Unit 7 Practice: Hard
Equilibrium · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
For:
A ⇌ B
Kc = 9.0.
Initially:
[A] = 1.00 M [B] = 0
Calculate the equilibrium concentrations.
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Answer: [A] = 0.10 M
[B] = 0.90 M
Work:
Let x react.
At equilibrium:
[A] = 1.00 − x
[B] = x
Then:
K = x/(1.00−x)
9.0 = x/(1.00−x)
9.0 − 9x = x
9.0 = 10x
x = 0.90
Therefore:
[B] = 0.90 M
[A] = 0.10 M
Question 2
For:
A ⇌ 2B
Kc = 1.00.
Initially:
[A] = 1.00 M [B] = 0
Set up the equilibrium equation.
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Answer:
Kc = (2x)² / (1.00 − x)
Therefore:
1.00 = 4x²/(1.00−x)
Explanation: From the reaction:
A decreases by x.
B increases by 2x.
The coefficient 2 also becomes an exponent in the K expression.
Question 3
Why should Q sometimes be calculated before setting up an ICE-table change direction?
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Answer: Because Q tells you whether the system must shift right or left.
Explanation: If all species are already present initially, you cannot automatically assume the reaction proceeds forward.
Use:
Q < K → shift right
Q > K → shift left
Q = K → already at equilibrium
Question 4
A system has K = 100 and Q = 0.010. What can you conclude?
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Answer: The reaction will strongly move toward products.
Explanation: Q is much smaller than K.
The system contains far too little product compared with the equilibrium ratio.
Question 5
A reaction has K = 5.0 × 10⁻¹². A student says, “No products exist at equilibrium.” Explain the mistake.
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Answer: A very small K means products are strongly disfavored, not necessarily absent.
Explanation: Equilibrium generally contains both reactants and products.
The amount of product may be extremely small, but it is usually not literally zero.
Question 6
Why does adding more of a pure solid usually not shift an equilibrium involving that solid?
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Answer: The activity of a pure solid remains constant as long as some solid is present.
Explanation: Pure solids are not included in equilibrium expressions.
Adding more changes the amount of solid, but not the equilibrium ratio of dissolved or gaseous species.
Question 7
For:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
predict the shift when volume is increased.
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Answer: Left
Explanation: Increasing volume lowers pressure.
The system shifts toward the side with more gas particles.
Left:
3 mol gas
Right:
2 mol gas
So equilibrium shifts left.
Question 8
For:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
what happens when pressure changes by changing volume?
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Answer: No major shift.
Explanation: There are:
2 mol gas on the left
and
2 mol gas on the right
So pressure changes do not favor either side based on gas-particle count.
Question 9
Explain why a catalyst cannot change the value of K.
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Answer: It changes reaction rates, not the relative thermodynamic stability of reactants and products.
Explanation: A catalyst lowers the activation energy for both directions.
The forward and reverse reactions both become faster.
Their equilibrium ratio remains unchanged.
Question 10
For an exothermic reaction, what happens to K when temperature increases?
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Answer: K decreases.
Explanation: For an exothermic reaction, heat is a product.
Increasing temperature shifts equilibrium toward reactants.
At the new temperature, the equilibrium ratio contains fewer products relative to reactants, so K becomes smaller.
Question 11
For an endothermic reaction, what happens to K when temperature increases?
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Answer: K increases.
Explanation: Heat behaves like a reactant.
Increasing temperature shifts equilibrium toward products, producing a larger equilibrium ratio and larger K.
Question 12
Write Ksp in terms of molar solubility, s, for:
MX(s) ⇌ M⁺ + X⁻
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Answer:
Ksp = s²
Explanation: If s mol/L dissolves:
[M⁺] = s
[X⁻] = s
Therefore:
Ksp = s × s = s²
Question 13
Write Ksp in terms of s for:
MX₂(s) ⇌ M²⁺ + 2X⁻
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Answer:
Ksp = 4s³
Work:
If solubility = s:
[M²⁺] = s
[X⁻] = 2s
Therefore:
Ksp = [M²⁺][X⁻]²
= s(2s)²
= 4s³
Question 14
Write Ksp in terms of s for:
M₂X₃(s) ⇌ 2M³⁺ + 3X²⁻
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Answer:
Ksp = 108s⁵
Work:
[M³⁺] = 2s
[X²⁻] = 3s
Therefore:
Ksp = (2s)²(3s)³
= 4s² × 27s³
= 108s⁵
Question 15
AgCl has:
Ksp = 1.8 × 10⁻¹⁰
Estimate its molar solubility in pure water.
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Answer: 1.34 × 10⁻⁵ M
Work:
For AgCl:
Ksp = s²
So:
s = √(1.8 × 10⁻¹⁰)
s ≈ 1.34 × 10⁻⁵ M
Question 16
Explain the common-ion effect using AgCl as an example.
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Answer: Adding Ag⁺ or Cl⁻ decreases the solubility of AgCl.
Explanation: For:
AgCl(s) ⇌ Ag⁺ + Cl⁻
adding one of the ions makes Qsp larger.
The equilibrium shifts left, causing more solid to form and decreasing the amount that can dissolve.
Question 17
What is Qsp used for?
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Answer: To predict whether a precipitate will form.
Explanation:
Compare Qsp with Ksp:
Qsp < Ksp: unsaturated, no precipitation required
Qsp = Ksp: saturated equilibrium
Qsp > Ksp: precipitation occurs
Question 18
Equal volumes of 0.010 M AgNO₃ and 0.010 M NaCl are mixed. After mixing, what are the initial concentrations of Ag⁺ and Cl⁻ before precipitation is considered?
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Answer: 0.0050 M each
Explanation: Mixing equal volumes doubles the total volume.
Each original concentration is therefore cut in half.
Question 19
Using Question 18, calculate Qsp for AgCl.
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Answer: 2.5 × 10⁻⁵
Work:
Qsp = [Ag⁺][Cl⁻]
= (0.0050)(0.0050)
= 2.5 × 10⁻⁵
If Ksp is much smaller than this value, AgCl precipitates.
Question 20
Explain the difference between changing equilibrium position and changing K.
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Answer: A system can shift left or right without changing K, but only temperature changes K.
Explanation: Changing concentration, pressure, or volume changes Q.
The system then shifts until Q once again equals the same K.
Temperature is different because it changes the actual equilibrium constant.
This distinction is one of the most important ideas in Unit 7.