Which of the following curve represents the first order reaction?
Show Hint
General relation for half-life: \( t_{1/2} \propto [R]_0^{(1-n)} \), where \( n \) is the order of reaction. For \( n=1 \), \( t_{1/2} \propto [R]_0^0 \).
Step 1: Understanding the Concept:
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one reactant. This means that the rate changes as the concentration changes. Furthermore, the half-life (\( t_{1/2} \)) of a first-order reaction remains constant and is independent of the initial concentration of the reactant. Step 2: Detailed Explanation:
The half-life of a first-order reaction is given by the equation \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant. This equation shows that the half-life is dependent only on the rate constant \( k \) and not on the initial concentration of the reactant \( [R]_0 \). As a result, the half-life remains constant regardless of how much reactant is present initially.
If you were to plot \( t_{1/2} \) (half-life) against the initial concentration \( [R]_0 \), the graph would be a straight, horizontal line, indicating that \( t_{1/2} \) does not change as \( [R]_0 \) changes.
Now, let's examine the options:
- Option (A) represents a zero-order reaction, where the half-life (\( t_{1/2} \)) is directly proportional to the initial concentration \( [R]_0 \), meaning as concentration increases, half-life increases as well. This is not characteristic of a first-order reaction.
- Option (C) is also related to a zero-order reaction, where the rate of reaction is independent of the concentration of the reactant.
- In a first-order reaction, the plot of rate versus concentration should yield a straight line that passes through the origin. Step 3: Final Answer:
Graph (B) accurately represents the behavior of a first-order reaction because, in such reactions, the half-life remains constant regardless of the initial concentration of the reactant.
