Consider the following general reaction and data: \(2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C} \longrightarrow \mathrm{D}+3 \mathrm{E}\) $$ \begin{array}{ccccc} \text { Expt } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [C] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 6.0 \times 10^{-6} & 0.024 & 0.085 & 0.032 \\ 2 & 9.6 \times 10^{-5} & 0.096 & 0.085 & 0.032 \\ 3 & 1.5 \times 10^{-5} & 0.024 & 0.034 & 0.080 \\ 4 & 1.5 \times 10^{-6} & 0.012 & 0.170 & 0.032 \end{array} $$ (a) What is the reaction order with respect to each reactant? (b) Calculate the rate constant. (c) Write the rate law for this reaction. (d) Express the rate in terms of changes in concentration with time for each of the components.

Step by step solution

01

Identify the Rate Law Form

The rate law for the reaction has the general form: \[\text{Rate} = k[A]^m[B]^n[C]^p\] where \(k\) is the rate constant and \(m, n, p\) are the orders of the reaction with respect to reactants A, B, and C respectively. We need to determine the values of \(m\), \(n\), and \(p\).

02

Determine the Order with Respect to A

Compare the rates of experiments 1 and 2 where \([A]\) changes and \([B]\) and \([C]\) are constant. \[\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[0.096]^m[0.085]^n[0.032]^p}{k[0.024]^m[0.085]^n[0.032]^p} = \frac{9.6 \times 10^{-5}}{6.0 \times 10^{-6}}\] Simplify this equation: \[\frac{9.6 \times 10^{-5}}{6.0 \times 10^{-6}} = \frac{(0.096)^m}{(0.024)^m}\] \[16 = 4^m\] Taking the logarithm on both sides, \[m = 2\] Thus, \(m = 2\).

03

Determine the Order with Respect to B

Compare the rates of experiments 1 and 4 where \([B]\) changes and \([A]\) and \([C]\) are constant. \[\frac{\text{Rate}_4}{\text{Rate}_1} = \frac{k[0.012]^m[0.170]^n[0.032]^p}{k[0.024]^m[0.085]^n[0.032]^p} = \frac{1.5 \times 10^{-6}}{6.0 \times 10^{-6}}\] Simplify this equation: \[\frac{1.5 \times 10^{-6}}{6.0 \times 10^{-6}} = \frac{(0.170)^n}{(0.085)^n}\] \[\frac{1}{4} = 2^n\] Taking the logarithm on both sides, \[n = 1\] Thus, \(n = 1\).

04

Determine the Order with Respect to C

Compare the rates of experiments 1 and 3 where \([C]\) changes and \([A]\) and \([B]\) are constant. \[\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[0.024]^m[0.034]^n[0.080]^p}{k[0.024]^m[0.085]^n[0.032]^p} = \frac{1.5 \times 10^{-5}}{6.0 \times 10^{-6}}\] Simplify this equation: \[\frac{1.5 \times 10^{-5}}{6.0 \times 10^{-6}} = \frac{(0.080)^p}{(0.032)^p}\] \[2.5 = 2.5^p\] Thus, \(p = 1\).

05

Calculate the Rate Constant (k)

Use the rate law and data from any experiment to calculate \(k\). Here, use experiment 1: \[\text{Rate}_1 = k[0.024]^2[0.085]^1[0.032]^1\] \[6.0 \times 10^{-6} = k(0.024)^2(0.085)(0.032)\] \[6.0 \times 10^{-6} = k(4.9152 \times 10^{-6})\] \[k = \frac{6.0 \times 10^{-6}}{4.9152 \times 10^{-6}}\] \[k \thickapprox 1.22 \text{ L}^2/(\text{mol}^2 \text{s})\]

06

Write the Rate Law

Based on the determined orders of reaction: \[\text{Rate} = 1.22 \text{ L}^2/(\text{mol}^2 \text{s}) [A]^2[B][C]\]

07

Express the Rate in Terms of Changes in Concentration

The rate law for the reaction can be expressed for each component: \[-\frac{1}{2}\frac{d[\text{A}]}{dt} = -\frac{1}{2}\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = \frac{d[\text{D}]}{dt} = \frac{1}{3}\frac{d[\text{E}]}{dt} = 1.22 [A]^2[B][C]\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

reaction order

The concept of reaction order is crucial in chemical kinetics. It tells us how the rate of reaction depends on the concentration of each reactant. In the given exercise, we use experimental data to find the order of reaction with respect to each reactant, denoted by \(m\), \(n\), and \(p\). By comparing the initial rates and concentrations in different experiments and applying logarithmic transformations, we determined the reaction orders.
The reaction order gives insights into:

• How doubling the concentration of a reactant changes the reaction rate.
• Predicting how the reaction rate will respond to changes in concentrations.

Understanding the reaction order helps us better control and optimize chemical processes.

rate constant

The rate constant, often denoted as \(k\), is a unique value that links the concentration of reactants with the rate of the reaction in the rate law expression. In the context of our exercise, we computed the rate constant by substituting the known concentrations and rate into the rate law expression.
The importance of the rate constant:

• Determines the speed of the reaction.
• Provides insights into the reaction mechanism.
• Varies with temperature, thus indicating the thermal sensitivity of the reaction.

The units of the rate constant depend on the overall reaction order and must be consistent to maintain the correct dimensions in the rate law.

chemical kinetics

Chemical kinetics is the study of reaction rates and the steps involved in a chemical reaction. It provides a detailed understanding of how different factors affect the rate, including concentration, temperature, and catalysts.
Key components of chemical kinetics include:

• Reaction mechanisms: The step-by-step pathways that lead to the products.
• Reaction rate: The speed at which reactants turn into products, normally measured in mol/L/s.
• Influence of conditions: How different conditions alter the rate and the mechanism.

By understanding chemical kinetics, chemists can predict and control the speed of reactions, essential for industrial processes, pharmaceuticals, and numerous other applications.

rate law expression

The rate law expression is a mathematical representation that relates the rate of reaction to the concentration of reactants. It is written in the form \[\text{Rate} = k[A]^m[B]^n[C]^p\],
where:

• \(k\) is the rate constant.
• \([A], [B], [C]\) are the concentrations of reactants.
• \(m, n, p\) represent the reaction orders.

In our exercise, we derived the specific rate law \[\text{Rate} = 1.22 \text{ L}^2/(\text{mol}^2 \text{s}) [A]^2[B][C]\],
The rate law expression:

• Helps predict how changes in reactant concentrations affect the reaction rate.
• Is determined experimentally.
• Serves as a foundation for understanding more complex reaction dynamics.

Understanding and utilizing the rate law expression allows for precise control over chemical reactions, maximizing efficiency and output.