From a Rate at an Instant to a Concentration at a Time
A rate law such as rate = k[A]² gives the rate at one instant, calculated from the concentrations at that instant.
It does not directly answer the question an experiment usually asks: how much reactant remains after a given time?
The integrated rate law answers that question. It combines the rate law with time and gives [A]t, the concentration of A at time t, for a reaction of the form A → products.
There is one integrated law for each order. Each one is a straight line equation, y = mx + b, written in a different variable, and that is what makes the order testable from data.
The Three Integrated Rate Laws
For a zero order reaction, [A]t = [A]₀ − kt. A plot of [A] against time is a straight line with slope −k and intercept [A]₀, and k has units M s⁻¹.
For a first order reaction, ln[A]t = ln[A]₀ − kt. A plot of ln[A] against time is a straight line with slope −k and intercept ln[A]₀, and k has units s⁻¹.
For a second order reaction, 1/[A]t = kt + 1/[A]₀. A plot of 1/[A] against time is a straight line with slope +k and intercept 1/[A]₀, and k has units M⁻¹ s⁻¹.
The second order form above applies to a rate law rate = k[A]² in a single reactant. The slope of the 1/[A] plot is positive because 1/[A] grows as [A] falls.
In every case the intercept hands you the starting concentration, in the form that plot uses, and the slope hands you k, with its sign handled as above.
Identifying the Order: Which Plot Is Straight
Given concentration against time data, construct all three plots. Exactly one of them is straight, and which one it is gives the order.
A straight [A] plot means zero order, a straight ln[A] plot means first order, and a straight 1/[A] plot means second order.
The same test can be run on the numbers before anything is plotted. Over equal time intervals, a zero order reaction removes the same amount of A each interval.
A first order reaction removes the same fraction each interval, so ln[A] falls by the same amount each interval. A second order reaction adds the same amount to 1/[A] each interval.
If none of the three plots is straight, the reaction does not follow simple zero, first or second order kinetics in A alone, and none of these three laws applies.
Half-Life and How It Changes
The half-life, t₁/₂, is the time taken for the concentration of a reactant to fall to half of its value at the start of that interval.
For zero order, t₁/₂ = [A]₀ ÷ 2k. For first order, t₁/₂ = ln 2 ÷ k = 0.693 ÷ k. For second order, t₁/₂ = 1 ÷ (k[A]₀).
Only the first order half-life is independent of concentration, because its expression contains no concentration term. It stays the same at every point in the reaction.
Successive half-lives show the difference. In a zero order reaction each half-life is half as long as the one before, because there is less reactant to consume at the same fixed rate.
In a second order reaction each half-life is twice as long as the one before, because the falling concentration slows the rate steeply. In a first order reaction every half-life is equal.
After n half-lives of a first order reaction, the fraction remaining is (1/2)ⁿ. Two half-lives leave one quarter, so 75% consumed always means two half-lives, whatever k is.
Reaching 10% remaining is not a whole number of half-lives. It takes ln 10 ÷ k, which is ln 10 ÷ ln 2 ≈ 3.32 half-lives.
Worked Examples
Example 1: order and k from a data table. The decomposition 2N₂O₅(g) → 4NO₂(g) + O₂(g) is first order in N₂O₅. The measured concentrations are 0.800 M at 0 s, 0.400 M at 20 s, 0.200 M at 40 s and 0.100 M at 60 s. The corresponding ln[A] values are −0.223, −0.916, −1.609 and −2.303, falling by equal steps, which confirms first order. The slope of the ln[A] plot is (−2.303 − (−0.223)) ÷ 60 = −0.0347 s⁻¹, so k = 0.0347 s⁻¹. The half-life is 0.693 ÷ 0.0347 ≈ 20 s, exactly the halving interval visible in the table.
