For a certain reaction \(R\rightarrow\) Product, the plot of concentration \([R]\) versus time has a negative slope as shown. The order of reaction is:
Show Hint
For zero order reaction, \([R]\) versus time gives a straight line with slope \(-k\).
Step 1: Understanding the Question:
In chemical kinetics, the relationship between concentration and time is unique for every reaction order. By observing the graphical representation of how the reactant concentration $[R]$ changes as time progresses, we can deduce the order of the reaction. The question provides a plot where $[R]$ is on the y-axis and time $t$ is on the x-axis, showing a straight line with a negative slope. Step 2: Key Formula or Approach:
We evaluate the integrated rate laws for various orders:
Zero Order: $[R]_t = -kt + [R]_0$ (This is a linear equation of the form $y = mx + c$).
First Order: $\ln[R]_t = -kt + \ln[R]_0$ (Linear only if $\ln[R]$ is on the y-axis).
Second Order: $\frac{1}{[R]_t} = kt + \frac{1}{[R]_0}$ (Linear only if $1/[R]$ is on the y-axis).
Step 3: Detailed Explanation:
Analysis of the Graph: The provided image shows a plot of $[R]$ vs $t$. It is a perfectly straight line. This implies that the concentration decreases at a constant rate, regardless of how much reactant is present.
Mathematical Correlation: The equation for a straight line is $y = mx + c$. In the zero-order rate law ($[R] = -kt + [R]_0$), $[R]$ acts as $y$, $t$ acts as $x$, $-k$ acts as the slope $m$, and $[R]_0$ is the y-intercept.
Conclusion for Zero Order: Since the graph of concentration versus time is linear, the reaction rate ($Rate = -d[R]/dt = k$) is independent of the reactant concentration. This is the definition of a zero-order reaction.
Comparison with First Order: If the reaction were first order, the $[R]$ vs $t$ plot would be an exponential decay curve. Only a plot of $\ln[R]$ vs $t$ would yield a straight line for first order.
Step 4: Final Answer:
Because the plot of $[R]$ versus time is a straight line, the order of the reaction is $0$.
