In a reaction, different species appear and disappear at different speeds — and the
speeds are locked together by the balanced equation. "The rate of the reaction" is a single number
that all of them must agree on, and you get that number by dividing each species' own speed by its
coefficient.
Story track
Think of a bicycle factory. To build one bicycle you need two wheels and one frame. Now
suppose the factory produces 10 bicycles per hour.
How fast are wheels being used up? Twenty per hour — two for each bicycle. How fast are
frames used up? Ten per hour. The wheels disappear twice as fast as the frames, because the
recipe demands two of them.
Now someone asks: "what is the rate of the factory?" You cannot answer "twenty" and "ten" at the
same time. So you agree on a convention: divide each item's own speed by how many of that item the
recipe needs. Wheels: 20 ÷ 2 = 10. Frames: 10 ÷ 1 = 10. Bicycles: 10 ÷ 1 = 10. Everyone now
agrees the factory rate is 10.
That is exactly what chemists do. Divide by the coefficient and everybody agrees.
The minus signs go with reactants. Their concentrations fall, so d[A]/dt is negative; the
minus makes the rate come out positive. Products get a plus.
You divide by the coefficient, never multiply. This is the step people get backwards.
The direction of the division, settled once. The species
with the bigger coefficient changes faster. For 2HI → H₂ + I₂, HI disappears twice
as fast as H₂ forms. So to bring HI's fast number down to the common rate you must divide it by 2 —
hence Rate = −½·d[HI]/dt. If you find yourself multiplying, you have it upside down.
Worked examples straight from the chapter
Case 1 — all coefficients are 1. Hg(l) + Cl₂(g) → HgCl₂(s)
Rate = −Δ[Hg]/Δt = −Δ[Cl₂]/Δt = +Δ[HgCl₂]/Δt
Everything happens at the same speed, so no division is needed. This is the easy case, and it is
why students forget the rule exists.
Case 2 — a coefficient of 2. 2HI(g) → H₂(g) + I₂(g)
Rate = −½·Δ[HI]/Δt = +Δ[H₂]/Δt = +Δ[I₂]/Δt
HI vanishes twice as fast as H₂ appears, because two HI are consumed to make one H₂.
Case 3 — the full workout. 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)
H⁺ disappears fastest of all, six times faster than BrO₃⁻. This reaction is a favourite source of
exam questions precisely because the coefficients are all different.
How the questions are actually set
Almost every question of this type has the same shape: you are given one species' rate and
asked for another's. The safe two-step method never fails:
Maths track
Convert the given species' rate to the reaction rate by dividing by its coefficient.
Convert the reaction rate to the wanted species' rate by multiplying by its coefficient.
rate of X = (coefficient of X / coefficient of Y) × rate of Y
Worked: for 2N₂O₅ → 4NO₂ + O₂, if N₂O₅ disappears at 1.4 × 10⁻³ mol L⁻¹ min⁻¹, then NO₂ forms at
(4/2) × 1.4 × 10⁻³ = 2.8 × 10⁻³, and O₂ forms at (1/2) × 1.4 × 10⁻³ = 7.0 × 10⁻⁴.
Average rate and instantaneous rate
The same coefficient rule applies to both, so this unit and the average-versus-instantaneous idea
belong together:
r_av = −Δ[R]/Δt over an interval · r_inst = −d[R]/dt at an instant = slope of the tangent
Average rate is a chord between two points on the concentration–time curve.
Instantaneous rate is a tangent at one point.
Units for both: mol L⁻¹ s⁻¹, or atm s⁻¹ for gases measured by pressure.
The unit-conversion trap. A rate of 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹
becomes 1.13 × 10⁻⁵ mol L⁻¹ s⁻¹ (divide by 60) and 4.07 × 10⁻² mol L⁻¹ h⁻¹ (multiply by 60). NCERT's
Example 3.2 asks for all three from one calculation, and the arithmetic is where marks go missing.
Beyond the textbook
Why the convention matters physically. Without dividing by
coefficients, "the rate" would be ambiguous — you would have to state which species you meant every
time. The division makes the rate a property of the reaction rather than of any one species,
which is what allows a single rate constant k to describe the whole process.
