Question

For first order reaction the slope of the graph of log\(_{10}\)[A] vs. time is equal to

• A. k
• B. \(-\frac{k}{2.303}\)
• C. \(-k\)
• D. \(\frac{k}{2.303}\)

Hint:

For first-order reactions, the slope of the graph of \(\log_{10}[A]\) vs. time is \(-\frac{k}{2.303}\). This is a key characteristic of first-order kinetics.

Answer 1

The Correct Option is B

Step 1: Understanding the first order reaction.
For a first-order reaction, the integrated rate law is given by: \[ \ln[A] = -kt + \ln[A_0]. \] By converting the natural logarithm to base 10, we get the equation: \[ \log_{10}[A] = -\frac{k}{2.303}t + \log_{10}[A_0]. \] This equation represents a straight line with a slope of \(-\frac{k}{2.303}\).
Step 2: Conclusion.
Thus, the slope of the graph of \(\log_{10}[A]\) vs. time is equal to \(-\frac{k}{2.303}\).