AP Chemistry Equilibrium — Worked Answer Explanations

Unit 7 · 12 questions explained

Below is a complete answer key for our AP Chemistry Equilibrium practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Equilibrium practice test and come back here to review, or head back to the Equilibrium unit overview.

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Question 1 · Easy
For the equilibrium $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , which expression correctly represents $K_{c}$ ?
- A
  $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ Correct
- B
  $K_{c} = \frac{[ N _{2} ] [ H _{2} ] ^{3}}{[ N H _{3} ] ^{2}}$
  Why not B: Products over reactants, not reactants over products.
- C
  $K_{c} = \frac{[ N H _{3} ]}{[ N _{2} ] [ H _{2} ]}$
  Why not C: Coefficients (2 and 3) must be used as exponents.
- D
  $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] ^{2} [ H _{2} ] ^{3}}$
  Why not D: N $_{2}$ has a coefficient of 1, so its exponent is 1, not 2.
Explanation
The equilibrium constant expression $K_{c}$ has products in the numerator and reactants in the denominator, each raised to the power of their stoichiometric coefficients: $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ .

Key takeaway
$K_c$ = [products]$^{\text{coeff}}$ / [reactants]$^{\text{coeff}}$; coefficients become exponents.
Question 2 · Easy
According to Le Châtelier's principle, what happens to the equilibrium position of $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ ( $Δ H = - 92 kJ$ ) when the temperature is increased?
- A
  Shifts forward; more NH $_{3}$ is produced.
  Why not A: Increasing temperature favors the endothermic direction, which is the reverse.
- B
  Shifts reverse; $K_{c}$ decreases.Correct
- C
  No effect; temperature does not affect $K_{c}$ .
  Why not C: $K_{c}$ is temperature-dependent; it changes when temperature changes.
- D
  Shifts forward; $K_{c}$ increases.
  Why not D: For an exothermic reaction, raising temperature decreases $K_{c}$ .
Explanation
The reaction is exothermic ( $Δ H < 0$ ), so heat can be viewed as a product. Increasing temperature is equivalent to adding a product, shifting the equilibrium toward reactants (reverse). Mathematically, $K_{c}$ decreases with increasing temperature for exothermic reactions.

Key takeaway
For exothermic reactions: increase $T$ → $K_c$ decreases, shifts reverse. For endothermic: increase $T$ → $K_c$ increases, shifts forward.
Question 3 · Easy
For a reaction with $K_{c} = 5.0 \times 1 0^{- 8}$ , which statement best characterizes the equilibrium?
- A
  At equilibrium, the mixture contains mostly products.
  Why not A: A very small $K_{c}$ means reactants are favored at equilibrium.
- B
  At equilibrium, reactants and products are present in roughly equal amounts.
  Why not B: Equal amounts occur when $K_{c} \approx 1$ .
- C
  At equilibrium, the mixture is predominantly reactants.Correct
- D
  The reaction does not reach equilibrium because $K_{c}$ is too small.
  Why not D: All reactions reach equilibrium; a small $K_{c}$ just means the equilibrium strongly favors reactants.
Explanation
A small $K_{c}$ ( $≪ 1$ ) means the numerator (products) is much smaller than the denominator (reactants) at equilibrium. The reaction goes only slightly in the forward direction. A large $K_{c}$ ( $≫ 1$ ) favors products; $K_{c} \approx 1$ gives a roughly equal mixture.

Key takeaway
$K_c \gg 1$: products favored. $K_c \approx 1$: comparable amounts. $K_c \ll 1$: reactants favored.
Question 4 · Easy
At equilibrium, increasing the pressure on a gas-phase reaction by reducing volume will shift the equilibrium in the direction that:
- A
  Decreases the total number of moles of gas.Correct
- B
  Increases the total number of moles of gas.
  Why not B: Increasing the number of moles of gas would increase pressure further, not relieve the stress.
- C
  Has no effect, since gases are compressible.
  Why not C: Compressibility describes volume change, not equilibrium position.
- D
  Increases temperature.
  Why not D: Compression can warm a gas slightly, but that is a physical effect; Le Châtelier's principle concerns equilibrium position.
Explanation
Le Châtelier's principle: an increase in pressure (by compression) stresses the equilibrium. To reduce the stress, the system shifts toward the side with fewer moles of gas, thereby reducing pressure. If both sides have equal moles of gas, there is no shift.