Rates in terms of pressure. For a gas at constant
temperature and volume, concentration is proportional to partial pressure. So every relation on this
page holds with p in place of [ ], and rates come out in atm s⁻¹ or bar min⁻¹.
See it move — 3 animations
The third animation uses the five-coefficient bromate reaction deliberately — if you can work that one comfortably, every other example in this unit is easier.
ANIM 1
Why dividing by the coefficient makes everyone agree
Set the coefficients and press run. Watch A vanish and C appear at visibly different speeds — then look at the readout, where dividing each by its own coefficient produces the same number every time. That common number is the rate of reaction. Try a = 1, c = 4 to see the effect at its most dramatic.
ANIM 2
Steeper curve, same reaction rate
The product curve gets steeper as you raise its coefficient, while the reactant curve never changes. It is tempting to read the steeper curve as a faster reaction — it is not. It is the same reaction producing more molecules per event. Dividing by the coefficient removes the illusion.
ANIM 3
The two-step converter, on NCERT's hardest example
This is the bromate reaction from page 65, the one with five different coefficients. Choose which species you are given, which you want, and the rate value — and the panel walks the two steps in full. Use it to build the habit: divide by the given coefficient, multiply by the wanted one. Never try to jump straight across.
Formula sheet
The starred rows do all the work. Everything else is a check or a special case of them.
Four graph questions and three assertion–reason questions are included, marked by their coloured left borders. Questions tagged PYQ pattern follow forms that have appeared in NEET/AIPMT papers or come directly from NCERT exercises — exact year attributions are deliberately omitted rather than guessed.
Q01PYQ pattern
For the reaction 2NH₃ → N₂ + 3H₂ on a platinum surface (zero order) with k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹, the rates of production of N₂ and H₂ respectively are:
(a) 2.5 × 10⁻⁴ and 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹
(b) 5.0 × 10⁻⁴ and 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹
(c) 2.5 × 10⁻⁴ and 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
(d) 1.25 × 10⁻⁴ and 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹
Given
2NH₃ → N₂ + 3H₂, zero order, k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
Asked
Rates of production of N₂ and H₂
Concept
For zero order the rate of reaction equals k; then multiply by each product's coefficient.
Formula
Rate = k = +d[N₂]/dt = (1/3)d[H₂]/dt
Baby steps
Zero order means Rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
N₂ has coefficient 1, so d[N₂]/dt = 1 × Rate = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
H₂ has coefficient 3, so d[H₂]/dt = 3 × Rate = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
Sanity check: three times as much H₂ is made as N₂, matching the equation.
Step 2 — NH₃ has coefficient 2, so d[NH₃]/dt = 2 × 2 × 10⁻³ = 4 × 10⁻³.
Or in one line: (2/3) × 6 × 10⁻³ = 4 × 10⁻³ mol L⁻¹ s⁻¹.
Sanity check: three H₂ make two NH₃, so NH₃ appears more slowly than H₂ vanishes. 4 < 6. Correct.
Answer · (a) 4 × 10⁻³ mol L⁻¹ s⁻¹
Shortcut · Always sanity-check with the coefficient ratio: the species with the smaller coefficient must have the smaller rate.
Q04PYQ pattern
For 2HI(g) → H₂(g) + I₂(g), the rate of reaction is correctly written as:
(a) −½·d[HI]/dt = +d[H₂]/dt = +d[I₂]/dt
(b) −2·d[HI]/dt = +d[H₂]/dt
(c) −d[HI]/dt = +½·d[H₂]/dt
(d) +½·d[HI]/dt = −d[H₂]/dt
Given
2HI → H₂ + I₂
Asked
Correct rate expression
Concept
Divide by coefficients; minus for reactants, plus for products.
Formula
Rate = −(1/a)d[A]/dt = +(1/c)d[C]/dt
Baby steps
HI has coefficient 2 and is a reactant, so its term is −½·d[HI]/dt.
H₂ and I₂ both have coefficient 1 and are products, so their terms are +d[H₂]/dt and +d[I₂]/dt.
