In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
| A/mol L−1 | 0.20 | 0.20 | 0.40 |
|---|---|---|---|
| B/mol L−1 | 0.30 | 0.10 | 0.05 |
| r0/mol L−1s−1 | 5.07×10−5 | 5.07×10−5 | 1.43×10−4 |
What is the order of the reaction with respect to A and B?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Average Rate Of Reaction
Reaction Rate Stoichiometry – From Intuition to Precision
Imagine you are watching a simple reaction:
2NO2→2NO+O2
As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.
That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.
The Intuitive Picture
Think of a factory assembly line. The balanced equation is like a recipe:
- 2 units of raw material A → 2 units of product B + 1 unit of byproduct C
If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).
The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.
The Precise Statement
For a general reaction:
aA+bB→cC+dD
The rate of reaction (often called the rate of the process, r) is defined as:
r=−a1dtd[A]=−b1dtd[B]=c1dtd[C]=d1dtd[D]
Here:
- dtd[X] is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
- The minus sign is used for reactants (their concentration decreases with time).
- The plus sign is used for products (their concentration increases with time).
- Dividing by the coefficient normalises the rate — so r is the same number for every species.
r=−a1dtd[A]=c1dtd[C]
This single value r is the intrinsic rate of the reaction, independent of which species you measure.
How to Use It (Step by Step)
Suppose you are given:
2H2+O2→2H2O
And you measure that dtd[H2]=−0.040 M/s (negative because H₂ is being consumed).
Step 1: Write the rate relation:
r=−21dtd[H2]=−11dtd[O2]=21dtd[H2O]
Step 2: Plug in the known value:
r=−21(−0.040)=+0.020 M/s
Step 3: Find the rate for O₂:
−dtd[O2]=r⇒dtd[O2]=−0.020 M/s
Step 4: Find the rate for H₂O:
21dtd[H2O]=r⇒dtd[H2O]=2r=0.040 M/s
A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.
--- …
Why this formula?
Average Rate of Reaction — Why the Formula Holds
Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.
1. What does "rate of reaction" mean physically?
A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.
- Rate = how fast this change happens.
- If you measure the change over a finite time interval, you get the average rate.
2. The core idea: change per unit time
For any quantity X that changes from X1 to X2 over time t1 to t2:
Average rate of change of X=ΔtΔX=t2−t1X2−X1
This is just the slope of the straight line connecting the two points on a concentration vs. time graph.
3. Applying this to a reaction
Consider a simple reaction:
A→B
- Reactant A is consumed: [A] decreases.
- Product B is formed: [B] increases.
For reactant A (disappearing):
Average rate=−ΔtΔ[A]
Why the minus sign?
Because Δ[A]=[A]2−[A]1 is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by −1.
For product B (appearing):
Average rate=+ΔtΔ[B]
Here Δ[B] is positive, so no minus sign needed.
4. The general formula for any reaction
For a balanced reaction:
aA+bB→cC+dD
The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.
We divide each ΔtΔ[species] by its stoichiometric coefficient:
Average rate=−a1ΔtΔ[A]=−b1ΔtΔ[B]=+c1ΔtΔ[C]=+d1ΔtΔ[D]
Why divide by the coefficient?
If 2 moles of A disappear for every 1 mole of C formed, then ΔtΔ[A] is twice as large as ΔtΔ[C]. Dividing by the coefficient normalises them to the same "per mole of reaction" rate.
5. Key exam point: the formula in one line
For any species X with stoichiometric coefficient νX (negative for reactants, positive for products):
Average rate=νX1ΔtΔ[X] …
Concept: Average Rate Of Reaction — we compare how the initial rate changes when only one concentration is varied.
Step 1: Order with respect to B
Compare experiments 1 and 2 (A constant at 0.20 M).
B changes from 0.30 to 0.10 M (factor of 3 decrease), but rate stays 5.07×10−5.
Since rate is unchanged, order with respect to B is zero.
Step 2: Order with respect to A
Compare experiments 2 and 3 (B changes, but B is zero-order so its change doesn’t matter).
A changes from 0.20 to 0.40 M (factor of 2 increase).
Rate changes from 5.07×10−5 to 1.43×10−4 — that’s a factor of 0.5071.43≈2.82. …
The reaction is order 1.5 with respect to A and zero order with respect to B, so the rate law is r0=k[A]1.5[B]0=k[A]1.5.
Let the rate law be r0=k[A]m[B]n. We find m and n by the method of initial rates, comparing experiments in which only one concentration changes.
Order with respect to B. Compare experiments 1 and 2, where [A] is held constant at 0.20 mol L−1 while [B] changes from 0.30 to 0.10 mol L−1:
r0,2r0,1=5.07×10−55.07×10−5=1=(0.100.30)n=3n⟹n=0
The rate is unchanged when [B] changes, so the reaction is zero order in B.
Order with respect to A. Since n=0, r0=k[A]m. Compare experiments 1 and 3 (where [A] doubles from 0.20 to 0.40): …
Method: Initial Rate Method (Comparing Experiments)
This method determines reaction order by comparing how the initial rate changes when the concentration of only one reactant is varied while keeping the other constant.
