Q.For the reaction:
2A+B→A2B
the rate =k[A][B]2 with k=2.0×10−6 mol−2L2s−1. Calculate the initial rate of the reaction when [A]=0.1 mol L−1, [B]=0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
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.
Common Pitfall to Avoid
Do not write dtd[reactant] as a positive number and then forget the minus sign. The rate of change of a reactant is negative (concentration falls). The minus sign in the definition flips it to a positive r. If you skip the sign, you will get the wrong magnitude for other species.
Why This Matters
In exams (JEE, NEET, etc.), you will often be given the rate for one species and asked to find the rate for another. The stoichiometric relation is the only tool you need — no extra formulas. It also appears in more advanced topics like the rate law (where the exponents are not the coefficients) — but that is a separate concept. Reaction rate stoichiometry is purely about the definition of the reaction rate itself.
Final takeaway: The coefficients in the balanced equation are the conversion factors between the rates of different species. Always normalise by dividing by the coefficient to get the universal reaction rate r.
Average rate of reaction is one of the first ideas introduced in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘average rate of reaction formula’ or ‘average vs instantaneous rate’ are common important-question topics in board exams and JEE Main chemistry. A solid grasp of this basic definition is also assumed in nearly every subsequent kinetics numerical asked in NEET and state CETs.
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]
- νX is negative for reactants → the minus sign is already built in.
- νX is positive for products.
6. Why this is the average rate (not instantaneous)
- Average rate uses a finite Δt — it’s the slope of the chord between two points.
- Instantaneous rate uses Δt→0 — it’s the slope of the tangent at a single point.
The average rate formula is just the discrete version of the derivative:
Instantaneous rate=νX1dtd[X]
Summary — the "why" in one sentence
The average rate formula holds because it measures change in concentration per unit time, uses a minus sign to keep rates positive for reactants, and divides by stoichiometric coefficients to give a single, comparable value for the whole reaction.
Always remember:
- Δ[reactant] is negative → minus sign makes it positive.
- Δ[product] is positive → no minus sign.
- Divide by coefficient → normalise to "per mole of reaction".
Concept: Average Rate Of Reaction — the rate law is given directly, so we simply substitute concentrations into r=k[A][B]2.
Step 1: Initial rate
r0=(2.0×10−6)(0.1)(0.2)2
=(2.0×10−6)(0.1)(0.04)
=8.0×10−9 mol L−1s−1
Step 2: Rate after [A] drops to 0.06 mol L−1
From stoichiometry 2A+B→A2B, when [A] falls by 0.04 mol L−1, [B] falls by half of that: 0.02 mol L−1.
So [B]=0.2−0.02=0.18 mol L−1.
Step 3: Substitute
r=(2.0×10−6)(0.06)(0.18)2
=(2.0×10−6)(0.06)(0.0324)
=3.888×10−9 mol L−1s−1
The initial rate is 8.0×10−9 mol L−1s−1 and the rate after [A] is reduced to 0.06 mol L−1 is 3.9×10−9 mol L−1s−1 (rounded to two significant figures).
Substituting into the rate law gives an initial rate of 8.0×10−9 mol L−1s−1. When [A] falls to 0.06 mol L−1, stoichiometry gives [B]=0.18 mol L−1, and the rate becomes 3.888×10−9 mol L−1s−1.
The rate law Rate=k[A][B]2 is first order in A and second order in B. For the second part, note that as A is consumed, B is consumed too, in the ratio set by the balanced equation 2A+B→A2B.
1. Initial rate. Substitute [A]=0.1, [B]=0.2:
Rate0=k[A][B]2=(2.0×10−6)(0.1)(0.2)2=(2.0×10−6)(0.1)(0.04)=8.0×10−9 mol L−1s−1
2. Amount of A reacted. Δ[A]=0.1−0.06=0.04 mol L−1.
3. Amount of B reacted. From 2A+B→A2B, one mole of B is consumed for every two moles of A:
Δ[B]=21Δ[A]=21(0.04)=0.02 mol L−1
[B]new=0.2−0.02=0.18 mol L−1
4. New rate with [A]=0.06, [B]=0.18:
Rate=(2.0×10−6)(0.06)(0.18)2=(2.0×10−6)(0.06)(0.0324)=3.888×10−9 mol L−1s−1
Initial rate =8.0×10−9 mol L−1s−1. After [A] falls to 0.06 mol L−1 (so [B]=0.18 mol L−1), rate =3.888×10−9 mol L−1s−1.
Method: Direct Substitution into Rate Law
This method uses the experimentally determined rate equation to calculate reaction rates at given concentrations.
Steps:
Step 1: Identify the rate law
Given:
Rate=k[A][B]2
where k=2.0×10−6 mol−2L2s−1.
