] at equilibrium?
- A
0.10 M
- B
0.50 M
- C
0.20 M
- Dcheck_circle
0.40 M
Why
2x = 0.20, so x = 0.10. [N2O4] = 0.50 - 0.10 = 0.40 M.
- A
Sample 3difficulty 3/5
For A ⇌ 2B with [A]₀ = 1.0 M and [B]₀ = 0, the equilibrium expressions are
- Acheck_circle
[A] = 1.0 − x; [B] = 2x
- B
[A] = 1.0; [B] = x
- C
[A] = 1.0 − 2x; [B] = x
- D
[A] = 1.0 + x; [B] = 2x
Why
Stoichiometry: A loses x, B gains 2x.
- A
Sample 4difficulty 3/5
Consider N2(g) + 3 H2(g) <-> 2 NH3(g) starting from the initial concentrations shown.
What value belongs in the H2 "Change" cell of the ICE table?
- A
+3x
- B
-x
- Ccheck_circle
-3x
- D
+x
Why
Stoichiometry: for every 1 mol N2 consumed, 3 mol H2 are consumed. If N2 changes by -x, then H2 changes by -3x.
- A
Sample 5difficulty 4/5
A ⇌ B with [A]₀ = 0.50 M and K = 1.0. At equilibrium [B] is
- A
0.40 M
- Bcheck_circle
0.25 M
- C
0.50 M
- D
0.10 M
Why
K = [B]/[A] = x/(0.50 − x) = 1 → x = 0.25 M.
- A