In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times.

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Q. In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?

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Concept understanding — Effect of Concentration

Effect of Concentration: The Intuition

Imagine you're in a large, empty hall with just one other person. The two of you are trying to bump into each other accidentally. It'll take a while, right? Now imagine the same hall packed with a thousand people. Bumping into someone becomes almost certain within seconds.

That's the core idea behind the effect of concentration on reaction rates. Concentration simply means how much of a substance is packed into a given space. Higher concentration = more particles in the same volume.

When particles are more crowded, they collide more frequently. And since chemical reactions happen only when particles collide with enough energy and the right orientation, more collisions mean more reactions per second. The reaction speeds up.

The Precise Statement

For most chemical reactions, the rate of a reaction is directly proportional to the molar concentration of the reactants (raised to some power, which we'll get to).

\text{Rate} \propto [\text{Reactant}]^n

Here, $[ \ ]$ means "concentration in moles per litre" (mol/L or M), and $n$ is the order of reaction with respect to that reactant.

What does "order" mean?

For a simple reaction like $A \rightarrow \text{Products}$ :

First order ( $n=1$ ): Double the concentration of A → double the rate.
Second order ( $n=2$ ): Double the concentration of A → quadruple the rate ( $2^2 = 4$ ).
Zero order ( $n=0$ ): Changing concentration has no effect on the rate. This happens when the reaction is limited by something else (like a catalyst surface that's already fully covered).

Watch out

The order $n$ is not the same as the stoichiometric coefficient from the balanced equation. It must be determined experimentally. For example, the reaction $2A \rightarrow B$ could be first order in A, not second order.

Why does this happen? The collision theory

The rate depends on two things:

Collision frequency — how often particles meet.
Fraction of effective collisions — how many of those collisions have enough energy (activation energy) and the right orientation.

Doubling the concentration doubles the number of particles per unit volume. This roughly doubles the collision frequency. For a first-order reaction, that directly doubles the rate. For higher orders, the effect compounds because multiple reactant particles must meet simultaneously.

A concrete example

Consider the reaction between hydrochloric acid and sodium thiosulphate:

$\text{Na}_2\text{S}_2\text{O}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{S}(s) + \text{SO}_2(g) + \text{H}_2\text{O}(l)$ …

Why this formula?

Effect of Concentration on Reaction Rate — The Reasoning

The Effect of Concentration is rooted in collision theory. The key idea is simple:

More particles in the same volume → more frequent collisions → higher reaction rate.

Let's break down why the mathematical relationships hold.

1. The Rate Law — Why $r = k[A]^m[B]^n$ ?

This is not derived from theory alone — it is empirical (found experimentally). But the reasoning behind its form comes from collision probability.

For an elementary reaction (one step):

Consider:

A + B \rightarrow \text{products}

The rate depends on how often A and B molecules meet.
In a given volume, the number of A molecules is proportional to $[A]$ , and the number of B molecules is proportional to $[B]$ .
The number of A–B collisions per second is proportional to the product:

\text{Collision frequency} \propto [A] \times [B]

Therefore:

r \propto [A][B]

r = k[A][B]

For a reaction with coefficient $aA + bB$ :

If the reaction is elementary, the stoichiometric coefficients become the exponents:

r = k[A]^a[B]^b

Why? Because for $2A$ to react, two A molecules must collide simultaneously — the probability of that happening is proportional to $[A] \times [A] = [A]^2$ .

2. The Integrated Rate Laws — Why These Forms?

These come from solving the differential equation $r = -\frac{d[A]}{dt} = k[A]^n$ .

Zero-order ( $n=0$ ):

-\frac{d[A]}{dt} = k

Reasoning: Rate is independent of concentration. This happens when the reaction is limited by something else (e.g., a saturated catalyst surface).
Integrate:

\int_{[A]_0}^{[A]} d[A] = -k \int_0^t dt

\Rightarrow [A] = [A]_0 - kt

First-order ( $n=1$ ):

-\frac{d[A]}{dt} = k[A]

Reasoning: Rate is directly proportional to $[A]$ . Each molecule has a constant probability of reacting per unit time (like radioactive decay).
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k \int_0^t dt

\Rightarrow \ln[A] = \ln[A]_0 - kt

[A] = [A]_0 e^{-kt}

Second-order ( $n=2$ ):

-\frac{d[A]}{dt} = k[A]^2

Reasoning: Rate depends on two molecules of A colliding. Doubling $[A]$ quadruples the collision frequency.
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k \int_0^t dt

\Rightarrow \frac{1}{[A]} = \frac{1}{[A]_0} + kt

3. The Half-Life — Why It Depends on Order …

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The reaction order is found by comparing how the rate changes when concentration changes. Here, tripling $[A]$ multiplies the rate by $27$ , so $3^n = 27$ , giving $n = 3$ . The reaction is third order with respect to A.

Why this approach works

The rate law for a reaction involving a single reactant A is:

\text{Rate} = k [A]^n

where $n$ is the order of the reaction with respect to A. When we change the concentration of A, the rate changes according to that exponent $n$ . If we triple $[A]$ , the new rate becomes:

\text{Rate}_\text{new} = k (3[A])^n = 3^n \cdot k [A]^n = 3^n \cdot \text{Rate}_\text{old}

The problem tells us that $\text{Rate}_\text{new} = 27 \times \text{Rate}_\text{old}$ . So we simply need to find $n$ such that $3^n = 27$ .

Step-by-step solution

Write the general rate law For a reaction where only the concentration of A affects the rate:

\text{Rate} = k [A]^n

Express the new rate after tripling $[A]$ New concentration: $[A]_\text{new} = 3[A]_\text{old}$

\text{Rate}_\text{new} = k (3[A]_\text{old})^n = 3^n \cdot k [A]_\text{old}^n = 3^n \cdot \text{Rate}_\text{old}

Use the given ratio of rates The problem states: $\text{Rate}_\text{new} = 27 \times \text{Rate}_\text{old}$ Therefore:

3^n = 27

Solve for $n$ Since $27 = 3^3$ , we have: …

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Here are the common mistakes students make when solving this type of problem, along with how to avoid each one.

Mistake 1: Confusing the factor change with the order

What students do wrong:

They see "tripled" and "twenty seven times" and guess the order is 3 because $3 \times 9 = 27$ or because $3^3 = 27$ . That part is correct, but they often skip writing the rate law and jump to the answer without checking the logic.

How to avoid:

Always write the general rate law first:

\text{Rate} = k [A]^n

Then apply the change:

Initial: $\text{Rate}_1 = k [A]^n$
After tripling: $\text{Rate}_2 = k (3[A])^n = k \cdot 3^n [A]^n$

Given $\text{Rate}_2 = 27 \times \text{Rate}_1$ , we have:

k \cdot 3^n [A]^n = 27 \cdot k [A]^n

Cancel $k[A]^n$ :

3^n = 27

Since $27 = 3^3$ , we get $n = 3$ .

Key takeaway: Always set up the ratio $\frac{\text{Rate}_2}{\text{Rate}_1} = (\text{factor})^n$ and solve for $n$ .

Mistake 2: Forgetting that concentration change applies to the entire reactant

What students do wrong:

They treat "tripled" as adding 3 to the concentration instead of multiplying by 3. For example, they might write $[A] + 3$ instead of $3[A]$ .

How to avoid:

Remember: "tripled" means multiply by 3, not add 3. The rate law uses powers, so:

\text{New rate} = k (3[A])^n = k \cdot 3^n \cdot [A]^n

Never write $k ([A] + 3)^n$ — that is incorrect.

Mistake 3: Misidentifying the order when the factor is not a perfect power

What students do wrong:

If the rate increases by a factor that is not a simple power (e.g., 8 times for doubling concentration), they might guess the order incorrectly. Here, 27 is $3^3$ , so it's clean — but students sometimes think $27 = 3^2 \times 3$ and get confused.

How to avoid:

Use logarithms if needed:

3^n = 27 \implies n = \log_3 27 = 3

For non-integer orders, take $\log$ of both sides:

n = \frac{\log(\text{factor of rate})}{\log(\text{factor of concentration})}

Mistake 4: Forgetting that the rate constant $k$ cancels out

What students do wrong:

They try to solve for $k$ or think they need its value. This wastes time and leads to errors.

How to avoid:

Remember: $k$ is constant at a fixed temperature. When you take the ratio of two rates for the same reaction at the same temperature, $k$ cancels. So you only need the concentration factor and the rate factor.

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Q.In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?

Concept understanding — Effect of Concentration

Effect of Concentration: The Intuition

The Precise Statement

What does "order" mean?

Why does this happen? The collision theory

A concrete example

Why this formula?

Effect of Concentration on Reaction Rate — The Reasoning

1. The Rate Law — Why $r = k[A]^m[B]^n$ ?

For an elementary reaction (one step):

For a reaction with coefficient $aA + bB$ :

2. The Integrated Rate Laws — Why These Forms?

Zero-order ( $n=0$ ):

First-order ( $n=1$ ):

Second-order ( $n=2$ ):

3. The Half-Life — Why It Depends on Order …

Why this approach works

Step-by-step solution

Method: Initial Rate Method (Order from Concentration-Rate Dependence)

Steps

Mistake 1: Confusing the factor change with the order

Mistake 2: Forgetting that concentration change applies to the entire reactant

Mistake 3: Misidentifying the order when the factor is not a perfect power

Mistake 4: Forgetting that the rate constant $k$ cancels out

Q.In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?

Why this approach works

Step-by-step solution

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