Animated Solution for Chemistry - Chemical Kinetics: The rate equation for the reaction $2A+B\to C$ is found to be rate $=k[A][B]$. The correct statement in relation to this reaction is that the

Imagine you are a chemical detective, and you've just been handed a case file. The file contains a chemical reaction: $2A+B\to C$. Along with it, you are given a crucial piece of evidence—the rate equation: $\text{Rate}=k[A][B]$.

Our mission is to interrogate four different statements and find out which one is telling the truth. To do this, we first need to understand the fundamental nature of this reaction. The rate law is our key. By looking at the exponents of the concentration terms in the rate law, we can determine the overall order of the reaction. The concentration of $A$ is raised to the power of $1$, and the concentration of $B$ is also raised to the power of $1$.

Therefore, the overall order is $1+1=2$. This is a second-order reaction. Now that we know its identity, let's put the options to the test.

Let's start with Option (a), which claims the unit of the rate constant $k$ must be $s^{-1}$. We know the general formula for the unit of $k$ for an $n$-th order reaction is $({\text{mol L}}^{-1}{)}^{1-n}{s}^{-1}$. Since our reaction is second-order ($n=2$), the unit becomes $({\text{mol L}}^{-1}{)}^{1-2}{s}^{-1}$, which simplifies to ${\text{L mol}}^{-1}{s}^{-1}$. The unit $s^{-1}$ belongs to a first-order reaction. So, Option (a) is false.

Next up is Option (b), stating that the half-life ($t_{1/2}$) is a constant. Is it? A constant half-life is the hallmark of a first-order reaction (like radioactive decay). However, for a second-order reaction, the half-life is inversely proportional to the initial concentration: $t_{1/2}\propto \frac{1}{[A{]}_0}$. If you change the starting amount, the half-life changes. Thus, Option (b) is also false.

Now let's examine Option (c). It says the rate of formation of $C$ is twice the rate of disappearance of $A$. Let's look at the stoichiometry of the reaction: $2A+B\to C$. For every $2$ moles of $A$ that disappear, only $1$ mole of $C$ is formed. Mathematically, we write this as $-\frac{1}{2}\frac{d[A]}{dt}=\frac{d[C]}{dt}$. This clearly shows that the rate of formation of $C$ is actually half the rate of disappearance of $A$. Option (c) is caught in a lie!

Finally, we arrive at Option (d), which states that the value of $k$ is independent of the initial concentrations of $A$ and $B$. Think about what the rate constant $k$ truly represents. It is a fundamental property of the reaction itself at a specific temperature.

According to the Arrhenius equation, $k=A{e}^{-E_a/RT}$, the rate constant depends only on the temperature ($T$) and the activation energy ($E_a$). It does not care whether you start with a bucket full of reactants or just a few drops. The initial concentrations dictate the initial rate, but the rate constant remains steadfast and unchanged.

Therefore, Option (d) is the absolute truth and the correct answer to our problem.