Animated Solution for Chemistry - Chemical Kinetics: For the following reaction $2X+Y\stackrel{k}{\to }P$ the rate of reaction is $\frac{d[P]}{dt}=k[X]$. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: $\ln 2=0.693$)

Chemical kinetics is a beautiful dance of molecules, but it often hides subtle traps for the unwary student. This problem from JEE Advanced is a classic example of how a simple rate law can become a minefield if you ignore the stoichiometry of the reaction. Let's break down the journey step-by-step and uncover the elegance behind the math.

Imagine you are standing in a laboratory, holding a one-liter flask. You mix in $2$ moles of reactant $X$ and $1$ mole of reactant $Y$. Because the volume is exactly $1.0\text{ L}$, our initial concentrations are beautifully simple:

The problem hands us the rate of the reaction on a silver platter: $\text{Rate}=k[X]$. This tells us immediately that the reaction is first-order with respect to $X$ and zero-order with respect to $Y$. But here is where the trap lies. The rate of the reaction is not necessarily the rate of disappearance of a specific reactant.

To avoid silly mistakes, we must always anchor ourselves to the stoichiometric coefficients. The balanced equation is $2X+Y\stackrel{k}{\to }P$. According to the fundamental principles of chemical kinetics, the overall rate of reaction is linked to the individual rates of disappearance and appearance as follows:

Since we are given that $\text{Rate}=k[X]$, we can equate the two expressions for the rate of disappearance of $X$:

This is a crucial revelation! The effective rate constant for the disappearance of $X$ is not $k$, but $2k$.

The problem states that at $t=50\text{ s}$, there is $0.5$ mole of $Y$ left. Let's trace the stoichiometry. If $Y$ started at $1.0\text{ M}$ and is now at $0.5\text{ M}$, exactly $0.5\text{ M}$ of $Y$ has reacted.

Because $2$ moles of $X$ react for every $1$ mole of $Y$, the amount of $X$ that reacted must be double that of $Y$:

Subtracting this from the initial concentration of $X$, we find the concentration of $X$ at $50\text{ s}$:

Look closely at what just happened. The concentration of $X$ dropped from $2.0\text{ M}$ to exactly $1.0\text{ M}$ in $50$ seconds. It halved! This intuitive observation directly gives us the half-life of $X$:

This confirms that Option (B) is absolutely correct.

Now that we know the half-life of $X$, we can calculate the rate constant $k$. Remember our master equation? The effective rate constant for $X$ is $2k$. For a first-order reaction, the half-life is given by $\frac{\ln 2}{\text{effective rate constant}}$.

Comparing this to Option (A), which claims $k=13.86\times 10^{-4}{\text{ s}}^{-1}$, we can confidently say Option (A) is incorrect.

Let's verify the remaining options by calculating the specific rates at given times.

For Option (C), we need the rate of disappearance of $X$ at $50\text{ s}$. We already established that $[X{]}_{50}=1.0\text{ M}$.

This perfectly matches Option (C), making it correct.

Finally, for Option (D), we need the rate of disappearance of $Y$ at $100\text{ s}$. Notice that $100\text{ s}$ is exactly two half-lives ($2\times 50\text{ s}$). In each half-life, the concentration of $X$ halves. So, it goes from $2.0\text{ M}\to 1.0\text{ M}\to 0.5\text{ M}$.

At $t=100\text{ s}$, $[X]=0.5\text{ M}$. The rate of disappearance of $Y$ is simply equal to the overall rate of reaction:

This matches Option (D), confirming it is also correct.

By carefully navigating the relationship between stoichiometry and the rate law, we've successfully decoded the entire problem. The correct options are indeed (B), (C), and (D).