Which one of the following graph is not applicable for a 1st order reaction (R \(\rightarrow\) P)?
Show Hint
Always remember the characteristic plots for different reaction orders.
For 1st order: \(\ln[R]\) vs \(t\) is linear (negative slope).
For 0th order: \([R]\) vs \(t\) is linear (negative slope).
Step 1: Understanding the Question:
The question asks to identify the graph that does NOT represent the kinetic behavior of a first-order reaction. Step 2: Key Formula or Approach:
For a first-order reaction $R \rightarrow P$, the integrated rate law is:
\[ \ln[R]_t - \ln[R]_0 = -kt \]
This can be rewritten in various forms for graphical analysis. Step 3: Detailed Explanation:
Let's analyze each type of graph for a first-order reaction:
1. Graph of \(\ln[R]\) vs \(t\):
The integrated rate law is \(\ln[R]_t = -kt + \ln[R]_0\).
This is in the form \(y = mx + c\), where \(y = \ln[R]_t\), \(m = -k\) (a negative slope), and \(c = \ln[R]_0\).
Therefore, a plot of \(\ln[R]\) vs \(t\) is a straight line with a negative slope. This matches Graph (2) and is applicable.
2. Graph of \(\log_{10}[R]\) vs \(t\):
Dividing the integrated rate law by 2.303 gives \(\log_{10}[R]_t = -\frac{k}{2.303}t + \log_{10}[R]_0\).
This is also a straight line with a negative slope. This matches Graph (3) and is applicable.
3. Graph of \(\log_{10}\left(\frac{[R]_0}{[R]}\right)\) vs \(t\):
From \(\ln\left(\frac{[R]_0}{[R]_t}\right) = kt\), we get \(\log_{10}\left(\frac{[R]_0}{[R]_t}\right) = \frac{k}{2.303}t\).
This is a straight line passing through the origin with a positive slope. This matches Graph (4) and is applicable.
4. Graph of \([R]\) vs \(t\):
The concentration of reactant as a function of time is given by \([R]_t = [R]_0 e^{-kt}\).
This represents an exponential decay, meaning the concentration of reactant decreases over time, but not linearly. It is never a constant value unless $k=0$ (no reaction) or $t=0$.
Graph (1) shows \([R]\) as a horizontal line, implying constant concentration over time. This is incorrect for a reaction where reactant is consumed. Step 4: Final Answer:
Graph (1) is not applicable for a 1st order reaction.
