Question:medium

The slope of a graph $\log[A]_t$ versus 't' for first order reaction is $-2.5\times10^{-3}s^{-1}$. Find rate constant of the reaction?

Show Hint

Always multiply slope by 2.303 if the graph uses $\log_{10}$.
Updated On: Jun 19, 2026
  • $1.263\times10^{-3}s^{-1}$
  • $3.471\times10^{-3}s^{-1}$
  • $5.757\times10^{-3}s^{-1}$
  • $8.125\times10^{-3}s^{-1}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We need to find the rate constant ($k$) of a first-order reaction from the slope of the integrated rate law graph.

Step 2: Key Formula or Approach:

For a first-order reaction: \[ k = \frac{2.303}{t} \log \frac{[\text{A}]_0}{[\text{A}]_t} \]
Rearranging into $y = mx + c$ form: \[ \log[\text{A}]_t = -\frac{k}{2.303}t + \log[\text{A}]_0 \]
The slope ($m$) of $\log[\text{A}]$ vs $t$ is $-\frac{k}{2.303}$.

Step 3: Detailed Explanation:

Given: Slope $= -2.5 \times 10^{-3} \text{ s}^{-1}$
Calculation: \[ -\frac{k}{2.303} = -2.5 \times 10^{-3} \]
\[ k = 2.303 \times 2.5 \times 10^{-3} \]
\[ k = 5.7575 \times 10^{-3} \text{ s}^{-1} \]

Step 4: Final Answer:

The rate constant is $5.757 \times 10^{-3} \text{ s}^{-1}$.
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