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?
Show 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.
To determine the correct plot for the given reaction \( R \longrightarrow P \) with the rate constant \( k = 3 \times 10^{-3} \) mol L\(^{-1}\) s\(^{-1}\), we first need to identify the order of the reaction.
Since the rate constant has units of mol L\(^{-1}\) s\(^{-1}\), this indicates a zero-order reaction. For a zero-order reaction, the rate is constant, and the concentration of reactant \( [R] \) decreases linearly with time.
The integrated rate law for a zero-order reaction is:
\([R] = [R]_0 - kt\)
where:
\([R]\) is the concentration of the reactant at time t,
\([R]_0\) is the initial concentration of the reactant,
\(k\) is the rate constant,
\(t\) is the time.
The plot of concentration versus time for a zero-order reaction is a straight line with a negative slope, which matches option A.