Key takeaway
Increased pressure shifts equilibrium toward the side with fewer moles of gas (Le Châtelier).
Question 5 · Medium
For the equilibrium $A (g) ⇌ 2 B (g)$ , an ICE table gives the following initial and change rows:

A B
Initial $0.600 M$ $0$
Change $- x$ $+ 2 x$
Equilibrium $0.600 - x$ $2 x$

If $K_{c} = 0.040$ , what is the equilibrium concentration of A?
- A
  $0.550 M$ Correct
- B
  $0.600 M$
  Why not B: This is the initial concentration; equilibrium [A] is less because reaction proceeds forward.
- C
  $0.200 M$
  Why not C: Arithmetic error solving the quadratic or wrong $K_{c}$ expression.
- D
  $0.0500 M$
  Why not D: Solved for $x$ rather than $[A] = 0.600 - x$ .
Explanation
$K_{c} = \frac{[ B ] ^{2}}{[ A ]} = \frac{( 2 x ) ^{2}}{0.600 - x} = \frac{4 x ^{2}}{0.600 - x} = 0.040$

$4 x^{2} = 0.040 (0.600 - x) = 0.024 - 0.040 x$

$4 x^{2} + 0.040 x - 0.024 = 0$

Using the quadratic formula: $x \approx 0.0500 M$

$[A] = 0.600 - 0.050 = 0.550 M$

Key takeaway
ICE tables + $K_c$ expression yield an algebraic equation; solve for $x$, then find equilibrium concentrations.
Question 6 · Medium
For the reaction $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$ , $K_{c} = 1.8 \times 1 0^{- 2}$ at a certain temperature. If a reaction mixture has $[PC l_{5}] = 0.50 M$ , $[PC l_{3}] = 0.15 M$ , and $[C l_{2}] = 0.15 M$ , in which direction will the reaction proceed?
- A
  Forward, toward products, because $Q < K_{c}$ .
  Why not A: $Q = (0.15) (0.15) / (0.50) = 0.045 > K_{c} = 0.018$ , so reverse is favored.
- B
  Reverse, toward reactants, because $Q > K_{c}$ .Correct
- C
  Neither; the system is at equilibrium because all concentrations are nonzero.
  Why not C: Having nonzero concentrations doesn't mean equilibrium; compare $Q$ to $K_{c}$ .
- D
  Forward, because the products have lower concentrations than the reactant.
  Why not D: Direction depends on $Q$ vs. $K_{c}$ , not simply which side has lower concentration.
Explanation
$Q = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]} = \frac{( 0.15 ) ( 0.15 )}{0.50} = \frac{0.0225}{0.50} = 0.045$

$Q = 0.045 > K_{c} = 0.018$ . Product concentrations are too high relative to equilibrium. The reaction shifts in reverse to consume products and form PCl $_{5}$ .

Key takeaway
If $Q > K_c$: shift reverse. If $Q < K_c$: shift forward. If $Q = K_c$: at equilibrium.
Question 7 · Medium
The solubility product of silver chloride is $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ at $25° C$ . What is the molar solubility (in mol/L) of AgCl in pure water?
- A
  $1.8 \times 1 0^{- 10} M$
  Why not A: This is $K_{s p}$ , not solubility; for a 1:1 salt, $s = K_{s p}$ .
- B
  $1.3 \times 1 0^{- 5} M$ Correct
- C
  $9.0 \times 1 0^{- 21} M$
  Why not C: Squared $K_{s p}$ rather than taking its square root.
- D
  $3.6 \times 1 0^{- 10} M$
  Why not D: Divided $K_{s p}$ by 2 rather than taking the square root.
Explanation
$AgCl (s) ⇌ A g^{+} (aq) + C l^{-} (aq)$

Let solubility = $s$ : $[A g^{+}] = [C l^{-}] = s$

$K_{s p} = s^{2} = 1.8 \times 1 0^{- 10}$

$s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M$

Key takeaway
For a 1:1 salt, molar solubility $s = \sqrt{K_{sp}}$.
Question 8 · Medium
Adding an inert gas to a gaseous equilibrium system at constant volume has what effect on the equilibrium position?
- A
  Shifts toward the side with more moles of gas.
  Why not A: This response is for a pressure change, not an inert gas addition at constant volume.
- B
  Shifts toward the side with fewer moles of gas.
  Why not B: Inert gas at constant volume does not change partial pressures of reacting gases.
- C
  No shift; the equilibrium position is unchanged.Correct
- D
  Shifts forward to consume the extra pressure.
  Why not D: Adding an inert gas increases total pressure but does not change the partial pressures or concentrations of the reacting species at constant volume.
Explanation
At constant volume, adding an inert gas increases total pressure but does not change the partial pressures (or molar concentrations) of any of the reacting gases. Since $K_{c}$ and $K_{p}$ depend only on reacting species, the equilibrium position remains unchanged.

Key takeaway
Inert gas at constant volume: no effect on equilibrium. At constant pressure (expanded container): shifts toward more moles of gas.
Question 9 · Medium
The relationship between $K_{c}$ and $K_{p}$ for a gas-phase equilibrium is $K_{p} = K_{c} (R T)^{Δ n}$ , where $Δ n$ is the change in moles of gas. For the reaction $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ , $Δ n =$ ?
- A
  $- 1$
  Why not A: Subtracted products from reactants instead of the reverse.
- B
  $+ 1$ Correct
- C
  $+ 2$
  Why not C: Counted only the SO $_{2}$ product, not the total change.
- D
  $0$
  Why not D: Reactants (2 mol gas) and products (3 mol gas) are not equal.
Explanation
Moles of gaseous products: $2 + 1 = 3$ . Moles of gaseous reactants: $2$ . $Δ n = 3 - 2 = + 1$ . Therefore $K_{p} = K_{c} (R T)^{1} = K_{c} R T$ , so $K_{p} > K_{c}$ at temperatures above $1/ R \approx 0.12 K$ .

