The following data were obtained for the reaction $A + 2B \rightarrow C$: Experiment 1: $[A]=0.2$ M, $[B]=0.3$ M, initial rate of formation of C $=4.2 \times 10^{-2}$ M min$^{-1}$; Experiment 2: $[A]=0.1$ M, $[B]=0.1$ M, rate $=6.0 \times 10^{-3}$ M min$^{-1}$; Experiment 3: $[A]=0.4$ M, $[B]=0.3$ M, rate $=1.68 \times 10^{-1}$ M min$^{-1}$; Experiment 4: $[A]=0.1$ M, $[B]=0.4$ M, rate $=2.40 \times 10^{-2}$ M min$^{-1}$. (a) Find the order of reaction with respect to A and B. (b) Write the rate law and overall order of reaction. (c) Calculate the rate constant (k).

> [!TLDR] > Order = 2 in $A$, 1 in $B$; rate law $= k[A]^2[B]$; overall order = 3; $k = 6.0\ \text{M}^{-2}\,\text{min}^{-1}$. Let rate $= k[A]^x[B]^y$. **(a) Orders.** Compare Exp 1 and Exp 3 (same $[B]=0.3$): $[A]$ doubles ($0.2 \to 0.4$) and the rate rises $4.2\times10^{-2} \to 1.68\times10^{-1}$, a factor of $4$. So $2^x = 4 \Rightarrow x = 2$. Compare Exp 2 and Exp 4 (same $[A]=0.1$): $[B]$ increases fourfold ($0.1 \to 0.4$) and the rate rises $6.0\times10^{-3} \to 2.40\times10^{-2}$, a factor of $4$. So $4^y = 4 \Rightarrow y = 1$. Order with respect to $A = 2$; with respect to $B = 1$. **(b) Rate law and overall order.** $$\text{Rate} = k[A]^2[B], \qquad \text{overall order} = 2 + 1 = 3$$ **(c) Rate constant.** Using Exp 2: $$6.0\times10^{-3} = k(0.1)^2(0.1) = k(1.0\times10^{-3}) \implies k = 6.0\ \text{M}^{-2}\,\text{min}^{-1}$$ Exp 4 confirms this ($k = \dfrac{2.40\times10^{-2}}{(0.1)^2(0.4)} = 6.0$). …

The following data were obtained for the reaction A + 2B arrow C: Experiment 1: [A]=0.2 M, [B]=0.3 M, initial

Question

A + 2B \rightarrow C

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of reaction.

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Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

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).

Note

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 \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ 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 = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

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:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

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 $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

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 \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow 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 $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . 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 $\nu_X$ (negative for reactants, positive for products):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

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Order = 2 in $A$ , 1 in $B$ ; rate law $= k[A]^2[B]$ ; overall order = 3; $k = 6.0\ \text{M}^{-2}\,\text{min}^{-1}$ .

Let rate $= k[A]^x[B]^y$ .

Orders. Compare Exp 1 and Exp 3 (same $[B]=0.3$ ): $[A]$ doubles ( $0.2 \to 0.4$ ) and the rate rises $4.2\times10^{-2} \to 1.68\times10^{-1}$ , a factor of $4$ . So $2^x = 4 \Rightarrow x = 2$ . Compare Exp 2 and Exp 4 (same $[A]=0.1$ ): $[B]$ increases fourfold ( $0.1 \to 0.4$ ) and the rate rises $6.0\times10^{-3} \to 2.40\times10^{-2}$ , a factor of $4$ . So $4^y = 4 \Rightarrow y = 1$ . Order with respect to $A = 2$ ; with respect to $B = 1$ .
Rate law and overall order. $\text{Rate} = k[A]^2[B], \qquad \text{overall order} = 2 + 1 = 3$
Rate constant. Using Exp 2: $6.0\times10^{-3} = k(0.1)^2(0.1) = k(1.0\times10^{-3}) \implies k = 6.0\ \text{M}^{-2}\,\text{min}^{-1}$ Exp 4 confirms this ( $k = \dfrac{2.40\times10^{-2}}{(0.1)^2(0.4)} = 6.0$ ). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set B1 mark
$Q. Write True or False: The unit of rate of reaction is mol lit^-1 sec^-1.$
›Reveal solutionSolution

Rate = (change in concentration)/(time), so its unit is mol L^-1 s^-1 (or mol dm^-3 s^-1) — the statement is correct.

Rate of a chemical reaction = -(1/stoichiometric coefficient) x d[reactant]/dt = +(1/stoichiometric coefficient) x d[product]/dt. Since concentration is expressed in mo …

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CBSE 2024Set ANNUAL1 mark
$Q. When does average rate become equal to instantaneous rate?$
›Reveal solutionSolution

Average rate is a rate measured over a finite time span; as that span shrinks to zero it becomes the instantaneous rate — so the two are equal in the limit $\Delta t \to 0$ .

