For the reaction, 2A + B - C, the values of initial rate at different reactant concentrations are g

Q: Why do we divide the equations instead of subtracting them?

Because the rate law involves multiplication ($k \times [A]^x \times [B]^y$), dividing the equations allows us to cancel out common terms like the rate constant $k$ and any concentration terms that are identical in both experiments. Subtracting them would leave us with a messy algebraic expression that is very hard to solve.

Q: What if no concentration is kept constant across any two experiments?

You can still solve it! You would divide two equations to get an expression with both $x$ and $y$. Then, you would divide another pair of equations to get a second expression with $x$ and $y$. You would then have a system of two equations with two variables, which you can solve simultaneously. However, JEE problems usually provide at least one constant concentration to save time.

Q: Does it matter which equations I choose to divide?

Mathematically, no. You will get the same answer regardless of which pair you choose. However, strategically, yes! Choosing a pair where one concentration is constant makes the math significantly faster and less prone to calculation errors.

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

Absolutely. While this specific problem resulted in simple integers ($1$ and $2$), reaction orders can be fractions (like $1/2$) or even negative numbers. A negative order simply means that increasing the concentration of that reactant actually slows down the reaction!

Q: How would I find the value of the rate constant $k$ after finding the rate law?

Once you have the complete rate law ($\text{Rate} = k[A][B]^2$), you can pick the data from ANY single experiment in the table. Plug in the Rate, $[A]$, and $[B]$ values for that specific experiment, and simply solve for $k$.

Question

For the reaction, $2A + B ightarrow C$, the values of initial rate at different reactant concentrations are given in the table below.

\begin{array}{|c|c|c|}
\hline
\mathbf{[A]} 	ext{ (mol L}^{-1}	ext{)} & \mathbf{[B]} 	ext{ (mol L}^{-1}	ext{)} & 	ext{	extbf{Initial rate}} 	ext{ (mol L}^{-1}	ext{s}^{-1}	ext{)} \
\hline
0.05 & 0.05 & 0.045 \
0.10 & 0.05 & 0.090 \
0.20 & 0.10 & 0.72 \
\hline
\end{array}

The rate law for the reaction is

Options:

(A) rate = $k[A][B]^2$

(B) rate = $k[A]^2[B]^2$

(C) rate = $k[A][B]$

(D) rate = $k[A]^2[B]$

Sigma Percentile · Accepted Answer

### The Mystery of the Rate Law

Imagine you are a detective trying to figure out exactly how a chemical reaction behaves. You know the reactants, $A$ and $B$, but you don't know how much influence each one has on the overall speed of the reaction. This influence is what we call the **order of the reaction** with respect to each reactant.

To solve this mystery, we use the **Initial Rate Method**. We perform the reaction multiple times, changing the starting concentrations of the reactants and measuring how fast the reaction kicks off. By comparing these different "experiments," we can deduce the exact mathematical relationship.

### Setting Up the Investigation

We start by writing a general rate law equation. We assume the rate depends on the concentrations of $A$ and $B$ raised to some unknown powers, $x$ and $y$:

$$	ext{Rate} = k[A]^x[B]^y$$

Our goal is to find the exact values of $x$ and $y$. Let's translate our experimental data into three distinct mathematical equations:

**Experiment 1:** $0.045 = k[0.05]^x[0.05]^y$
**Experiment 2:** $0.090 = k[0.10]^x[0.05]^y$
**Experiment 3:** $0.72 = k[0.20]^x[0.10]^y$

Notice a crucial clue? In experiments 1 and 2, the concentration of $B$ is kept perfectly constant at $0.05 	ext{ mol L}^{-1}$. This is our way in!

### Cracking the Code for Reactant A

To find $x$, we need to eliminate $y$. We can do this by dividing equation (i) by equation (ii). Because the concentration of $B$ is the same in both, the $k$ and the terms with $y$ will cancel out beautifully.

$$\frac{0.045}{0.090} = \frac{k[0.05]^x[0.05]^y}{k[0.10]^x[0.05]^y}$$

This simplifies down to:

$$\frac{1}{2} = \left(\frac{0.05}{0.10}ight)^x$$

$$\left(\frac{1}{2}ight)^1 = \left(\frac{1}{2}ight)^x$$

Clearly, **$x = 1$**. The reaction is first order with respect to $A$.

### Uncovering the Order for Reactant B

Next, let's find $y$. We'll divide equation (ii) by equation (iii). This time, neither concentration is constant, but we have a secret weapon: we already know that $x = 1$!

$$\frac{0.090}{0.72} = \frac{k[0.10]^x[0.05]^y}{k[0.20]^x[0.10]^y}$$

Substituting $x = 1$, we simplify the expression:

$$\frac{1}{8} = \left(\frac{0.10}{0.20}ight)^1 \left(\frac{0.05}{0.10}ight)^y$$

$$\frac{1}{8} = \left(\frac{1}{2}ight) \left(\frac{1}{2}ight)^y$$

Dividing both sides by $\frac{1}{2}$ gives us:

$$\frac{1}{4} = \left(\frac{1}{2}ight)^y$$

Since $\frac{1}{4}$ is simply $\left(\frac{1}{2}ight)^2$, it must be true that **$y = 2$**. The reaction is second order with respect to $B$.

### The Final Verdict

We have successfully found both orders! Substituting $x = 1$ and $y = 2$ back into our general rate law, we get the final expression:

**$$	ext{Rate} = k[A][B]^2$$**

This tells us that doubling the concentration of $A$ will double the rate, but doubling the concentration of $B$ will quadruple the rate! The mystery is solved.

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Visualized Solution

$G e n er a l R a t e L a w E q u a t i o n$

$F or m u l a t in g E q u a t i o n s f r o m D a t a$

$D e t er minin g O r d er w i t h r es p ec tt o A$

$D e t er minin g O r d er w i t h r es p ec tt o B$

$F ina l R a t e L a w$

$O v er a l l O r d er an d U ni t s$

The Sigma Insight: Order and Molecularity

The Mystery of the Rate Law

Setting Up the Investigation

Cracking the Code for Reactant A

Uncovering the Order for Reactant B

The Final Verdict

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Visualized Solution

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

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

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