Example 2: a second order prediction. A → products is second order with k = 0.50 M⁻¹ s⁻¹ and [A]₀ = 0.20 M, so 1/[A]₀ = 5.0 M⁻¹. At t = 30 s, 1/[A]t = (0.50)(30) + 5.0 = 20 M⁻¹, giving [A]t = 0.050 M. The first half-life is 1 ÷ (k[A]₀) = 1 ÷ (0.50 × 0.20) = 10 s, yet reaching one quarter of [A]₀ takes 30 s rather than 20 s, because the second half-life, from 0.10 M to 0.050 M, lasts 20 s.
Example 3: a zero order prediction. A → products is zero order with k = 0.040 M s⁻¹ and [A]₀ = 1.20 M. At t = 12 s, [A]t = 1.20 − (0.040)(12) = 1.20 − 0.48 = 0.72 M. The half-life is [A]₀ ÷ 2k = 1.20 ÷ 0.080 = 15 s. The same law reaches [A] = 0 at 30 s, and it cannot be used beyond that point, because a concentration cannot fall below zero.
Example 4: time to a given fraction remaining. For a first order reaction with k = 0.0231 s⁻¹, how long until only 10% of the reactant remains? Writing the law as ln([A]t ÷ [A]₀) = −kt with [A]t ÷ [A]₀ = 0.100 gives t = ln 10 ÷ k = 2.303 ÷ 0.0231 = 99.7 s. Questions about a percentage remaining are always questions about the ratio [A]t ÷ [A]₀, so [A]₀ itself is never needed.
Traps That Cost Marks in Topic 5.3
Losing the sign of the slope. The [A] and ln[A] plots have slope −k, while the 1/[A] plot has slope +k. Reporting k itself as a negative number, or taking the slope of a 1/[A] plot as −k, are the two standard errors.
Reading [A]₀ straight off an ln[A] plot. The intercept of an ln[A] plot is ln[A]₀, not [A]₀. The initial concentration is e raised to the intercept, so an intercept of −1.61 means [A]₀ = 0.20 M, not −1.61 M.
Assuming every half-life is constant. Only the first order half-life is constant. Applying 0.693 ÷ k to a zero or second order reaction gives a wrong time and usually the wrong units as well.
Swapping the half-life formulas. The zero order half-life is [A]₀ ÷ 2k and the second order half-life is 1 ÷ (k[A]₀). They are not interchangeable, and each one depends on [A]₀ in a different way.
Confusing fraction consumed with fraction remaining. Consuming 75% leaves 25%, which is two half-lives for a first order reaction. Reading 75% as the fraction remaining turns the answer into a non whole number of half-lives and the wrong time.
Running a zero order law past zero. The equation [A]t = [A]₀ − kt happily returns negative concentrations after the reactant is exhausted. Those values are meaningless: the law describes the reaction only while reactant remains.
Key Relationships
- Integrated rate laws: zero order [A]t = [A]₀ − kt; first order ln[A]t = ln[A]₀ − kt; second order 1/[A]t = kt + 1/[A]₀
- Linear plots: [A] against t for zero order, ln[A] against t for first order, 1/[A] against t for second order
- Slopes and intercepts: slope −k and intercept [A]₀ for zero order; slope −k and intercept ln[A]₀ for first order; slope +k and intercept 1/[A]₀ for second order
- Half-lives: zero order t₁/₂ = [A]₀ ÷ 2k; first order t₁/₂ = 0.693 ÷ k, independent of [A]₀; second order t₁/₂ = 1 ÷ (k[A]₀)
- Successive half-lives: each one halves for zero order, stays constant for first order, and doubles for second order
- After n half-lives of a first order reaction the fraction remaining is (1/2)ⁿ; the time to reach a fraction f remaining is ln(1 ÷ f) ÷ k
- Units of k: M s⁻¹ for zero order, s⁻¹ for first order, M⁻¹ s⁻¹ for second order
1 thought on “Topic 5.3: Concentration Changes Over Time Revision Notes”