Question

For a reaction R \(\longrightarrow\) P with a rate constant of 3 \(\times 10^{-3}\) mol L\(^{-1}\) s\(^{-1}\), which one of the following plots is correct?

• A. A
• B. B
• C. C
• D. D

Hint:

Always check the units of the rate constant \(k\) first:
- \(\text{mol L}^{-1}\text{ s}^{-1}\) \(\to\) Zero order \(\to\) \([\text{R}]\) vs \(t\) is linear.
- \(\text{s}^{-1}\) \(\to\) First order \(\to\) \(\ln[\text{R}]\) vs \(t\) is linear.
This basic diagnostic step avoids any confusion.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify the correct concentration-time plot for a given chemical reaction \(\text{R} \rightarrow \text{P}\) with a rate constant \(k = 3 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}\).

Step 2: Key Formula or Approach:
We must first determine the order of the reaction by looking at the units of the rate constant \(k\):
1. The unit of a rate constant is given by:
\[ \text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1} \]
where \(n\) is the order of the reaction.
2. For the given rate constant, the unit is \(\text{mol L}^{-1}\text{ s}^{-1}\), which corresponds to \(n = 0\) (a zero-order reaction).

Step 3: Detailed Explanation:
Let's analyze the integrated rate equation for a zero-order reaction:
- The rate of reaction is independent of the reactant concentration:
\[ -\frac{d[\text{R}]}{dt} = k \]
- Integrating both sides with respect to time \(t\):
\[ [\text{R}] = [\text{R}]_0 - kt \]
where \([\text{R}]_0\) is the initial concentration of reactant R at \(t = 0\), and \([\text{R}]\) is the concentration at time \(t\).
- This equation represents a straight line of the form \(y = mx + c\):
- \(y\)-axis: \([\text{R}]\)
- \(x\)-axis: \(t\)
- Slope (\(m\)): \(-k\) (negative constant slope)
- Intercept (\(c\)): \([\text{R}]_0\)
- Therefore, a plot of \([\text{R}]\) versus \(t\) must be a straight line with a downward slope.
- Comparing with the given choices:
- Plot (a) shows a straight line with a negative slope for \([\text{R}]\) vs \(t\). This matches the zero-order kinetics perfectly.
- Plot (b), (c), and (d) represent first-order reaction graphs (exponential decay, \(\ln[\text{R}]\) vs \(t\) linear, and \(\log([\text{R}]_0/[\text{R}])\) vs \(t\) linear, respectively).

Step 4: Final Answer:
Plot (a) represents the correct graph for this zero-order reaction, which corresponds to option (A).