Option (b) multiplies rather than divides — the classic inversion error.
Option (d) has the signs reversed.
Answer · (a) −½·d[HI]/dt = +d[H₂]/dt = +d[I₂]/dt
Q05PYQ pattern
For 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O, which species disappears fastest?
(a) H⁺
(b) Br⁻
(c) BrO₃⁻
(d) All disappear at the same rate
Given
The balanced equation shown
Asked
Fastest-disappearing species
Concept
The species with the largest coefficient changes fastest.
Formula
rate of X = coefficient of X × rate of reaction
Baby steps
Coefficients of the reactants are: Br⁻ = 5, BrO₃⁻ = 1, H⁺ = 6.
Each species' own rate equals its coefficient multiplied by the reaction rate.
H⁺ has the largest coefficient at 6, so it disappears fastest.
It disappears six times faster than BrO₃⁻.
Answer · (a) H⁺
Q06PYQ pattern
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. The average rate in mol L⁻¹ s⁻¹ is:
(a) 6.66 × 10⁻⁶
(b) 4.0 × 10⁻⁴
(c) 2.5 × 10⁻⁵
(d) 1.2 × 10⁻³
Given
[R] falls 0.03 → 0.02 M in 25 min, coefficient 1
Asked
Average rate in mol L⁻¹ s⁻¹
Concept
Compute in minutes first, then convert the time unit.
Formula
r_av = −Δ[R]/Δt
Baby steps
Δ[R] = 0.02 − 0.03 = −0.01 mol L⁻¹.
r_av = 0.01/25 = 4.0 × 10⁻⁴ mol L⁻¹ min⁻¹.
Convert to seconds: divide by 60.
4.0 × 10⁻⁴/60 = 6.66 × 10⁻⁶ mol L⁻¹ s⁻¹.
Answer · (a) 6.66 × 10⁻⁶ mol L⁻¹ s⁻¹
Shortcut · Option (b) is the per-minute answer left unconverted — check which time unit the options use before selecting.
Q07PYQ pattern
In the reaction 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. The rate of reaction during this interval is:
(a) 0.005 mol L⁻¹ min⁻¹
(b) 0.01 mol L⁻¹ min⁻¹
(c) 0.02 mol L⁻¹ min⁻¹
(d) 0.05 mol L⁻¹ min⁻¹
Given
2A → Products; [A] falls 0.5 → 0.4 M in 10 min
Asked
Rate of reaction
Concept
The coefficient 2 must be divided out to get the reaction rate.
Formula
Rate = −(1/2)·Δ[A]/Δt
Baby steps
Δ[A] = 0.4 − 0.5 = −0.1 mol L⁻¹ over 10 minutes.
Rate of disappearance of A = 0.1/10 = 0.01 mol L⁻¹ min⁻¹.
The instantaneous rate of a reaction is obtained from a concentration–time graph by:
(a) drawing a tangent at that time and taking its slope
(b) drawing a chord between two points
(c) measuring the area under the curve
(d) reading the intercept on the concentration axis
Given
Concentration versus time plot
Asked
How to obtain the instantaneous rate
Concept
Instantaneous rate is a derivative, which geometrically is the slope of a tangent.
Formula
r_inst = −d[R]/dt = slope of the tangent
Baby steps
The instantaneous rate is defined as the derivative −d[R]/dt at a particular instant.
A derivative at a point corresponds geometrically to the slope of the tangent there.
A chord between two points gives the average rate instead, so (b) is wrong.
NCERT's Fig. 3.2 shows a tangent drawn at t = 600 s for exactly this purpose.
Answer · (a) drawing a tangent at that time and taking its slope
Q12
Which of the following expressions gives the rate of reaction for 4NH₃ + 5O₂ → 4NO + 6H₂O?
(a) −(1/5)d[O₂]/dt
(b) −(1/4)d[O₂]/dt
(c) −5·d[O₂]/dt
(d) +(1/5)d[O₂]/dt
Given
4NH₃ + 5O₂ → 4NO + 6H₂O
Asked
Rate expression in terms of O₂
Concept
Divide by O₂'s own coefficient, with a minus sign since it is a reactant.
Formula
Rate = −(1/b)d[B]/dt
Baby steps
O₂ has coefficient 5 and is a reactant.
So its term is −(1/5)d[O₂]/dt.
Option (b) uses NH₃'s coefficient by mistake; option (c) multiplies instead of dividing.
Answer · (a) −(1/5)d[O₂]/dt
Q13
For the reaction 2SO₂ + O₂ → 2SO₃, the ratio of the rate of disappearance of SO₂ to that of O₂ is:
(a) 2 : 1
(b) 1 : 2
(c) 1 : 1
(d) 3 : 2
Given
2SO₂ + O₂ → 2SO₃
Asked
Ratio of disappearance rates
Concept
Rates are proportional to coefficients.
Formula
rate of X ∝ coefficient of X
Baby steps
Coefficient of SO₂ is 2, of O₂ is 1.
Since each species' rate equals its coefficient times the reaction rate, the ratio equals the coefficient ratio.
Ratio = 2 : 1.
Answer · (a) 2 : 1
Q14PYQ pattern
A gaseous reaction is monitored by pressure. The units of the rate are then:
(a) atm s⁻¹
(b) atm⁻¹ s⁻¹
(c) mol L⁻¹ s⁻¹ only
(d) dimensionless
Given
Gas-phase reaction monitored by partial pressure
Asked
Units of rate
Concept
Rate is always the measured quantity per unit time.
Formula
rate = Δp/Δt
Baby steps
For a gas at constant temperature and volume, concentration is proportional to partial pressure.
So the rate may be expressed as the change in pressure per unit time.
That gives units of atm s⁻¹ (or bar min⁻¹, depending on the units chosen).
NCERT states this explicitly when introducing rate units.
Answer · (a) atm s⁻¹
Q15
For A → B, the concentration of B rises from 0 to 0.6 M in 30 s. The average rate of the reaction is:
(a) 0.02 mol L⁻¹ s⁻¹
(b) 0.6 mol L⁻¹ s⁻¹
(c) 0.01 mol L⁻¹ s⁻¹
(d) 18 mol L⁻¹ s⁻¹
Given
[B] rises 0 → 0.6 M in 30 s, coefficient 1
Asked
Average rate
Concept
For a product, the rate is the rise in concentration divided by time, with a plus sign.
Formula
r_av = +Δ[B]/Δt
Baby steps
Δ[B] = 0.6 − 0 = 0.6 mol L⁻¹.
Δt = 30 s.
r_av = 0.6/30 = 0.02 mol L⁻¹ s⁻¹.
Coefficient is 1, so no division is needed.
Answer · (a) 0.02 mol L⁻¹ s⁻¹
Q16Graph
On the concentration–time plot shown, the quantity marked by the chord between t₁ and t₂ represents:
(a) the average rate between t₁ and t₂
(b) the instantaneous rate at t₁
(c) the rate constant
(d) the half-life
Given
A chord drawn between two points on the concentration–time curve
Asked
What it represents
Concept
A chord's slope is a change over an interval, which is the average rate.
Formula
r_av = −Δ[R]/Δt
Baby steps
The chord connects the concentrations at two separate times.
Its slope is the total change in concentration divided by the total time, which is exactly −r_av.
An instantaneous rate would require a tangent at a single point, not a chord.
So the chord gives the average rate over that interval.
Answer · (a) the average rate between t₁ and t₂
Shortcut · Chord = average. Tangent = instantaneous. Two words, two geometries.
Q17Graph
Two curves are plotted for the reaction A → 2C. Which statement about their slopes is correct?
(a) The C curve rises twice as steeply as the A curve falls
(b) Both curves have equal slope magnitudes
(c) The A curve falls twice as steeply as the C curve rises
(d) The slopes are unrelated
Given
A → 2C with concentration–time curves for both species
Asked
Relation between the slopes
Concept
Slope magnitudes are proportional to stoichiometric coefficients.