Key takeaway
$\Delta n = $ (moles gas products) $-$ (moles gas reactants); used in $K_p = K_c(RT)^{\Delta n}$.
Question 10 · Hard
The common-ion effect predicts that the molar solubility of AgCl ( $K_{s p} = 1.8 \times 1 0^{- 10}$ ) in a $0.10 M$ NaCl solution is:
- A
  Greater than $1.3 \times 1 0^{- 5} M$
  Why not A: Adding a common ion (Cl $^{-}$ ) decreases solubility, not increases it.
- B
  $1.3 \times 1 0^{- 5} M$ (unchanged)
  Why not B: The presence of 0.10 M Cl $^{-}$ suppresses AgCl dissolution.
- C
  $1.8 \times 1 0^{- 9} M$ Correct
- D
  $1.8 \times 1 0^{- 10} M$
  Why not D: This is $K_{s p}$ itself, not the solubility in NaCl.
Explanation
In 0.10 M NaCl, $[C l^{-}] \approx 0.10$ M (common ion). Let $s$ = solubility of AgCl:

$K_{s p} = s (0.10 + s) \approx s \times 0.10 = 1.8 \times 1 0^{- 10}$

$s = 1.8 \times 1 0^{- 9} M$ ( $≪ 0.10$ , so approximation valid)

Solubility drops from $1.3 \times 1 0^{- 5}$ to $1.8 \times 1 0^{- 9} M$ — reduced by ~10,000×.

Key takeaway
Common-ion effect: adding a shared ion suppresses solubility. Approximate by assuming the added ion concentration dominates.
Question 11 · Hard
For the dissolution of calcium fluoride, $Ca F_{2} (s) ⇌ C a^{2 +} (aq) + 2 F^{-} (aq)$ , the $K_{s p} = 3.9 \times 1 0^{- 11}$ . What is the molar solubility of $Ca F_{2}$ ?
- A
  $2.1 \times 1 0^{- 4} M$ Correct
- B
  $3.9 \times 1 0^{- 11} M$
  Why not B: This is $K_{s p}$ ; need to solve the equilibrium expression for $s$ .
- C
  $6.2 \times 1 0^{- 6} M$
  Why not C: Used $s = K_{s p}$ as for a 1:1 salt, ignoring the 2:1 F $^{-}$ :Ca $^{2 +}$ ratio.
- D
  $1.3 \times 1 0^{- 4} M$
  Why not D: Used $K_{s p} = 4 s^{2}$ instead of $K_{s p} = 4 s^{3}$ .
Explanation
$[C a^{2 +}] = s$ ; $[F^{-}] = 2 s$

$K_{s p} = (s) (2 s)^{2} = 4 s^{3} = 3.9 \times 1 0^{- 11}$

$s^{3} = \frac{3.9 \times 1 0 ^{- 11}}{4} = 9.75 \times 1 0^{- 12}$

$s = (9.75 \times 1 0^{- 12})^{1/3} \approx 2.1 \times 1 0^{- 4} M$

Key takeaway
For CaF$_2$ type salts: $K_{sp} = 4s^3$. Set up the ICE table and use stoichiometry correctly.
Question 12 · Hard
Removing a product from a reaction system at equilibrium will:
- A
  Shift the equilibrium to the left (toward reactants) to replenish what was removed.
  Why not A: The system shifts to replenish the removed product, which means shifting right (toward products), not left.
- B
  Shift the equilibrium to the right (toward products) to partially restore the product concentration.Correct
- C
  Have no effect because $K_{c}$ does not change.
  Why not C: While $K_{c}$ is unchanged, the system is no longer at equilibrium ( $Q < K_{c}$ ) and will shift to re-establish it.
- D
  Decrease $K_{c}$ to account for the lower product concentration.
  Why not D: $K_{c}$ depends only on temperature and does not change when concentrations change.
Explanation
When a product is removed, $Q$ decreases below $K_{c}$ . Le Châtelier's principle: the system shifts to increase $Q$ back to $K_{c}$ , meaning more product is formed (forward shift). This is exploited industrially to drive reactions to completion.

Key takeaway
Removing a product ($Q < K_c$) shifts equilibrium forward; removing a reactant ($Q > K_c$) shifts it in reverse.

	A	B
Initial	$0.600 M$	$0$
Change	$- x$	$+ 2 x$
Equilibrium	$0.600 - x$	$2 x$