Average and Instantaneous Rate of Reaction from Data
“Rate of reaction” sounds like one number, but in Class 12 chemical kinetics there are two kinds: the average rate over a time interval and the instantaneous rate at one moment. Board papers ask you to calculate both from a table or graph, and also to link the rate to the stoichiometric equation. This page shows the method step by step with one clear set of data.
The formulas
Instantaneous rate = −d[R]/dt = slope of the tangent to the concentration–time curve
For aA + bB → cC + dD: rate of reaction = −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt
[R] is the concentration of a reactant and [P] of a product, in mol/L. Δt is the time interval in seconds (or minutes). The unit of rate is mol L−1 s−1. A minus sign is used for a reactant because its concentration falls, and we want the rate to be a positive number. For a product the concentration rises, so no minus sign is needed.
Average rate versus instantaneous rate
A reaction slows down as reactants are used up. So the rate in the first minute is higher than in the fifth minute. The average rate tells you what happened over a whole interval; it is the slope of the straight line joining two points on the concentration–time graph. The instantaneous rate tells you the speed at one exact time; it is the slope of the tangent at that point. As the interval Δt becomes very small, the average rate becomes the instantaneous rate.
Data used in the examples
These numbers are made up for practice. They describe a reaction A → products in which the concentration of A falls as follows.
| Time (s) | [A] (mol/L) |
|---|---|
| 0 | 0.400 |
| 100 | 0.280 |
| 200 | 0.196 |
| 300 | 0.137 |
Worked example 1: average rates in each interval
Interval 0–100 s. Rate = −(0.280 − 0.400)/(100 − 0) = 0.120/100 = 1.20 × 10−3 mol L−1 s−1.
Interval 100–200 s. Rate = (0.280 − 0.196)/100 = 0.084/100 = 8.4 × 10−4 mol L−1 s−1.
Interval 200–300 s. Rate = (0.196 − 0.137)/100 = 0.059/100 = 5.9 × 10−4 mol L−1 s−1.
Whole 0–300 s. Rate = (0.400 − 0.137)/300 = 0.263/300 = 8.77 × 10−4 mol L−1 s−1.
The three interval rates fall steadily (1.20, 0.84, 0.59 × 10−3), which matches the idea that the reaction slows as A is used up. The 0–300 s value is only an overall average and hides this slowing.
Worked example 2: instantaneous rate from the tangent
The data above come from a first-order reaction, so rate = k[A] at every moment. From the first 100 s: k = ln(0.400/0.280)/100 = 0.3567/100 = 3.57 × 10−3 s−1.
Instantaneous rate at t = 0: k[A] = 3.567 × 10−3 × 0.400 = 1.43 × 10−3 mol L−1 s−1.
Instantaneous rate at t = 100 s: k[A] = 3.567 × 10−3 × 0.280 = 1.00 × 10−3 mol L−1 s−1.
Check against the average. The average rate over 0–100 s was 1.20 × 10−3, which lies between the instantaneous values at the two ends (1.43 and 1.00 × 10−3). This is what you expect for a curve that is steadily flattening.
The graph method. In an exam you are usually given a graph, not k. Draw the tangent at t = 100 s and measure its slope. For this curve, the tangent at t = 100 s cuts the time axis at t = 100 + 1/k = 100 + 280 = 380 s. So the slope is −0.280/280 = −1.00 × 10−3 mol L−1 s−1. The rate is the size of this slope, 1.00 × 10−3. It matches the value from k[A].
Worked example 3: linking the rate to the equation
Question A. For 2A → B + C, the concentration of A falls at 1.20 × 10−3 mol L−1 s−1. Find the rate of reaction and the rate of formation of B.
Rate of reaction = −(1/2) d[A]/dt = 0.5 × 1.20 × 10−3 = 6.0 × 10−4 mol L−1 s−1. B is formed with coefficient 1, so d[B]/dt = 6.0 × 10−4 as well.
Question B. In N2 + 3H2 → 2NH3, ammonia forms at 4.0 × 10−3 mol L−1 s−1. Find the rates at which N2 and H2 are used.
Rate of reaction = (1/2)(4.0 × 10−3) = 2.0 × 10−3. So −d[N2]/dt = 1 × 2.0 × 10−3 = 2.0 × 10−3, and −d[H2]/dt = 3 × 2.0 × 10−3 = 6.0 × 10−3 mol L−1 s−1. Check: the mole ratio N2 : H2 : NH3 = 1 : 3 : 2 gives 2.0 : 6.0 : 4.0. Same.
Common mistakes
- A negative rate. Rate is never negative. For a reactant, include the minus sign in the formula so that the final number is positive.
- Ignoring the coefficients. Rate of disappearance of H2 is not the rate of reaction when its coefficient is 3. Divide by the coefficient to get the rate of reaction.
- Calling an average rate “the rate at that time”. The average over 0–100 s is not the rate at 50 s or at 100 s.
- Wrong tangent. A tangent touches the curve at one point; do not join two far-apart points and call that the instantaneous rate.
- Mixing time units. If time is in minutes, the rate is in mol L−1 min−1. Do not report it in seconds.
Exam relevance
| Question type | Method |
|---|---|
| Rate between two times | −Δ[R]/Δt |
| Rate at one time from a graph | Slope of the tangent at that point |
| Rate of formation of a product from reactant data | Use coefficients: divide by one, multiply by the other |
| Unit of rate | mol L−1 time−1 (for gases sometimes atm time−1) |
The data in this page are only for practice. Your textbook or board paper will give its own data; the method is the same.
This topic has no dedicated calculator tool, but you can use the Scientific Calculator keys (ln, division) to check your slopes and rate constants.
Open the Chemistry Calculator Suite →Reaction mechanisms stick only with practice: attempt previous-year CSIR-NET questions and a timed quiz.
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