Step 1 – Write the general rate law
The rate law is:
r0=k[A]m[B]n
We need to find m (order w.r.t. A) and n (order w.r.t. B).
Step 2 – Find order with respect to B
Compare experiments 1 and 2 (where [A] is constant at 0.20 mol L⁻¹):
| Experiment | [A] | [B] | r0 |
|---|---|---|---|
| 1 | 0.20 | 0.30 | 5.07×10−5 |
| 2 | 0.20 | 0.10 | 5.07×10−5 |
Since [A] is constant, the ratio of rates is:
r0,2r0,1=([B]2[B]1)n
Substitute values:
5.07×10−55.07×10−5=(0.100.30)n
1=(3)n
Therefore: n=0
The reaction is zero order with respect to B.
Step 3 – Find order with respect to A
Now compare experiments 2 and 3 (where [B] changes, but we already know n=0, so B does not affect rate).
| Experiment | [A] | [B] | r0 |
|---|---|---|---|
| 2 | 0.20 | 0.10 | 5.07×10−5 |
| 3 | 0.40 | 0.05 | 1.43×10−4 |
Common Mistakes Students Make on This Concept (Average Rate of Reaction & Order Determination)
Mistake 1: Confusing "Average Rate" with "Initial Rate"
The error: Students often try to calculate an average rate over a time interval, but the table gives initial rates (r0) — the rate at the very start of the reaction. These are not the same as average rates over time.
How to avoid:
- Read the problem carefully: "initial rate of reaction (r0)" means the instantaneous rate at t=0.
- The data is already in rate form — no need to compute ΔtΔ[product] from concentration vs. time data.
Mistake 2: Assuming the Order is an Integer Without Checking
The error: Jumping to conclusions like "order = 1" or "order = 2" without systematically comparing experiments.
How to avoid:
- Use the method of initial rates: Compare two experiments where only one reactant's concentration changes.
- Write the rate law:
r0=k[A]m[B]n
- Take ratios to eliminate k:
For order w.r.t. B:
Compare Expt 1 and Expt 2 (both have [A]=0.20):
r0,2r0,1=5.07×10−55.07×10−5=1=(0.100.30)n=(3)n
So 3n=1⇒n=0 — order w.r.t. B is zero.
For order w.r.t. A:
No pair of experiments holds [B] constant while changing [A] alone. That is not a problem: we have just established n=0, so [B] does not affect the rate at all. Compare Expt 1 and Expt 3 directly (the change in [B] between them is irrelevant because B is zero order):
- Expt 1: [A]=0.20, [B]=0.30, r0=5.07×10−5
- Expt 3: [A]=0.40, [B]=0.05, r0=1.43×10−4
With [B] playing no role:
r0,1r0,3=5.07×10−51.43×10−4≈2.82=(0.200.40)m=(2)m
So 2m=2.82⇒m≈1.5 — order w.r.t. A is 1.5 (not an integer!).
Mistake 3: Forgetting to Check if B Affects the Rate at All
The error: Assuming both reactants affect the rate, then getting confused when the ratio for B equals 1.
How to avoid:
- When the rate doesn't change despite changing [B], order w.r.t. B = 0 immediately. …
- AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQQ.Observe the following reaction 2N2O5(g)→4NO2(g)+O2(g) At T(K), the concentration of N2O5(g) changed from 2 mol L−1 to 1.5 mol L−1 in 100 min. What is the average rate (in mol L−1min−1) of this reaction? (A) 5×10−3 (B) 2.5×10−3 (C) 2.5×103 (D) 1.25×10−3
›Reveal solutionSolution
This tests the definition of "rate of reaction" that is normalised by stoichiometric coefficients so it gives the same value regardless of which species you track. The rate works out to 2.5×10−3 molL−1min−1.
Concept and Intuition
For a general reaction aA→bB+cC, different species are consumed/produced at different numerical speeds proportional to their coefficients. To get one unambiguous rate of reaction, each species' rate of change is divided by its own coefficient (with a minus sign for reactants, since their concentration falls): Rate=−a1dtd[A]=b1dtd[B]=c1dtd[C].
Step-by-Step Solution
- Reaction: 2N2O5(g)→4NO2(g)+O2(g), so the coefficient of N2O5 is 2.
- Concentration of N2O5 changes from 2 to 1.5 molL−1, so Δ[N2O5]=1.5−2=−0.5 molL−1, over Δt=100 min.
- Average rate of disappearance of N2O5=−ΔtΔ[N2O5]=−100−0.5=5×10−3 molL−1min−1. …
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.A→B is a first order reaction. The concentration of A is decreased from x mol L−1 to y mol L−1 in 100 min. What is the average velocity of the reaction in mol L−1 min−1 ? (A) 100∣x−y∣ (B) 100∣y−x∣2 (C) ∣x−y∣100 (D) ∣x+y∣100
›Reveal solutionSolution
This tests the basic definition of average rate of reaction as the change in concentration of a reactant over the time interval.
Concept and Intuition
The average rate of a reaction over a time interval is simply how much the concentration of a reactant (or product) changes, divided by how long it took — with a sign convention so the rate is always reported as a positive quantity.