Formula
Rate = −d[A]/dt = +(1/2)d[C]/dt
Baby steps
From the rate expression, d[C]/dt = 2 × (−d[A]/dt).
So the magnitude of the C slope is twice that of the A slope.
Physically: each A that disappears produces two C, so C accumulates twice as fast.
The rate of reaction itself is still a single number — obtained by halving C's slope or taking A's directly.
Answer · (a) The C curve rises twice as steeply as the A curve falls
Shortcut · Bigger coefficient means steeper curve. The reaction rate is what you get after dividing that steepness out.
Q18Graph
The graph shows [C₄H₉Cl] against time for its hydrolysis. The instantaneous rate at t = 600 s is found by:
(a) taking the slope of the tangent drawn at t = 600 s
(b) dividing the initial concentration by 600
(c) taking the slope of the line joining t = 0 and t = 600 s
(d) taking the area under the curve up to 600 s
Given
Concentration–time curve with a tangent drawn at 600 s
Asked
Method for the instantaneous rate
Concept
Instantaneous rate is the tangent slope, as NCERT demonstrates for this exact reaction.
Formula
r_inst = −d[R]/dt = −(slope of tangent)
Baby steps
The instantaneous rate at a given moment is the derivative at that moment.
Geometrically that is the slope of the tangent line touching the curve there.
Option (c) describes an average rate over the whole interval, not an instantaneous one.
NCERT works this example and obtains 5.12 × 10⁻⁵ mol L⁻¹ s⁻¹ at 600 s.
Answer · (a) taking the slope of the tangent drawn at t = 600 s
Q19Graph
For a reaction A → P, the concentration–time curve for A is steep at first and flattens later. This shows that:
(a) the rate decreases as the reaction proceeds
(b) the rate constant decreases with time
(c) the reaction is zero order
(d) the rate increases with time
Given
A curve that is steep early and flat later
Asked
What this indicates
Concept
Slope magnitude is the instantaneous rate, so a flattening curve means a falling rate.
Formula
r_inst = −d[R]/dt
Baby steps
The steep early portion means a large slope magnitude, hence a high rate.
The flat later portion means a small slope magnitude, hence a low rate.
So the rate falls as the reaction proceeds, because the reactant concentration is falling.
The rate constant k does not change with time at constant temperature — only the rate does. So (b) is wrong.
Zero order would give a straight line, not a curve, so (c) is wrong.
Answer · (a) the rate decreases as the reaction proceeds
Shortcut · Rate falls, rate constant does not. Distinguishing these two is worth more marks than any calculation in this unit.
Q20Assertion–Reason
Assertion (A): For 2HI → H₂ + I₂, the rate of disappearance of HI is twice the rate of formation of H₂. Reason (R): Two moles of HI are consumed for every one mole of H₂ produced.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about HI decomposition
Asked
Truth values and explanation
Concept
Stoichiometry directly fixes the ratio of species rates.
Formula
Rate = −½·d[HI]/dt = +d[H₂]/dt
Baby steps
Check A: rearranging the rate expression gives −d[HI]/dt = 2 × d[H₂]/dt. A is true.
Check R: the balanced equation shows two HI producing one H₂. R is true.
Does R explain A? Yes — the 2:1 consumption ratio is precisely why the rates stand in the ratio 2:1.
Note that the rate of reaction is still a single number, obtained after dividing HI's rate by 2.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q21Assertion–Reason
Assertion (A): The rate of a reaction is always expressed as a positive quantity. Reason (R): The change in concentration of a reactant is negative, so a minus sign is included in its rate expression.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about the sign convention
Asked
Truth values and explanation
Concept
The minus sign exists purely to keep the rate positive.
Formula
Rate = −Δ[R]/Δt with Δ[R] negative
Baby steps
Check A: a rate describes how fast something happens, and is conventionally reported as positive. A is true.
Check R: reactant concentrations fall, so Δ[R] is negative, and multiplying by −1 makes the result positive. R is true.
Does R explain A? Yes — NCERT states explicitly that Δ[R] is multiplied by −1 to make the rate positive.
