From the concentrations of C4H9Cl (butyl chloride) at different times given below, calculate the average rate of the reaction:
C4H9Cl+H2O→C4H9OH+HCl
during different intervals of time.
| t/s | 0 | 50 | 100 | 150 | 200 | 300 | 400 | 700 | 800 |
|---|---|---|---|---|---|---|---|---|---|
| [C4H9Cl]/mol L−1 | 0.100 | 0.0905 | 0.0820 | 0.0741 | 0.0671 | 0.0549 | 0.0439 | 0.0210 | 0.017 |
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 change in concentration of a reactant (or product) per unit time over a finite interval.
For the reaction C4H9Cl+H2O→C4H9OH+HCl, the average rate is:
Average rate=−ΔtΔ[C4H9Cl]
The negative sign is used because the concentration of butyl chloride decreases with time.
Step 1: Choose an interval.
For example, between t=50 s and t=100 s:
- Δt=100−50=50 s
- Δ[C4H9Cl]=0.0820−0.0905=−0.0085 mol L−1
Step 2: Apply the formula.
Average rate=−50(−0.0085)=500.0085=1.7×10−4 mol L−1s−1
Step 3: Repeat for other intervals as needed.
Here are the average rates for all consecutive intervals:
| Interval (s) | Δ[C4H9Cl] (mol L−1) | Average rate (mol L−1 s−1) |
|---|---|---|
| 0–50 | –0.0095 | 1.9×10−4 |
| 50–100 | –0.0085 | 1.7×10−4 |
| 100–150 | –0.0079 | 1.58×10−4 |
| 150–200 | –0.0070 | 1.4×10−4 |
| 200–300 | –0.0122 | 1.22×10−4 |
| 300–400 | –0.0110 | 1.1×10−4 |
| 400–700 | –0.0229 | 7.63×10−5 |
| 700–800 | –0.0040 | 4.0×10−5 |
Note: NCERT's printed Table 3.1 shows a 400–500 s row (rate 1.04×10−4) computed from a t=500 s concentration (0.0335) that does not appear in the question's own data listing; the 400–700 s row above is computed from the data exactly as printed in the question.
The average rate of reaction decreases from 1.9×10−4 to 4.0×10−5 mol L−1 s−1 over the time intervals from 0–50 s to 700–800 s.
The average rate of a reaction is the change in concentration of a reactant (or product) divided by the time interval. For butyl chloride, we compute Δt−Δ[C4H9Cl] for each consecutive pair of time points. The average rates (in mol L−1s−1) are: 0–50 s: 1.90×10−4; 50–100 s: 1.70×10−4; 100–150 s: 1.58×10−4; 150–200 s: 1.40×10−4; 200–300 s: 1.22×10−4; 300–400 s: 1.10×10−4; 400–700 s: 7.63×10−5; 700–800 s: 4.0×10−5.
The average rate of a reaction tells us how fast the concentration of a reactant or product changes over a specific time interval. For the reaction
C4H9Cl+H2O→C4H9OH+HCl,
butyl chloride (C4H9Cl) is consumed. Since water is in large excess, its concentration hardly changes, so we track only [C4H9Cl].
The average rate over a time interval Δt=t2−t1 is defined as:
Average rate=−ΔtΔ[C4H9Cl]=−t2−t1[C4H9Cl]t2−[C4H9Cl]t1
The negative sign ensures the rate is positive (concentration of reactant decreases). We’ll compute this for each consecutive time interval given in the table.
-
Interval 0 s to 50 s
Δt=50−0=50 s
Δ[C4H9Cl]=0.0905−0.100=−0.0095 mol L−1
Rate =−50−0.0095=500.0095=1.90×10−4 mol L−1 s−1
-
Interval 50 s to 100 s
Δt=100−50=50 s
Δ[C4H9Cl]=0.0820−0.0905=−0.0085 mol L−1
Rate =−50−0.0085=500.0085=1.70×10−4 mol L−1 s−1
-
Interval 100 s to 150 s
Δt=150−100=50 s
Δ[C4H9Cl]=0.0741−0.0820=−0.0079 mol L−1
Rate =−50−0.0079=500.0079=1.58×10−4 mol L−1 s−1
-
Interval 150 s to 200 s
Δt=200−150=50 s
Δ[C4H9Cl]=0.0671−0.0741=−0.0070 mol L−1
Rate =−50−0.0070=500.0070=1.40×10−4 mol L−1 s−1
-
Interval 200 s to 300 s
Δt=300−200=100 s
Δ[C4H9Cl]=0.0549−0.0671=−0.0122 mol L−1
Rate =−100−0.0122=1000.0122=1.22×10−4 mol L−1 s−1
-
Interval 300 s to 400 s
Δt=400−300=100 s
Δ[C4H9Cl]=0.0439−0.0549=−0.0110 mol L−1
Rate =−100−0.0110=1000.0110=1.10×10−4 mol L−1 s−1
-
Interval 400 s to 700 s
Δt=700−400=300 s
Δ[C4H9Cl]=0.0210−0.0439=−0.0229 mol L−1
Rate =−300−0.0229=3000.0229=7.63×10−5 mol L−1 s−1
NCERT's printed Table 3.1 shows a 400–500 s row at this point (rate 1.04×10−4 mol L−1 s−1), computed from a t=500 s concentration (0.0335 mol L−1) that does not appear in the question's own data listing, which jumps from 400 s to 700 s. The 400–700 s rate above (7.63×10−5 mol L−1 s−1) is computed directly from the data actually given — the book's own question and solution are inconsistent on this row.
- Interval 700 s to 800 s Δt=800−700=100 s Δ[C4H9Cl]=0.017−0.0210=−0.0040 mol L−1 Rate =−100−0.0040=1000.0040=4.0×10−5 mol L−1 s−1
A common mistake is to forget the negative sign or to use the wrong order of subtraction. Always do [C]2−[C]1 and then apply the minus sign in front — this guarantees a positive rate. Also, ensure the time interval is in seconds (it is here) and that units are consistent.
Notice that the average rate decreases as time increases. This is typical for a first-order reaction: the rate depends on concentration, so as the reactant is used up, the reaction slows down. The numbers here are consistent with a first-order decay.
The average rates of the reaction during the given time intervals are 1.90×10−4, 1.70×10−4, 1.58×10−4, 1.40×10−4, 1.22×10−4, 1.10×10−4, 7.63×10−5, and 4.0×10−5 mol L−1 s−1, respectively.
Method: Average Rate from Concentration–Time Data
The average rate of reaction over a time interval is defined as the negative of the change in concentration of a reactant divided by the change in time.
Steps
-
Identify the reactant whose concentration is given.
Here, it is C4H9Cl. Since it is a reactant, its concentration decreases with time.
-
Write the formula for average rate:
Average rate=−ΔtΔ[reactant]
The negative sign makes the rate positive (since Δ[reactant] is negative).
-
Choose a time interval (t1,t2) and read the corresponding concentrations [C4H9Cl]1 and [C4H9Cl]2 from the table.
-
Calculate:
Average rate=−t2−t1[C4H9Cl]2−[C4H9Cl]1
Example Calculation
Take the interval t=0 s to t=50 s:
- [C4H9Cl]1=0.100 mol L−1 at t1=0 s
- [C4H9Cl]2=0.0905 mol L−1 at t2=50 s
Average rate=−50−00.0905−0.100=−50−0.0095=500.0095=1.9×10−4 mol L−1s−1
Result for 0–50 s: 1.9×10−4 mol L−1s−1
General Note
Repeat the same steps for any other interval (e.g., 50–100 s, 100–150 s, etc.) using the corresponding concentration values. The average rate will generally decrease as time increases because the reactant concentration becomes smaller.
Here are the most common mistakes students make when calculating the average rate of reaction from concentration-time data, along with how to avoid each.
1. Confusing “average rate” with “instantaneous rate”
Mistake:
Students try to draw a tangent or use calculus concepts when the question explicitly asks for the average rate over an interval.
How to avoid:
Remember:
- Average rate = Δt−Δ[reactant] over a finite time interval.
- Use the formula directly:
Average rate=−t2−t1[C4H9Cl]t2−[C4H9Cl]t1
2. Forgetting the negative sign (or misinterpreting it)
Mistake:
Students write the rate as positive without the negative sign, or they put the negative sign in the wrong place.
How to avoid:
- The concentration of reactant decreases with time, so Δ[reactant] is negative.
- The rate must be positive, so we use:
Rate=−ΔtΔ[reactant]
- Example: For t=0 to 50 s: Δ[C4H9Cl]=0.0905−0.100=−0.0095 Rate =−50−0.0095=+1.9×10−4 mol L−1s−1
3. Using the wrong time interval (mixing up t1 and t2)
Mistake:
Students calculate Δt as t1−t2 instead of t2−t1, leading to a negative denominator.
How to avoid:
- Always use: Δt=tlater−tearlier
- The order in the numerator must match: [conc]later−[conc]earlier
4. Not using the correct units
Mistake:
Students forget to include units or write them incorrectly (e.g., mol L−1 without time, or s−1 without concentration).
How to avoid:
- The unit of average rate is always:
mol L−1s−1(or mol L−1min−1,etc.)
