Animated Solution for Chemistry - Chemical Kinetics: The results given in the below table were obtained during kinetic studies of the following reaction : $2A+B\to C+D$ $\begin{array}{cccc}\text{Experiment}& [A]/{\text{mol L}}^{-1}& [B]/{\text{mol L}}^{-1}& \text{Initial rate}/{\text{mol L}}^{-1}{\text{min}}^{-1}\\ \text{I}& 0.1& 0.1& 6.00\times 10^{-3}\\ \text{II}& 0.1& 0.2& 2.40\times 10^{-2}\\ \text{III}& 0.2& 0.1& 1.20\times 10^{-2}\\ \text{IV}& X& 0.2& 7.20\times 10^{-2}\\ \text{V}& 0.3& Y& 2.88\times 10^{-1}\end{array}$ X and Y in the given table are respectively

The study of chemical kinetics is like being a detective. You are given a set of clues—how fast a reaction proceeds under different conditions—and your job is to uncover the hidden rules governing the molecules. In this problem, we are tasked with finding the missing concentrations, $X$ and $Y$, from a table of experimental data. To do this, we must first decipher the rate law of the reaction.

Every chemical reaction has a rate law that describes how the speed of the reaction depends on the concentration of its reactants. For the reaction $2A+B\to C+D$, we can write the general rate law as:

Here, $r$ is the initial rate, $k$ is the rate constant, and $a$ and $b$ are the orders of the reaction with respect to reactants $A$ and $B$, respectively. Our first mission is to find these exponents, $a$ and $b$.

The Method of Initial Rates

To find the order with respect to one reactant, we need to see how the rate changes when we vary its concentration while keeping everything else constant. This is the core principle of the method of initial rates.

Let's look at Experiment I and Experiment II. In Experiment I, $[A]=0.1\text{ M}$ and $[B]=0.1\text{ M}$. The rate is $6.00\times 10^{-3}{\text{ M min}}^{-1}$. In Experiment II, $[A]$ remains $0.1\text{ M}$, but $[B]$ is doubled to $0.2\text{ M}$. The rate becomes $2.40\times 10^{-2}{\text{ M min}}^{-1}$.

By dividing the rate of Experiment II by the rate of Experiment I, the constant terms ($k$ and $[A{]}^a$) cancel out:

Since $2^2=4$, we can confidently conclude that $b=2$. The reaction is second-order with respect to $B$.

Now, we apply the same logic to find $a$. We need two experiments where $[B]$ is constant but $[A]$ changes. Experiment I and Experiment III are perfect for this.

In Experiment III, $[A]$ is doubled to $0.2\text{ M}$, while $[B]$ remains $0.1\text{ M}$. The rate is $1.20\times 10^{-2}{\text{ M min}}^{-1}$.

Dividing the rate of Experiment III by the rate of Experiment I:

This simply means $a=1$. The reaction is first-order with respect to $A$.

With the orders discovered, our rate law is now fully formed:

Before we can find $X$ and $Y$, we need the value of the rate constant, $k$. We can use the data from any complete experiment to find it. Let's use Experiment I:

Now for the grand finale. We have the complete rate equation: $r=6[A][B{]}^2$. We can use this to find the missing concentrations in Experiments IV and V.

Finding X (Experiment IV): We are given $r=7.20\times 10^{-2}$ and $[B]=0.2$. We need to find $[A]$, which is $X$.

Finding Y (Experiment V): We are given $r=2.88\times 10^{-1}$ and $[A]=0.3$. We need to find $[B]$, which is $Y$.

Taking the square root, we get:

By systematically breaking down the experimental data, we deduced the rate law, calculated the rate constant, and finally solved for the missing concentrations. The values are $X=0.3$ and $Y=0.4$, which corresponds to option (c). This problem beautifully illustrates the logical progression required in chemical kinetics!