The average rate of a reaction over an interval $\Delta t = t_2 - t_1$ is:

$\text{Average rate} = -\dfrac{\Delta[R]}{\Delta t} = -\dfrac{[R]_2-[R]_1}{t_2-t_1}$

This is the mean rate over that whole time span and can mask how quickly the rate is actually changing within the interval.

The instantaneous rate at a specific time $t$ is the slope of the concentration-vs-time curve at that exact point:

$\text{Instantaneous rate} = -\dfrac{d[R]}{dt}$

Mathematically, the instantaneous rate is defined as the limit of the average rate as the time interval shrinks to zero:

$\text{Instantaneous rate} = \lim_{\Delta t \to 0} \left(-\dfrac{\Delta[R]}{\Delta t}\right) = -\dfrac{d[R]}{dt}$ …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. To express the rate at a particular moment of time we determine the ________.$
(a) Initial rate
(b) Instantaneous rate
(c) Average rate
(d) Standard rate
›Reveal solutionSolution

The rate 'at a particular moment' (not averaged over an interval) is called the instantaneous rate.

Average rate is calculated over a finite time interval (Delta[R]/Delta t) and changes depending on which interval is chosen. To get the rate AT one specific instant, we let the time interval shrink to zero:

…

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CBSE 2022Set E1 markMCQ
$Q. The rate of a chemical reaction$
(a) increases with time
(b) decreases with time
(c) may increase or decrease with time
(d) remains constant with time
›Reveal solutionSolution

Rate depends on reactant concentration; as reactants are consumed, concentration and hence rate fall with time.

For a typical reaction, Rate = k[reactant]^n. As the reaction proceeds the reactant concentration continuously decreases, so the rate also decreases with time (it is maximum at the start and appr …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. In the reaction 3A -> 2B, the rate of production of B is +d[B]/dt when the rate of reaction of A is$
(a) -(1/2) d[A]/dt
(b) -(2/3) d[A]/dt
(c) +2 d[A]/dt
(d) -(3/2) d[A]/dt
›Reveal solutionSolution

For a reaction $3A \to 2B$ , the rate of the reaction expressed via each species must be divided by its own stoichiometric coefficient, so rates in terms of A and B are related by the ratio of those coefficients.

Setting up the rate expression: For $3A \to 2B$ , the overall rate of reaction is defined uniquely (independent of which species you track) as:

$\text{Rate}=-\frac{1}{3}\frac{d[A]}{dt}=+\frac{1}{2}\frac{d[B]}{dt}$

Solving for d[B]/dt in terms of d[A]/dt:

Given the rate of production of B is $+\dfrac{d[B]}{dt}$ , equate:

$\frac12\frac{d[B]}{dt}=-\frac13\frac{d[A]}{dt}$ …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. For a gaseous reaction, the units of rate of reaction are -$
(a) Latm s⁻¹
(b) atm s⁻¹
(c) atm mol⁻¹ s⁻¹
(d) mol s⁻¹
›Reveal solutionSolution

For gaseous reactions, concentration is expressed as partial pressure, so rate has units of pressure/time.

Rate of reaction = (change in concentration)/(time). For reactions involving gases, concentration is conveniently measured as partial pressure (atm) …

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CBSE 2019Set ANNUAL1 mark
$Q. When does the average rate of a reaction become equal to instantaneous rate?$
›Reveal solutionSolution

Average rate is measured over a finite $\Delta t$ ; as that interval is shrunk toward zero, the average rate converges exactly to the instantaneous rate at that moment.

The average rate of a reaction over an interval is $-\dfrac{\Delta[R]}{\Delta t}$ (or $+\dfrac{\Delta[P]}{\Delta t}$ ), computed using the change in concentration over a finite time interval $\Delta t$ . The instantaneous rate is the rate at one particular instant, given by the derivative $-\dfrac{d[R]}{dt}$ .

…

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CBSE 2018Set ANNUAL1 mark
$Q. What is instantaneous rate of reaction?$
›Reveal solutionSolution

Instantaneous rate is the rate at one exact moment, found as $-d[R]/dt$ — the slope of the tangent drawn to the concentration-time plot at that instant (Δt → 0).

Whereas the average rate is measured over a finite time interval, the instantaneous rate is the rate of the reaction at one particular instant of time. It is obtained mathematically by making the time interval Δt infinitesimally small (Δt → 0), turning the average-rate expression into a derivative:

$\text{Instantaneous rate} = \lim_{\Delta t \to 0} \left(-\frac{\Delta[R]}{\Delta t}\right) = -\frac{d[R]}{dt} = \frac{d[P]}{dt}$ …

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CBSE 2016Set ANNUAL1 mark
$Q. What is average rate of a reaction?$
›Reveal solutionSolution

Average rate = (change in concentration) / (time interval) over a finite interval $\Delta t$ .

For a reaction R → P, the average rate over the time interval $t_1$ to $t_2$ is

$\text{Average rate} = -\frac{[R]_2-[R]_1}{t_2-t_1} = \frac{[P]_2-[P]_1}{t_2-t_1} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}.$ …

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