Chemical Kinetics · Lesson 4 of 10
Zero-Order Reactions
“Derive zero-order concentration and half-life relationships and explain why surface saturation can make rate independent of concentration.”
• Interpret concentration independence over a stated kinetic range. • Integrate the zero-order disappearance equation step by step. • Extract k from a concentration–time graph and calculate times. • Derive the half-life and identify when a zero-order model stops applying.
Constant rate does not mean no reaction
A process can consume reactant at nearly the same rate even while its concentration changes. Imagine a catalytic surface with every active site occupied: supplying more reactant does not create more sites. If each site processes reactant at a fixed rate under the conditions, the total throughput is approximately constant. Such behaviour can be zero order with respect to the bulk reactant concentration over a saturation range.
The zero exponent says that rate does not respond to concentration changes within the observed range, not that molecules or reactant are unnecessary. If the concentration becomes too low to maintain saturation, the same law can cease to apply. Temperature, catalyst amount, surface area and other variables can still change the rate even when the partial order in R is zero.
Integrate the rate equation with its initial condition
Multiply the differential equation by dt to write d[R]=−k₀dt. Add the concentration changes from the initial state [R]₀ at time zero to the state [R]t at time t. Because k₀ is constant over the assumed interval, its integral is simply k₀t. Using definite limits automatically incorporates the initial condition.
An indefinite integration would instead give [R]=−k₀t+I. At t=0, [R]=[R]₀, so I=[R]₀ and the same law follows. On a plot of concentration against time, the intercept is [R]₀ and the slope is −k₀. A straight line by itself should be judged over a meaningful concentration range and against experimental uncertainty rather than inferred from only two points.
Problem
A zero-order disappearance process starts at 0.200 M with k₀=0.00200 M s⁻¹. Find concentration after 30 s and the formal time to zero.
- 1.After 30 s, [R]t=0.200−0.00200×30=0.140 M. The fall is 0.060 M, consistent with constant disappearance.
- 2.The linear model reaches zero at t=[R]₀/k₀=0.200/0.00200=100 s.
- 3.At 120 s it would give a negative concentration, which is physically impossible. The model must stop or change regime before that extrapolation.
- 4.A catalytic surface often loses saturation before exhaustion, so the formal zero time need not equal an actual complete-conversion time.
Derive the zero-order half-life
Substitute half the initial concentration into [R]t=[R]₀−k₀t. Rearranging gives k₀t=[R]₀/2 and division by k₀ gives the result. Starting with twice as much concentration at the same constant rate takes twice as long to consume the first half. Successive halving intervals in the same zero-order run become shorter because each interval removes a smaller absolute amount at the same rate.
Problem
For the 0.200 M, k₀=0.00200 M s⁻¹ process, compare the times to reach 0.100 M and then 0.050 M.
- 1.Time from 0.200 to 0.100 M is 0.100/0.00200=50.0 s, the first half-life.
- 2.The next fall is only 0.050 M, taking 0.050/0.00200=25.0 additional seconds. The total time to 0.050 M is 75.0 s.
- 3.Successive half-lives are not equal for a zero-order process. This contrasts with the constant halving intervals of a first-order process.
Surface examples and stoichiometric rate conventions
Ammonia decomposition on hot platinum can become approximately zero order in ammonia at sufficiently high pressure, when the surface is saturated. An illustrative set of conditions is about 1130 K with platinum present: 2NH₃(g) → N₂(g)+3H₂(g). Thermal HI decomposition on a gold surface and some enzyme-saturated processes provide other examples of concentration-insensitive regimes. The conditions are part of the statement; a change in ammonia concentration, temperature or surface coverage can change the kinetic behaviour.
Problem
For 2NH₃→N₂+3H₂, the zero-order ammonia disappearance constant is explicitly defined as k_d=2.5×10⁻⁴ M s⁻¹. Find product appearance rates.
- 1.The stated definition means −d[NH₃]/dt=k_d, not necessarily r=k_d. Normalised r=k_d/2=1.25×10⁻⁴ M s⁻¹.
- 2.Nitrogen coefficient is one, so d[N₂]/dt=r=1.25×10⁻⁴ M s⁻¹. Hydrogen coefficient is three, so d[H₂]/dt=3r=3.75×10⁻⁴ M s⁻¹.
- 3.If a constant were instead defined as the normalised r, product rates would be k_r and 3k_r. State the definition before using a numerical value.
- 4.A zero-order constant must carry concentration/time units. Inverse-concentration units would belong to a different order, not to this disappearance law.
The integrated law assumes constant k₀ and the same zero-order conditions throughout. It cannot predict negative concentrations, unlimited product growth without reactant, or a constant rate after a surface is no longer supplied.
Quick Quiz
Quiz
What does zero order in R mean within the valid range?
Explanation: The partial exponent in R is zero; rate can be nonzero and can still depend on temperature or surface properties.
Which plot is linear for zero-order disappearance?
Explanation: [R]t=[R]₀−k₀t is a straight line with slope −k₀.
If initial concentration doubles at unchanged k₀, its zero-order first half-life:
Explanation: t₁/₂=[R]₀/(2k₀), so it is directly proportional to the initial concentration.
What can explain zero-order surface kinetics at high reactant supply?
Explanation: Once sites are occupied, extra bulk reactant need not increase the surface processing rate.
A linear extrapolation gives [R]=−0.020 M. What is the right conclusion?
Explanation: Concentrations cannot be negative; the assumed regime must end or change before that extrapolation.
Practice Problems
- A zero-order process falls from 0.300 to 0.180 M in 40 s. Find k₀ and the first half-life for an initial 0.300 M.
- Derive [R]t=[R]₀−k₀t using the integration constant and identify the slope and intercept.
- Explain why doubling ammonia pressure need not double its decomposition rate on a saturated platinum surface.
- If the normalised zero-order rate for 2NH₃→N₂+3H₂ is explicitly k_r=2.5×10⁻⁴ M s⁻¹, find all three species rates.
- For k₀=0.0010 M s⁻¹ and initial 0.100 M, compare the time to consume 10% with the time to consume 90%.
The disappearance is 0.120 M in 40 s, so k₀=0.120/40=0.00300 M s⁻¹. Substitute into t₁/₂=[R]₀/(2k₀)=0.300/0.00600=50.0 s. The units are concentration/time because the rate is constant.
Key Takeaways
• Zero order means concentration independence within a stated range, not absence of reaction. • Integrating −d[R]/dt=k₀ gives [R]t=[R]₀−k₀t. • The concentration–time slope is −k₀ and the intercept is [R]₀. • A zero-order disappearance constant has units concentration/time. • The first half-life is [R]₀/(2k₀) and depends on initial concentration. • Surface or enzyme saturation can generate approximate zero-order regimes. • Do not extend the linear model into negative concentrations or switch rate conventions silently.