Q.Identify the reaction order from each of the following rate constants.
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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] …
The key idea is that the units of the rate constant k directly reveal the overall order of the reaction. For a general reaction of order n, the rate has units of mol L−1s−1, and since rate=k[conc]n, the units of k are (mol L−1)1−ns−1. …
The units of the rate constant k directly reveal the overall reaction order. For (i), units of L mol−1s−1 indicate second order; for (ii), units of s−1 indicate first order.
The rate constant k is not just a number — its units carry a fingerprint of the reaction’s overall order. This works because the rate of a reaction always has units of concentration per time (typically mol L−1s−1). When you write the rate law Rate=k[A]n, the units of k must adjust so that the right-hand side matches the units of rate.
For a reaction of order n, the general unit of k is:
Units of k=(concentration)1−n×time−1
In molar terms, concentration is in mol L−1, so:
Units of k=Ln−1mol1−ns−1
Let’s apply this to each case.
-
Case (i): k=2.3×10−5 L mol−1s−1
- The units are L1mol−1s−1.
- Compare with the general form Ln−1mol1−ns−1.
- For the exponent of mol: 1−n=−1⟹n=2.
- Check the exponent of L: n−1=1, which matches.
- So the reaction is second order overall.
-
Case (ii): k=3×10−4 s−1
- The units are simply s−1, with no concentration term.
- From the general form, this means (concentration)1−n must be dimensionless, so 1−n=0⟹n=1.
- The reaction is first order overall. …
Method: Unit Analysis of Rate Constant
This method uses the units of the rate constant (k) to determine the reaction order. The logic is based on the general rate law:
Rate=k[A]n
where:
- Rate has units of mol L−1s−1 (or concentration⋅time−1)
- [A] has units of mol L−1
- k has units that depend on n
Steps
- Write the general unit equation Units of Rate = Units of k × (Units of concentration)n
mol L−1s−1=[units of k]×(mol L−1)n
- Solve for units of k
units of k=(mol L−1)nmol L−1s−1=(mol L−1)1−n⋅s−1
- Match given units to find n Compare the given units of k with the expression above.
(i) k=2.3×10−5 L mol−1s−1
-
Given units: L mol−1s−1
This can be rewritten as: (mol L−1)−1⋅s−1
-
From step 2: (mol L−1)1−n⋅s−1
-
Equate exponents:
1−n=−1⇒n=2
Answer: Second order reaction
(ii) k=3×10−4 s−1
-
Given units: s−1 (no concentration term)
-
From step 2: (mol L−1)1−n⋅s−1 …
Here are the most common mistakes students make when identifying reaction order from the units of the rate constant (k), along with how to avoid each.
Mistake 1: Forgetting the General Formula for Units of k
Students often try to memorize the units for each order individually (e.g., "first order is s−1") but fail to connect them to a single, logical formula.
How to avoid:
Always derive the units using the general rate law:
Rate=k[Reactant]n
Where:
- Rate has units of concentration⋅time−1 (usually mol L−1s−1).
- [Reactant] has units of concentration (usually mol L−1).
Rearrange to solve for the units of k:
Units of k=(Units of Concentration)nUnits of Rate=(mol L−1)nmol L−1s−1
Simplify:
Units of k=(mol L−1)1−n⋅s−1
For your specific problems:
-
(i) k=2.3×10−5 L mol−1s−1
This is (mol L−1)−1s−1.
Set 1−n=−1⟹n=2. Second order.
-
(ii) k=3×10−4 s−1
This is (mol L−1)0s−1.
Set 1−n=0⟹n=1. First order.
Mistake 2: Confusing L mol−1s−1 with mol L−1s−1
A very common slip: reading the unit as "mol per liter per second" (which is actually the unit of rate, not k).
How to avoid:
Read the unit backwards or in words:
- L mol−1s−1 = "liters per mole per second" → This is second order.
- mol L−1s−1 = "moles per liter per second" → This is zero order (rate = k).
Quick check: If the unit has a negative exponent on concentration (like L mol−1), the order is greater than 1. If it has a positive exponent (like mol L−1), the order is less than 1 (zero order).
Mistake 3: Ignoring the Time Unit (e.g., s−1 vs min−1)
Students sometimes think that any k with s−1 must be first order. But s−1 alone is not enough — you must check the concentration unit too.
How to avoid:
- First order: k has units of only time−1 (e.g., s−1, min−1). No concentration unit appears.
- Second order: k has units of concentration−1⋅time−1 (e.g., L mol−1s−1). …
- 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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