Products need no minus sign because their Δ is already positive.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q22Assertion–Reason
Assertion (A): The average rate and the instantaneous rate of a reaction are generally different. Reason (R): The average rate is constant over the chosen interval, whereas the rate actually changes continuously as the reaction proceeds.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about average versus instantaneous rate
Asked
Truth values and explanation
Concept
Averaging over an interval smooths out a continuously changing quantity.
Formula
r_av = −Δ[R]/Δt over an interval ; r_inst = −d[R]/dt at a point
Baby steps
Check A: NCERT's Table 3.1 shows the average rate falling from 1.90 × 10⁻⁴ to 0.4 × 10⁻⁴ across intervals, so an average over any one interval differs from the rate at most instants within it. A is true.
Check R: the average rate is a single number assigned to the whole interval, while the true rate falls continuously as reactant is consumed. R is true.
Does R explain A? Yes — the discrepancy arises precisely because a single averaged number is being used to describe a changing quantity.
This is exactly why NCERT introduces the instantaneous rate at all.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q23
For a reaction 3A → 2B, the rate of appearance of B is 1.0 × 10⁻³ mol L⁻¹ s⁻¹. The rate of disappearance of A is:
(a) 1.5 × 10⁻³ mol L⁻¹ s⁻¹
(b) 0.67 × 10⁻³ mol L⁻¹ s⁻¹
(c) 1.0 × 10⁻³ mol L⁻¹ s⁻¹
(d) 3.0 × 10⁻³ mol L⁻¹ s⁻¹
Given
d[B]/dt = 1.0 × 10⁻³ mol L⁻¹ s⁻¹ for 3A → 2B
Asked
−d[A]/dt
Concept
Apply the coefficient ratio.
Formula
rate of A = (3/2) × rate of B
Baby steps
Coefficient of A is 3, of B is 2.
Rate of A = (3/2) × 1.0 × 10⁻³.
= 1.5 × 10⁻³ mol L⁻¹ s⁻¹.
Sanity check: A has the larger coefficient, so it must change faster. 1.5 > 1.0. Correct.
Answer · (a) 1.5 × 10⁻³ mol L⁻¹ s⁻¹
Q24
Which statement about the rate of a reaction is INCORRECT?
(a) The rate of reaction equals the rate of disappearance of any reactant, regardless of coefficients
(b) The rate of reaction is always positive
(c) Rate has units of concentration per unit time
(d) The rate generally decreases as the reaction proceeds
Given
Four statements about reaction rate
Asked
The incorrect one
Concept
The rate of reaction equals a species' own rate only when that species' coefficient is 1.
Formula
Rate = −(1/a)d[A]/dt
Baby steps
Statements (b), (c) and (d) are all standard and correct.
Statement (a) omits the division by the coefficient.
For 2A → products, the rate of disappearance of A is twice the rate of reaction, not equal to it.
So (a) is incorrect.
Answer · (a) The rate of reaction equals the rate of disappearance of any reactant, regardless of coefficients
Q25
A rate of 3.0 × 10⁻³ mol L⁻¹ min⁻¹ expressed in mol L⁻¹ h⁻¹ is:
(a) 0.18
(b) 5.0 × 10⁻⁵
(c) 3.0 × 10⁻³
(d) 1.8 × 10⁻¹ × 10⁻³
Given
Rate = 3.0 × 10⁻³ mol L⁻¹ min⁻¹
Asked
Rate in mol L⁻¹ h⁻¹
Concept
A per-hour figure is 60 times the per-minute figure.
Formula
rate per hour = rate per minute × 60
Baby steps
One hour contains 60 minutes, so more happens per hour than per minute.
Rate = 3.0 × 10⁻³ × 60.
= 0.18 mol L⁻¹ h⁻¹.
Answer · (a) 0.18 mol L⁻¹ h⁻¹
Shortcut · Going to a larger time unit multiplies; going to a smaller one divides. Ask yourself 'is more happening?' before choosing.
Q26
In the reaction 2A + 3B → 4C, if A is consumed at 0.02 mol L⁻¹ s⁻¹, then C is produced at: