For the following reaction,
2A + B - products
When concentration of both (A and B) becomes double,

Q: Why do we divide the rate equations?

Dividing the rate equations allows us to cancel out the rate constant $k$ and the initial concentrations, leaving only the ratios of the changes and the unknown powers $a$ and $b$.

Q: What does it mean when the order is 1 or 2?

An order of 1 (first order) means the rate is directly proportional to the concentration. An order of 2 (second order) means the rate is proportional to the square of the concentration.

Q: Can the order of a reaction be negative or a fraction?

Yes, reaction orders are determined experimentally and can be integers, zero, fractions, or even negative numbers, depending on the reaction mechanism.

Q: Why didn't we need the exact initial concentrations of A and B?

Because we are given the relative changes (e.g., 'concentration becomes double'), the actual initial values cancel out when we take the ratio of the rates.

Q: How do we find the overall order of the reaction?

The overall order of a reaction is simply the sum of the individual orders with respect to each reactant. In this case, it's $a + b$.

Question

For the following reaction,
$2A + B \longrightarrow 	ext{products}$
When concentration of both ($A$ and $B$) becomes double, then rate of reaction increases from $0.3 	ext{ mol L}^{-1}	ext{ s}^{-1}$ to $2.4 	ext{ mol L}^{-1}	ext{ s}^{-1}$.
When concentration of only $A$ is doubled, the rate of reaction increases from $0.3 	ext{ mol L}^{-1}	ext{ s}^{-1}$ to $0.6 	ext{ mol L}^{-1}	ext{ s}^{-1}$.
Which of the following is true?

Options:

(A) The whole reaction is of 4th order

(B) The order of reaction w.r.t. $B$ is one

(C) The order of reaction w.r.t. $B$ is 2

(D) The order of reaction w.r.t. $A$ is 2

Sigma Percentile · Accepted Answer

Imagine you are a chemist in a lab, carefully observing a reaction unfold. You have two reactants, $A$ and $B$, and you want to understand exactly how their concentrations dictate the speed of the reaction. This is the essence of chemical kinetics, and determining the rate law is our primary mission.

### The Master Equation

Every kinetic analysis begins with a fundamental assumption: the rate law. We express the rate $r$ as being proportional to the concentrations of the reactants raised to some unknown powers.

Mathematically, we write this as:
$$r = k[A]^a[B]^b$$

Here, $k$ is the rate constant, $a$ is the order with respect to reactant $A$, and $b$ is the order with respect to reactant $B$. Our entire goal is to act as chemical detectives and deduce the values of $a$ and $b$ using the experimental data provided.

### Analyzing the First Experiment

The problem gives us a fantastic starting point. Initially, the rate is $0.3 	ext{ mol L}^{-1}	ext{ s}^{-1}$. Let's call this $r_1$:
$$r_1 = 0.3 = k[A]^a[B]^b$$

Then, we are told that when the concentrations of **both** $A$ and $B$ are doubled, the rate skyrockets to $2.4 	ext{ mol L}^{-1}	ext{ s}^{-1}$. Let's call this new rate $r_2$:
$$r_2 = 2.4 = k[2A]^a[2B]^b$$

To find a relationship between our unknown powers, we can divide the second equation by the first. This is a classic algebraic trick in kinetics because it elegantly cancels out the rate constant $k$ and the initial concentrations $[A]$ and $[B]$:

$$\frac{r_2}{r_1} = \frac{2.4}{0.3} = \frac{k[2A]^a[2B]^b}{k[A]^a[B]^b}$$

Simplifying the left side gives us $8$. On the right side, we can pull out the factors of $2$:
$$8 = 2^a \cdot 2^b$$

Using the laws of exponents, we combine the terms on the right:
$$8 = 2^{a+b}$$

Since we know that $8$ is simply $2^3$, we can equate the exponents to reveal a beautiful linear relationship:
$$a + b = 3$$

### Analyzing the Second Experiment

Now, we need a second piece of evidence to solve for the individual values of $a$ and $b$. The problem states that when **only** the concentration of $A$ is doubled (meaning $B$ is kept constant), the rate increases to $0.6 	ext{ mol L}^{-1}	ext{ s}^{-1}$. Let's call this $r_3$:
$$r_3 = 0.6 = k[2A]^a[B]^b$$

Once again, we divide this new rate equation by our initial rate equation $r_1$:

$$\frac{r_3}{r_1} = \frac{0.6}{0.3} = \frac{k[2A]^a[B]^b}{k[A]^a[B]^b}$$

This time, the simplification is even more dramatic. The concentration of $B$ completely cancels out!
$$2 = 2^a \cdot 1^b$$

Since $1$ raised to any power is just $1$, we are left with:
$$2^1 = 2^a$$

This immediately tells us that the order with respect to $A$ is exactly $1$ ($a = 1$).

### The Final Calculation

We are now in the endgame. We have our master relationship $a + b = 3$, and we just discovered that $a = 1$. Substituting this value in is trivial:
$$1 + b = 3$$
$$b = 2$$

**Conclusion:**
The order of the reaction with respect to $A$ is $1$, and the order with respect to $B$ is $2$. The overall order of the reaction is $1 + 2 = 3$.

Looking at the given options, the statement "The order of reaction w.r.t. $B$ is 2" is the correct one. You have successfully decoded the kinetics of this reaction!

Select Answer:

Visualized Solution

$Rate Law Expression$

$Experiment 1: Both Concentrations Doubled$

$Analyzing Experiment 1$

$Experiment 2: Only A Doubled$

$Analyzing Experiment 2$

$Finding the Order with respect to B$

$The Way Forward$

The Sigma Insight: Order and Molecularity

The Master Equation

Analyzing the First Experiment

Analyzing the Second Experiment

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate Law Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Experiment 1: Both Concentrations Doubled

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Experiment 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Experiment 2: Only A Doubled

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Experiment 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Order with respect to B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward