Question

R\(\rightarrow\)P is a first order reaction. For this reaction a graph of ln[R] (on y-axis) and time (on x-axis) gave a straight line with negative slope. The intercept on y-axis is equal to (k=rate constant)

• A. ln[R]\(_0\)
• B. [R]\(_0\)
• C. k \(\times\) 2.303
• D. \( \frac{k}{2.303} \)

Hint:

Memorize the graphical representations of the integrated rate laws: - \textbf{Zero Order:} A plot of \([R]\) vs. time is linear with slope \(-k\). - \textbf{First Order:} A plot of \(\ln[R]\) vs. time is linear with slope \(-k\). - \textbf{Second Order:} A plot of \(1/[R]\) vs. time is linear with slope \(+k\).

Answer 1

The Correct Option is A

For a first-order reaction R \(\rightarrow\) P, the integrated rate law is given by:
\( \ln[\text{R}]_t = -kt + \ln[\text{R}]_0 \).
Here:
- \([\text{R}]_t\) is the concentration of the reactant R at time t.
- \([\text{R}]_0\) is the initial concentration of the reactant R at time t=0.
- k is the first-order rate constant.
This equation is in the form of a straight line, \( y = mx + c \).
If we plot a graph with \(y = \ln[\text{R}]_t\) on the y-axis and \(x = t\) on the x-axis, we get:
\( \underbrace{\ln[\text{R}]_t}_{y} = \underbrace{(-k)}_{m} \underbrace{t}_{x} + \underbrace{\ln[\text{R}]_0}_{c} \).
From this comparison, we can see that:
- The slope of the line (m) is \(-k\). This matches the description of a straight line with a negative slope.
- The y-intercept (c), which is the value of y when x=0, is \( \ln[\text{R}]_0 \).
Therefore, the intercept on the y-axis is equal to \( \ln[\text{R}]_0 \).