- Check that both concentration and time units are consistent.
5. Computing one overall 0–800 s average instead of interval-by-interval rates
Mistake:
Some students compute a single average rate over the entire 0–800 s span and present that one number as “the” average rate of the reaction.
Why it's wrong:
The question asks for the average rate during different intervals of time — the whole point is to see the rate fall as the reactant is consumed. NCERT's own Solution (Table 3.1) computes the rate for every consecutive interval of the data; that is the expected approach here.
How to avoid:
- Compute −ΔtΔ[C4H9Cl] for each consecutive pair of readings, exactly as in Table 3.1.
- Present the results as a table so the decreasing trend is visible.
6. Forgetting that rate changes with time
Mistake:
Students assume the average rate is constant and report only one value.
How to avoid:
- The average rate decreases as the reaction proceeds (because reactant concentration falls).
- Compute every consecutive interval (as NCERT's Table 3.1 does) so the decreasing trend is explicit.
7. Arithmetic errors in subtraction or division
Mistake:
Simple calculation mistakes, especially when dealing with numbers like 0.0905−0.100=−0.0095 (some write −0.0095 as −0.095).
How to avoid:
- Write each step clearly.
- Double-check decimal places.
- Use a calculator carefully.
✓ Quick checklist to avoid all mistakes
| Step | What to do |
|---|---|
| 1 | Identify t1, t2 and corresponding concentrations |
| 2 | Compute Δ[reactant]=[conc]t2−[conc]t1 |
| 3 | Compute Δt=t2−t1 |
| 4 | Apply: Rate=−ΔtΔ[reactant] |
| 5 | Write answer with correct units (mol L−1s−1) |
| 6 | Repeat for every consecutive interval, as in Table 3.1 |
Final tip:
Always show the formula in your answer before plugging numbers — examiners love seeing that you know the concept, not just the calculation.
- 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.
- Rate of the reaction =21×(rate of disappearance of N2O5)=21×5×10−3=2.5×10−3 molL−1min−1.
Common Mistakes
- Reporting the raw rate of disappearance of N2O5 (5×10−3) without dividing by its coefficient 2 -- that gives option (A), a common trap.
- Sign errors: forgetting the negative sign convention for a reactant, which doesn't change the magnitude here but can confuse students on which direction is positive.
✓Final answerThe correct option is (B) -- 2.5×10−3.
ANSWER: B
- 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∣.
- This has correct units of concentration/time (mol L−1 min−1), unlike the other options which either invert the ratio or square a term.
Common Mistakes
- Inverting the fraction (time over concentration change) — this gives the wrong units.
- Squaring the concentration difference, which is dimensionally and physically meaningless here.
✓Final answerThe correct option is (A) — 100∣x−y∣.
ANSWER: A
- 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.
- 24000.01=4.1667×10−6 molL−1s−1.
- This matches option (B).
Common Mistakes
- Leaving time in minutes instead of converting to seconds, which would give 2.5×10−4 (option A) — a common careless-unit trap built into the distractors.
- Forgetting the negative sign convention or misplacing a decimal, landing on 2.5×10−5 (option D) instead of the correctly converted value.
✓Final answerThe correct option is (B) — 4.167×10−6.
ANSWER: B
- 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.
Common Mistakes
- Forgetting to convert minutes to seconds (would give 4×10−4, a distractor option).
- Confusing this simple average-rate calculation with computing the first-order rate constant k (which needs the log form) — the question only asks for the rate, not k.
✓Final answerThe correct option is (A) — 6.667×10−6.
ANSWER: A
- 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).
- The mention of 'first order' is a distractor here — the rate-from-tangent relationship on a [A] vs t graph does not depend on reaction order; that information would only matter if the graph or question asked for the rate constant k.
- So the instantaneous rate at C is simply m, with no 2.303 factor needed.
Common Mistakes
- Applying the 2.303 conversion factor here — that factor belongs to log[A] vs t plots used for finding k in first-order kinetics, not to reading the rate off a plain [A] vs t curve.
- Being distracted by 'first order' into thinking the answer must involve the rate constant relation k=t2.303log[A][A]0, which is unrelated to reading a tangent slope.
✓Final answerThe correct option is (B) — m.
ANSWER: B
- 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.
- So statement (D), which claims Ms−1 for a first-order k, is false — that unit actually belongs to a zero-order reaction (n=0⇒M1s−1).
Common Mistakes
- Confusing the zero-order rate-constant unit (Ms−1) with the first-order one (s−1).
- Thinking rate constant units are always the same as rate units.
✓Final answerThe correct option is (D) — Unit of rate constant k for a first order reaction is Ms−1.
ANSWER: D
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