Question

For a first-order reaction with rate constant k, the slope of the plot of log(reactant concentration) against time is:

• A. k/2.303
• B. k
• C. −k/2.303
• D. -k

Hint:

For first-order reactions, remember that the slope of the log(concentration) vs. time plot is always \(\frac{-k}{2.303}\).

Answer 1

The Correct Option is C

For a first-order reaction, the integrated rate law is:

\( \ln[A] = -kt + \ln[A]_0 \)

where [A] is the concentration at time t, k is the rate constant, and \([A]_0\) is the initial concentration. Taking the logarithm to base 10:

\( \log[A] = \frac{-k}{2.303}t + \log[A]_0 \)

This equation is of the form \(y = mx + c\), where:

- \(y = \log[A]\),
- \(x = t\),
- \(m = \frac{-k}{2.303}\),
- \(c = \log[A]_0\).

Thus, the slope of the plot of log(reactant concentration) against time is \(\frac{-k}{2.303}\).