Step 2: Calculate the initial rate
Substitute the initial concentrations [A]0=0.1 mol L−1 and [B]0=0.2 mol L−1:
Rate0=(2.0×10−6)×(0.1)×(0.2)2
First, (0.2)2=0.04
Then, 0.1×0.04=0.004
Finally, 2.0×10−6×0.004=8.0×10−9
Initial rate=8.0×10−9 mol L−1s−1
Step 3: Find [B] when [A] drops to 0.06 mol L−1
From the stoichiometry: 2A+B→A2B
- Change in [A] = 0.10−0.06=0.04 mol L−1 consumed
- For every 2 moles of A consumed, 1 mole of B is consumed
- So, [B] consumed = 20.04=0.02 mol L−1
- Remaining [B]=0.20−0.02=0.18 mol L−1
Step 4: Calculate the rate after change
Substitute [A]=0.06 and [B]=0.18:
Rate=(2.0×10−6)×(0.06)×(0.18)2
(0.18)2=0.0324
0.06×0.0324=0.001944
2.0×10−6×0.001944=3.888×10−9
Rate after change=3.89×10−9 mol L−1s−1
Key Exam Tip:
- Always check stoichiometric ratios when finding changed concentrations — the rate law exponents are not the stoichiometric coefficients.
Here are the common mistakes students make with this exact problem, and how to avoid each.
Mistake 1: Forgetting the stoichiometric link between [A] and [B]
The error:
When [A] drops from 0.1 to 0.06 mol L−1, students often plug [B]=0.2 into the rate law again — as if [B] didn’t change.
Why it’s wrong:
The reaction consumes A and B together. From the equation
2A+B→A2B,
for every 2 moles of A used, 1 mole of B is used. So when [A] falls, [B] must also fall.
How to avoid:
Always calculate the change in [A], then use stoichiometry to find the change in [B].
- Change in [A]=0.10−0.06=0.04 mol L−1
- Since 2 A consumed per 1 B, Δ[B]=21×0.04=0.02 mol L−1
- So [B]new=0.20−0.02=0.18 mol L−1
Mistake 2: Confusing the rate law with the stoichiometric coefficient
The error:
Some students write the rate as
Rate=−21dtd[A]=k[A][B]2
and then try to use the −21 factor when plugging numbers.
Why it’s wrong:
The given rate law Rate=k[A][B]2 already defines the rate in terms of appearance of product or disappearance of reactant in the standard way. You don’t adjust k or the concentrations with coefficients unless the problem explicitly asks for −dtd[A].
How to avoid:
Read the problem carefully. If it says “the rate =k[A][B]2”, use that expression exactly as given. Do not insert stoichiometric factors unless asked.
Mistake 3: Using the wrong units or forgetting units entirely
The error:
Students compute the numeric value but write the rate with wrong units (e.g., s−1 or mol L−1).
Why it’s wrong:
The rate of reaction always has units of concentration per time. Here, k has units mol−2L2s−1, and [A][B]2 has units mol3L−3. Multiplying gives:
Units=(mol−2L2s−1)×(mol3L−3)=molL−1s−1
How to avoid:
Always do a unit check after every calculation. If the final unit isn’t mol L−1s−1 (or similar), you’ve made an error.
Mistake 4: Forgetting to square [B] in the rate law
The error:
Plugging [B]=0.2 instead of [B]2=0.04 in the initial rate calculation.
Why it’s wrong:
The rate law explicitly says [B]2. Missing the square changes the answer by a factor of 5 in this case.
How to avoid:
Write the rate law in full before substituting numbers:
Rate=k×[A]×[B]2
Then substitute step-by-step.
Mistake 5: Not checking if the reaction is elementary (over-interpreting the rate law)
The error:
Assuming the rate law matches the stoichiometric coefficients (i.e., thinking order = coefficient for every reactant).
Why it’s wrong:
Here it happens to match (order 1 in A, order 2 in B), but that’s not guaranteed in general. The rate law is experimental, not derived from the balanced equation.
How to avoid:
Treat the given rate law as fixed data. Never change the exponents based on the balanced equation unless told the reaction is elementary.
✓ Quick summary — correct calculations
Initial rate:
Rate0=(2.0×10−6)×(0.1)×(0.2)2
=2.0×10−6×0.1×0.04
=8.0×10−9 mol L−1s−1
Rate after [A] drops to 0.06:
[A]=0.06,[B]=0.18
Rate=(2.0×10−6)×(0.06)×(0.18)2
=2.0×10−6×0.06×0.0324
=3.888×10−9 mol L−1s−1
Final tip: Always write the stoichiometric link between reactants before doing the second part — it’s the most commonly missed step.
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