Question

Which one of the following graph is not applicable for a 1st order reaction (R \(\rightarrow\) P)?

• A. A graph of [R] vs t with a downward sloping curve.
• B. A graph of ln[R] vs t with a downward sloping straight line.
• C. A graph of log\(_{10}\)[R] vs t with an upward sloping straight line.
• D. A graph of log\(_{10}\)([R]\(_0\)/[R]) vs t with an upward sloping straight line through the origin.

Hint:

Memorize the linear plots for different reaction orders:

• \textbf{Zero Order:} [R] vs. t is a straight line with slope = -k.
• \textbf{First Order:} ln[R] vs. t is a straight line with slope = -k.
• \textbf{Second Order:} 1/[R] vs. t is a straight line with slope = k.

Knowing these plots is the fastest way to solve such graphical questions.

Answer 1

The Correct Option is C

Step 1: Key Formula - The Integrated Rate Law for a First-Order Reaction.
The integrated rate law for a first-order reaction can be expressed in several forms:

• \( [R] = [R]_0 e^{-kt} \)
• \( \ln[R] = \ln[R]_0 - kt \)
• \( \log_{10}[R] = \log_{10}[R]_0 - \frac{k}{2.303}t \)
• \( k = \frac{2.303}{t} \log_{10}\left(\frac{[R]_0}{[R]}\right) \implies \log_{10}\left(\frac{[R]_0}{[R]}\right) = \frac{k}{2.303}t \)

where [R] is the concentration of reactant at time t, [R]\(_0\) is the initial concentration, and k is the rate constant.
Step 2: Analyze each graph based on the integrated rate laws.
(A) [R] vs t: From equation (1), \( [R] = [R]_0 e^{-kt} \), the concentration of the reactant decreases exponentially with time. This corresponds to a downward sloping curve. This graph is applicable. (The image shows an upward arrow from [R], which is confusing, but the concept of a curve is correct).
(B) ln[R] vs t: From equation (2), \( \ln[R] = -kt + \ln[R]_0 \). This is in the form of a linear equation y = mx + c, where y = ln[R], x = t, the slope m = -k, and the y-intercept c = ln[R]\(_0\). Since the slope (-k) is negative, this is a downward sloping straight line. This graph is applicable.
(C) log\(_{10}\)[R] vs t: From equation (3), \( \log_{10}[R] = -\left(\frac{k}{2.303}\right)t + \log_{10}[R]_0 \). This is also in the form y = mx + c, where y = log\([R]\), x = t, and the slope m = \(-\frac{k}{2.303}\). The slope is negative. The graph should be a downward sloping straight line. The option shows an upward sloping line. This graph is not applicable.
(D) log\(_{10}\)([R]\(_0\)/[R]) vs t: From equation (4), \( \log_{10}\left(\frac{[R]_0}{[R]}\right) = \left(\frac{k}{2.303}\right)t \). This is in the form y = mx, where y = log([R]\(_0\)/[R]), x = t, and the slope m = \(\frac{k}{2.303}\). Since the slope is positive, this is an upward sloping straight line passing through the origin. This graph is applicable.
Step 3: Final Answer.
The graph of log\([R]\) vs t should be a straight line with a negative slope. The graph shown in option (C) has a positive slope and is therefore not applicable to a first-order reaction.