Step-by-Step Solution
- Average rate of disappearance of A =−ΔtΔ[A]=−t[A]final−[A]initial.
- Here [A] decreases from x to y over t=100 min, so Δ[A]=y−x (negative, since y<x).
- Average rate =−100y−x=100x−y, and writing it generally (regardless of which is larger) as a positive quantity: 100∣x−y∣. …
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.R⟶P is a first order reaction. The concentration of R changed from 0.04 to 0.03 mol L−1 in 40 min. What is the average velocity of the reaction in mol L−1 s−1? (A) 2.5×10−4 (B) 4.167×10−6 (C) 4.167×106 (D) 2.5×10−5
›Reveal solutionSolution
Average rate is simply the change in concentration of reactant divided by the time elapsed (converted to seconds); the 'first order' label is not even needed for this particular calculation. Answer: (B).
Concept and Intuition
The average rate of a reaction over a finite time interval is defined purely from the measured change in concentration and elapsed time — it does not require knowing the rate law or order of the reaction (order only matters for finding the instantaneous rate constant k via the integrated rate law). Here we are only asked for the average velocity, so a direct Δ[conc]/Δt calculation suffices, with careful unit conversion from minutes to seconds since the answer must be in molL−1s−1.
Step-by-Step Solution
- Change in concentration of R: Δ[R]=0.04−0.03=0.01 molL−1 (a decrease, since R is being consumed).
- Time elapsed: 40 min =40×60=2400 s.
- Average rate =Δt−Δ[R]=24000.01 molL−1s−1. …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.For a first order reaction, the concentration of reactant was reduced from 0.03 mol L−1 to 0.02 mol L−1 in 25 min. What is its rate (in mol L−1s−1)? (A) 6.667×10−6 (B) 4×10−4 (C) 6.667×10−4 (D) 4×10−6
›Reveal solutionSolution
Rate is simply the drop in concentration divided by the elapsed time (converted to seconds) — 6.667×10−6 molL−1s−1, answer (A).
Concept and Intuition
The (average) rate of a reaction over a time interval is defined as the change in concentration of a reactant divided by the time elapsed (with a negative sign convention for reactants, but the magnitude is what's asked here). Care must be taken to convert all times to consistent units (here, minutes to seconds, since the rate is required in s−1).
Step-by-Step Solution
- Concentration drop: Δ[reactant]=0.03−0.02=0.01 molL−1.
- Time elapsed: 25 min=25×60=1500 s.
- Average rate =15000.01=6.667×10−6 molL−1s−1. …
- AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQQ.A→P is a first order reaction. The following graph is obtained for this reaction. (x-axis = time; y-axis = conc. of A). [FIGURE] (a concentration-vs-time curve for a first-order reaction, with a point C marked on the curve and a tangent line drawn through C labelled 'slope = m'). The instantaneous rate of the reaction at point C is (A) m1 (B) m (C) 2.303 m (D) 2.303m1
›Reveal solutionSolution
On a plain concentration-vs-time graph, the instantaneous rate at any point is just the magnitude of the tangent's slope there — here that is simply m, regardless of the reaction being first order.
Concept and Intuition
The instantaneous rate of a reaction A→P is defined purely graphically as Rate=−dtd[A], which is exactly the negative of the slope of the concentration-vs-time curve at that instant. This definition holds for a reaction of ANY order — it is a direct consequence of the definition of rate, not something that depends on the rate law. The factor of 2.303 only enters when working with log10[A] vs t plots (used to extract the first-order rate constant k from the slope −k/2.303) — it is irrelevant here since the axes are plain concentration and time, not logarithmic concentration.
Step-by-Step Solution
- The graph plots [A] (concentration of A) on the y-axis against time on the x-axis — a decaying curve, as expected since A is being consumed.
- At point C, a tangent line is drawn with slope magnitude m (the curve is decreasing, so the true signed slope is −m, but this magnitude labelled m represents how fast concentration is falling at that instant).
- By definition, instantaneous rate =−dtd[A]C=m (taking the tangent's slope magnitude as the rate). …
- AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQQ.Which statement among the following is incorrect? (A) Unit of rate of disappearance is M s−1 (B) Unit of rate of reaction is M s−1 (C) Unit of rate constant k depends upon order of reaction (D) Unit of rate constant k for a first order reaction is M s−1
›Reveal solutionSolution
This tests the general rule "unit of k = (molL−1)1−ns−1" for a reaction of order n. The false statement is (D): first order k has units s−1, not Ms−1.
Concept and Intuition
For a reaction of order n: rate=k[A]n, and rate always has units Ms−1. Rearranging for k:
k=[A]nrate⇒units of k=(M)1−ns−1
So the units of k change with order — this is exactly why chemists use the units of an experimentally measured k to identify the reaction order.
Step-by-Step Solution
- Rate of reaction / rate of disappearance of reactant: both are ΔtΔ[conc], units Ms−1 — (A), (B) correct.
- Units of k depend on order: for order n, units are M1−ns−1 — (C) correct.
- For n=1 (first order): units of k=M1−1s−1=M0s−1=s−1, not Ms−1. …
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