Problem 8 - Chemistry The Central Science

Question

Which of the following linear plots do you expect for a reaction \(A \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4\(]\)

Step-by-Step Solution

Answer

For (a) zero-order, (b) first-order, and (c) second-order kinetics, the linear plots are: a) \( [A] \) vs. \( t \), with a straight line and a negative slope equal to \( -k \) b) \( ln[A] \) vs. \( t \), with a straight line and a negative slope equal to \( -k \) c) \( \frac{1}{[A]} \) vs. \( t \), with a straight line and a positive slope equal to \( k \)

Step 1: Review the mathematical expressions for each type of kinetics

First, let's review the mathematical expressions for each type of kinetics. 1. Zero-order kinetics: The reaction rate is constant. The rate equation is: \[r = k[A]^0 = k\] 2. First-order kinetics: The reaction rate is proportional to the concentration of the single reactant A. The rate equation is: \[r = k[A]\] 3. Second-order kinetics: The reaction rate is proportional to the square of the concentration of the single reactant A. The rate equation is: \[r = k[A]^2\] Now that we have the rate equations for each type of kinetics, we can proceed to determine their linear plots.

Step 2: Identify the linear plots for each type of kinetics

We will now find the linear plots for each type of kinetics by integrating the corresponding rate equations and looking for linear relationships. 1. Zero-order kinetics: For a zero-order reaction, after integrating the rate equation, we get: \[[A] = -kt + [A]_0\] This equation represents a linear plot; the concentration [A] decreases linearly with time t. Thus, a graph of [A] vs. t gives a straight line with a negative slope equal to -k. 2. First-order kinetics: For a first-order reaction, after integrating the rate equation, we get: \[ln[A] = -kt + ln[A]_0\] In this case, a linear plot can be achieved by plotting the natural logarithm of the concentration ln[A] vs. time t. This plot gives a straight line with a negative slope equal to -k. 3. Second-order kinetics: For a second-order reaction, after integrating the rate equation, we get: \[\frac{1}{[A]} = kt + \frac{1}{[A]_0}\] Here, a linear plot can be obtained by plotting the reciprocal of the concentration 1/[A] vs. time t. This plot results in a straight line with a positive slope equal to k. In summary, for reactions with (a) zero-order, (b) first-order, and (c) second-order kinetics, the expected linear plots are: a) [A] vs. t, with a straight line and a negative slope equal to -k b) ln[A] vs. t, with a straight line and a negative slope equal to -k c) 1/[A] vs. t, with a straight line and a positive slope equal to k

Key Concepts

Zero-Order Kinetics

In zero-order kinetics, the reaction rate is independent of the concentration of the reactant. This means that the rate remains constant throughout the reaction. The rate equation for zero-order kinetics can be expressed as:\[r = k[A]^0 = k\]Here, \(k\) is the rate constant and [A] is the concentration of the reactant. Since the concentration term has an exponent of zero, it does not affect the rate, which stays constant.The integrated rate equation for zero-order kinetics is:\[[A] = -kt + [A]_0\]This equation leads to a linear graph when plotting [A] (concentration) versus time \(t\). The line will have a negative slope equal to \(-k\), starting at the initial concentration \([A]_0\).**Characteristics:**

• The plot of [A] vs. \(t\) is a straight line with a negative slope.
• The reaction rate does not change as the reaction progresses.
• Reactions often become zero-order when the reactant concentration is high or constant, such as in saturated solutions or reactions catalyzed by enzymes where the catalyst is saturated.

First-Order Kinetics

First-order kinetics involve a reaction whose rate depends linearly on the concentration of one reactant. The rate equation for a first-order reaction is:\[r = k[A]\]In this case, the reaction rate is proportional to the concentration of reactant \(A\). This means that as the concentration of \(A\) decreases over time, so does the reaction rate.The integrated rate law for first-order kinetics is represented as:\[\ln[A] = -kt + \ln[A]_0\]This equation suggests a linear relationship if you plot the natural logarithm of the concentration \(\ln[A]\) over time \(t\). The straight line obtained will have a negative slope equivalent to the rate constant \(-k\).**Key Points:**

• Plotting \(\ln[A]\) vs. \(t\) yields a straight line with a negative slope.
• Common in radioactive decay and some simple chemical reactions.
• The half-life of a first-order reaction is constant and does not change as the concentration decreases.

Second-Order Kinetics

For reactions following second-order kinetics, the rate depends on the square of the concentration of one reactant. The general rate equation is:\[r = k[A]^2\]This implies the reaction rate increases quadratically with the concentration of \(A\). As the concentration increases, the rate increases much more rapidly than in first-order kinetics.The integrated rate law for second-order reactions is:\[\frac{1}{[A]} = kt + \frac{1}{[A]_0}\]To depict a linear graphical relationship, plot the reciprocal of the concentration \(\frac{1}{[A]}\) against time \(t\). The resulting line will have a positive slope equal to \(k\).**Understanding Second-Order Kinetics:**

• The plot of \(\frac{1}{[A]}\) vs. \(t\) is a straight line with a positive slope.
• Often occurs in bimolecular reactions where two molecules collide to form products.
• Unlike first-order, half-life is not constant and decreases as the concentration decreases.