Question

For a first order reaction with rate constant 'k', the slope of the line obtained by plotting log ([R\(_0\)] [R]) vs time is

• A. (k 2.303)
• B. k \(\times\) 2.303
• C. (-k 2.303)
• D. (2.303 k)
• E. (-2.303 k)

Hint:

For a first-order reaction, a plot of log([R\(_0\)]/[R]) vs. time gives a straight line with a positive slope of k/2.303.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The question asks for the slope of a specific graphical representation of a first-order reaction. This requires using the integrated rate law for first-order reactions.
Step 2: Key Formula or Approach
The integrated rate law for a first-order reaction R → P can be expressed in terms of natural logarithm (ln):
\[ \ln\left(\frac{[\text{R}]_0}{[\text{R}]}\right) = kt \] where:
\( [\text{R}]_0 \) = initial concentration of the reactant
\( [\text{R}] \) = concentration of the reactant at time 't'
\( k \) = rate constant
Step 3: Detailed Explanation
The question specifies a plot of \( \log \left(\frac{[\text{R}]_0}{[\text{R}]}\right) \) vs. time (t). We need to convert the integrated rate law from natural logarithm (ln) to base-10 logarithm (log). The relationship between ln and log is:
\[ \ln(x) = 2.303 \log(x) \] Applying this to our rate law:
\[ 2.303 \log\left(\frac{[\text{R}]_0}{[\text{R}]}\right) = kt \] To get the equation for the plot, we need to arrange it in the form of a straight line, \(y = mx + c\), where \(y = \log\left(\frac{[\text{R}]_0}{[\text{R}]}\right)\) and \(x = t\).
Rearranging the equation:
\[ \log\left(\frac{[\text{R}]_0}{[\text{R}]}\right) = \left(\frac{k}{2.303}\right) t \] Comparing this to \(y = mx + c\):
\( y = \log\left(\frac{[\text{R}]_0}{[\text{R}]}\right) \)
\( x = t \)
\( m = \text{slope} = \frac{k}{2.303} \)
\( c = \text{y-intercept} = 0 \)
The slope of the line is therefore \( \frac{k}{2.303} \).
Step 4: Final Answer
The slope of the line obtained by plotting \( \log\left(\frac{[\text{R}]_0}{[\text{R}]}\right) \) vs. time is \( \frac{k}{2.303} \).