Problem 94 - Chemistry The Central Science In Si Units

Question

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: \begin{tabular}{llll} \hline Experiment & {\([\mathrm{X}]_{0}(M)\)} & {\([\mathrm{Z}]_{0}(\mathrm{M})\)} & Rate \((\mathrm{M} / \mathrm{s})\) \\ \hline 1 & 0.25 & 0.25 & \(4.0 \times 10^{1}\) \\ 2 & 0.50 & 0.50 & \(3.2 \times 10^{2}\) \\ 3 & 0.50 & 0.75 & \(7.2 \times 10^{2}\) \\ \hline \end{tabular} (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{X}\) is \(0.75 \mathrm{M}\) and that of \(\mathrm{Z}\) is \(1.25 \mathrm{M} ?\)

Step-by-Step Solution

Answer

The rate law is \( \text{Rate} = k[ ext{X}]^3[ ext{Z}]^2 \). The rate constant is \( 4.10 \times 10^4 \mathrm{M}^{-4} \mathrm{s}^{-1} \). Rate at given concentrations is \( 2.70 \times 10^4 \, \mathrm{M/s} \).

Step 1: Identify the Rate Law Form

The general rate law for a reaction can be expressed as \[ ext{Rate} = k[ ext{X}]^m[ ext{Z}]^n \] where \(k\) is the rate constant, \(m\) is the order of the reaction with respect to \( ext{X}\), and \(n\) is the order with respect to \( ext{Z}\). Our task is to determine \(m\), \(n\), and \(k\).

Step 2: Determine Reaction Orders from Experiment 1 and 2

Examine experiments 1 and 2 to deduce the order of reaction with respect to \([ ext{X}]\):\( [ ext{Z}]_{0} \) remains constant, while \( [ ext{X}]_{0} \) is doubled. The rate changes by a factor of \(\frac{3.2 \times 10^2}{4.0 \times 10^1} = 8\).To find the order of \( ext{X}\), set up the equation: \[ rac{k(0.50)^m(0.25)^n}{k(0.25)^m(0.25)^n} = 8 \]Simplify to find \[ \left(\frac{0.50}{0.25}\right)^m = 8 \] \[ 2^m = 8 \] So, \( m = 3 \).

Step 3: Determine Reaction Order with Respect to Z

Use experiments 2 and 3 to find the order of reaction with respect to \([ ext{Z}]\):\( [ ext{X}]_{0} \) remains constant, while \( [ ext{Z}]_{0} \) is increased by \(1.5\) times. The rate increases by a factor of \(\frac{7.2 \times 10^2}{3.2 \times 10^2} = 2.25\).To find \( n \): \[ \frac{k(0.50)^3(0.75)^n}{k(0.50)^3(0.50)^n} = 2.25 \] \[ \left(\frac{0.75}{0.50}\right)^n = 2.25 \] \[ 1.5^n = 2.25 \] So, \( n = 2 \).

Step 4: Write the Rate Law Equation

Now that we've determined the reaction orders, the rate law is: \[ ext{Rate} = k[ ext{X}]^3[ ext{Z}]^2 \]

Step 5: Calculate the Rate Constant k

Substitute the known values from experiment 1 into the rate law to solve for \( k \):\[ 4.0 \times 10^1 = k(0.25)^3(0.25)^2 \]Simplify and solve for \( k \):\[ 4.0 \times 10^1 = k(0.25)^5 = k(0.0009765625) \]\[ k = \frac{4.0 \times 10^1}{0.0009765625} = 40960 \]Thus, \( k = 4.10 \times 10^4 \, \mathrm{M}^{-4} \mathrm{s}^{-1} \).

Step 6: Calculate Reaction Rate for New Concentrations

Using \([ ext{X}] = 0.75\, \mathrm{M}\) and \([ ext{Z}] = 1.25\, \mathrm{M}\) from the given condition:\[ ext{Rate} = 4.10 \times 10^4 (0.75)^3(1.25)^2 \]Calculate step-by-step:\[ (0.75)^3 = 0.421875 \]\[ (1.25)^2 = 1.5625 \]\[ ext{Rate} = 4.10 \times 10^4 \times 0.421875 \times 1.5625 = 26953.125 \]Thus, the reaction rate is \( 2.70 \times 10^4 \, \mathrm{M/s} \).

Key Concepts

Reaction Order

Understanding the order of a reaction is crucial when analyzing the rate law. The reaction order indicates how the rate of a chemical reaction depends on the concentration of the reactants. In our example, we found two reaction orders: one with respect to [X] and the other with respect to [Z].

• Order with respect to X: By doubling the concentration of X, we observed the rate increased by a factor of 8. This led us to deduce that the reaction is of the third order concerning reactant X, since \(2^3 = 8\).
• Order with respect to Z: For Z, increasing its concentration by 1.5 times and observing the rate increased by 2.25 times, we determined Z's reaction order as second order, because \(1.5^2 = 2.25\).

Knowing these orders helps form the rate law, which describes how the concentration of reactants influences the reaction rate.

Rate Constant

The rate constant, denoted as \( k \), is vital for quantifying the speed of the reaction under specific conditions. This constant is determined by plugging known values into the rate law equation.
Steps to Find the Rate Constant:

• Use experimental data to substitute values of reactants and initial rate.
• In our case, using values from experiment 1: \( k \) was isolated by the equation: \[4.0 \times 10^{1} = k(0.25)^5\].
• Simplifying gave us: \[k = \frac{4.0 \times 10^{1}}{0.0009765625}\]
• Calculating further, we determined \( k = 4.10 \times 10^4 \, \mathrm{M^{-4} \mathrm{s^{-1}}} \).

The rate constant could vary with temperature or other environmental factors.

Kinetic Data Analysis

Kinetic data analysis involves obtaining reaction specifics from experimental data to predict reaction behavior. It allows scientists to calculate reaction rates under various conditions.
Key Steps:

• Collect Experimental Data: Gather initial concentrations and rates from different experiments.
• Determine Reaction Orders: Compare how changes in concentrations affect the rate to find reaction orders.
• Develop Rate Law: Use the reaction orders and known data to establish the rate law formula.
• Calculate Rate Constant: Plug values into the rate law to solve for \( k \).
• Predict Reaction Rate: Use the rate law for new concentration values to project how the system behaves.

Understanding these steps aids in predicting and controlling chemical reactions effectively.