Unit 5 Practice: Medium
Kinetics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
For:
2A → B
B forms at a rate of 0.020 M/s. At what rate is A disappearing?
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Answer: 0.040 M/s
Explanation: The coefficients show:
2 mol A : 1 mol B
So A disappears twice as quickly as B appears.
0.020 × 2 = 0.040 M/s
Question 2
For:
Rate = k[A]²
[A] is tripled. By what factor does the rate change?
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Answer: 9 times
Explanation:
3² = 9
Since the reaction is second order in A, tripling A increases the rate by a factor of 9.
Question 3
For:
Rate = k[A][B]²
[A] doubles while [B] is cut in half. What happens to the rate?
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Answer: The rate becomes half as large.
Work:
Effect of A:
2¹ = 2
Effect of B:
(1/2)² = 1/4
Combined:
2 × 1/4 = 1/2
Question 4
In an initial-rates experiment, doubling [A] causes the reaction rate to quadruple. What is the order with respect to A?
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Answer: Second order
Explanation: We want:
2^m = 4
Since:
2² = 4
m = 2
Question 5
Doubling [B] causes no change in reaction rate. What is the order with respect to B?
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Answer: Zero order
Explanation: If changing concentration has no effect:
Rate ∝ [B]⁰
because:
2⁰ = 1
Question 6
Determine the units of k for:
Rate = k[A]²[B]
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Answer: M⁻²s⁻¹
Explanation: Overall order:
2 + 1 = 3
Rate units:
M/s
So:
k = (M/s)/M³
k = M⁻²s⁻¹
Question 7
A first-order reaction has:
k = 0.200 s⁻¹
Calculate its half-life.
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Answer: 3.47 s
Work:
t₁/₂ = 0.693/k
= 0.693/0.200
= 3.465 s
≈ 3.47 s
Question 8
What fraction of a first-order reactant remains after three half-lives?
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Answer: 1/8
Explanation:
After each half-life:
1 → 1/2 → 1/4 → 1/8
Question 9
A first-order reaction has:
[A]₀ = 1.00 M
k = 0.200 s⁻¹
Find [A] after 5.00 s.
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Answer: 0.368 M
Work:
Use:
ln[A]t = ln[A]₀ − kt
ln[A]t = ln(1.00) − (0.200)(5.00)
ln[A]t = −1.00
Take the inverse natural log:
[A]t = e⁻¹
= 0.368 M
Question 10
A second-order reaction has:
[A]₀ = 0.500 M
k = 0.400 M⁻¹s⁻¹
t = 5.00 s
Find [A].
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Answer: 0.250 M
Work:
Use:
1/[A]t = 1/[A]₀ + kt
1/[A]t = 1/0.500 + (0.400)(5.00)
= 2.00 + 2.00
= 4.00
Therefore:
[A]t = 1/4.00
= 0.250 M
Question 11
A graph of ln[A] vs. time is linear with slope −0.0350 s⁻¹. Determine the reaction order and k.
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Answer: First order
k = 0.0350 s⁻¹
Explanation: A linear ln[A] vs. time graph identifies first-order kinetics.
For first order:
slope = −k
So:
k = 0.0350 s⁻¹
Question 12
Write the rate law for the elementary reaction:
A + B → C
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Answer:
Rate = k[A][B]
Explanation: Because this is explicitly an elementary step, the stoichiometric coefficients can be used directly as exponents.
Question 13
Write the rate law for the elementary reaction:
2A → products
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Answer:
Rate = k[A]²
Explanation: The elementary step involves two A particles.
Therefore the step is second order in A.
Question 14
Identify the intermediate in this mechanism:
Step 1:
A + B → C
Step 2:
C + D → E
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Answer: C
Explanation: C is produced in Step 1 and consumed in Step 2.
When the steps are added, C cancels.
Question 15
How is a catalyst different from an intermediate?
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Answer: A catalyst is consumed and later regenerated, while an intermediate is produced and later consumed.
Explanation: A catalyst exists before the reaction begins and is recovered overall.
An intermediate forms during the mechanism and does not exist as an original reactant.
Question 16
What is the rate-determining step?
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Answer: The slow step that has the greatest effect on the overall reaction rate.
Explanation: In a multistep reaction, the overall reaction cannot proceed faster than its slowest important step.
Question 17
Why does increasing temperature usually increase reaction rate?
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Answer: A larger fraction of particles have enough energy to overcome the activation-energy barrier.
Explanation: Heating increases the kinetic-energy distribution.
More collisions therefore occur with:
collision energy ≥ Ea
Question 18
Does increasing temperature lower the activation energy?
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Answer: No.
Explanation: Temperature changes the distribution of particle energies.
It does not normally change the activation-energy barrier itself.
A catalyst lowers Ea.
Question 19
Write the Arrhenius equation.
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Answer:
k = Ae^(−Ea/RT)
Explanation: The equation relates the rate constant to:
activation energy
temperature
collision/orientation factor A
Question 20
At constant temperature, what happens to k if activation energy decreases?
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Answer: k increases.
Explanation: From:
k = Ae^(−Ea/RT)
a smaller Ea makes the exponent less negative.
That produces a larger value of k and therefore a faster reaction.