Question

For a first order reaction ($R \rightarrow P$), a plot of $\ln \frac{[R]_0}{[R]}$ on y-axis and time on x-axis gave a straight line passing through the origin. The slope of straight line is 'x'. The rate constant of this reaction is

• A. $2.303 x$
• B. $x^{-1}$
• C. $x$
• D. $2.303$

Hint:

Be careful with log bases. - Plot of $\ln(R_0/R)$ vs $t$: Slope = $k$. - Plot of $\log_{10}(R_0/R)$ vs $t$: Slope = $k/2.303$.

Answer 1

The Correct Option is C

Step 1: Integrated Rate Equation:
For a first-order reaction $R \rightarrow P$, the rate law is: \[ \ln[R] = \ln[R]_0 - kt \] Rearranging terms: \[ \ln[R]_0 - \ln[R] = kt \] \[ \ln \frac{[R]_0}{[R]} = kt \]
Step 2: Linear Plot Analysis:
This equation is of the form $y = mx$, where:

• $y = \ln \frac{[R]_0}{[R]}$
• $x = t$ (time)
• $m = k$ (slope)

The graph of $\ln \frac{[R]_0}{[R]}$ versus $t$ is a straight line passing through the origin with slope equal to the rate constant $k$.
Step 3: Conclusion:
The problem states the slope is 'x'. Therefore, Rate constant $k = x$.