AP Chemistry Practice Quiz: Concentration Changes Over Time
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 14 questions to check your progress.
Question 1 of 14
All Questions (14)
A) Reactant concentration vs. time
B) ln(reactant concentration) vs. time
C) 1/(reactant concentration) vs. time
D) (Reactant concentration)² vs. time
Correct Answer: B
According to the provided content, for a first-order reaction, a plot of ln(reactant concentration) vs. time is linear. This is a direct application of the integrated rate law for first-order reactions.
A) Zeroth-order
B) First-order
C) Second-order
D) Third-order
Correct Answer: C
The content states that for a second-order reaction, a plot of 1/(reactant concentration) vs. time is linear. Therefore, if this plot yields a straight line, the reaction must be second-order.
A) Zeroth-order
B) First-order
C) Second-order
D) The order cannot be determined from half-life alone.
Correct Answer: B
A key characteristic outlined in the content is that the half-life (t₁/₂) is constant for first-order reactions. It is independent of the initial reactant concentration.
A) slope = k
B) slope = -k
C) slope = 1/k
D) slope = 0.693/k
Correct Answer: B
The integrated rate law for a first-order reaction is ln[A] = -kt + ln[A]₀. This equation is in the form y = mx + b, where the slope (m) is equal to -k. The content states that the slopes of linear plots are related to the rate constant (k).
A) The rusting of iron
B) The combustion of methane
C) The neutralization of an acid with a base
D) Radioactive decay
Correct Answer: D
The provided content specifically identifies that 'Radioactive decay is an important example of first-order kinetics.'
A) Rate = k
B) Rate = k[reactant]
C) Rate = k[reactant]²
D) The rate law cannot be determined from this information.
Correct Answer: C
The rate law expression can be identified using data that show how concentrations change over time. A linear plot of 1/[reactant] vs. time indicates a second-order reaction. A curved plot of ln[reactant] vs. time confirms it is not first-order. Therefore, the rate law is Rate = k[reactant]².
A) The rate constant, k
B) The negative of the rate constant, -k
C) The half-life, t₁/₂
D) The initial concentration, [A]₀
Correct Answer: A
The integrated rate law for a second-order reaction is 1/[A] = kt + 1/[A]₀. This equation follows the linear form y = mx + b, where the slope (m) is equal to the rate constant, k.
A) 0.40 M
B) 0.20 M
C) 0.10 M
D) 0.00 M
Correct Answer: B
For a first-order reaction, the half-life is constant. 100 minutes is equal to two half-lives (100 min / 50 min = 2). After the first half-life (50 min), the concentration will be half of the initial value: 0.80 M / 2 = 0.40 M. After the second half-life (another 50 min), the concentration will be halved again: 0.40 M / 2 = 0.20 M.
A) t₁/₂ = k / 0.693
B) t₁/₂ = 0.693 / k
C) t₁/₂ = k * [A]₀
D) t₁/₂ = 1 / (k * [A]₀)
Correct Answer: B
The provided content gives the explicit formula for the half-life of a first-order reaction: t₁/₂ = 0.693/k. This shows that the half-life is inversely proportional to the rate constant.
A) By finding the slope of the tangent line on a concentration vs. time graph at t=0.
B) By determining which function of concentration ([A], ln[A], or 1/[A]) yields a linear plot when graphed against time.
C) By measuring the final concentration of the product after the reaction goes to completion.
D) By calculating the activation energy from the y-intercept of the linear plot.
Correct Answer: B
The content states that 'The order of a reaction can be inferred from a graph of concentration of reactant versus time.' This is achieved by testing which plot ([A], ln[A], or 1/[A] vs. time) is linear, as the linear plot is characteristic of a specific reaction order.
A) Rate = k
B) Rate = k[X]
C) Rate = k[X]²
D) The rate law cannot be determined without knowing k.
Correct Answer: B
The data show that the concentration is halved every 20 seconds (from 1.0 M to 0.5 M in 20s, and from 0.5 M to 0.25 M in the next 20s). This indicates a constant half-life. A constant half-life is the defining characteristic of a first-order reaction, for which the rate law is Rate = k[X].
A) The half-life of Reaction 1 is greater than the half-life of Reaction 2.
B) The half-life of Reaction 1 is less than the half-life of Reaction 2.
C) The half-lives of both reactions are equal.
D) The relationship cannot be determined without the initial concentrations.
Correct Answer: B
The half-life of a first-order reaction is given by t₁/₂ = 0.693/k. This shows an inverse relationship between the half-life and the rate constant. A larger rate constant (k) means the reaction is faster, which corresponds to a shorter (lesser) half-life. Since k₁ > k₂, the half-life of Reaction 1 must be less than the half-life of Reaction 2.
A) 0.693 / k
B) k / 0.693
C) 2 * (0.693 / k)
D) 0.75 / k
Correct Answer: C
A decay of 75% means that 25% (or 1/4) of the original sample remains. For a first-order reaction, the time for the sample to decrease to 1/2 is one half-life (t₁/₂). The time for it to decrease to 1/4 (half of 1/2) is a second half-life. Therefore, the total time is two half-lives, or 2 * t₁/₂. Since t₁/₂ = 0.693/k, the time required is 2 * (0.693 / k).
A) The change in temperature of the system over time.
B) The change in pressure for a gaseous reaction at constant volume.
C) The concentrations of reaction species as they change over time.
D) The initial rates of reaction at different initial concentrations.
Correct Answer: C
The first point of the provided content states that the rate law expression is identified 'using data that show how the concentrations of reaction species change over time.' The graphical method involves plotting these concentrations (or functions of them) against time.