Problem 21 - Chemistry: Principles and Reactions

Question

The reaction $$\mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{NOBr}(g)$$ is second-order in nitrogen oxide and first-order in bromine. The rate of the reaction is \(1.6 \times 10^{-8} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\) when the nitrogen oxide concentration is \(0.020 \mathrm{M}\) and the bromine concentration is \(0.030 \mathrm{M}\). (a) What is the value of \(k\) ? (b) At what concentration of bromine is the rate \(3.5 \times 10^{-7} \mathrm{~mol} / \mathrm{L} \cdot \min\) and \([\mathrm{NO}]=0.043 \mathrm{M} ?\) (c) At what concentration of nitrogen oxide is the rate \(2.0 \times\) \(10^{-6} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\) and the bromine concentration one fourth of the nitrogen oxide concentration?

Step-by-Step Solution

Answer

Answer: The concentration of bromine required is $0.033 \mathrm{M}$.

Step 1: 1. Write down the given information

We have a reaction of the type: \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{NOBr}(g)\) The rate law for this reaction is given as: \(rate = k[\mathrm{NO}]^{2}[\mathrm{Br}_{2}]\) The given rate is \(1.6 \times 10^{-8} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\), the concentration of nitrogen oxide is \(0.020 \mathrm{M}\), and the concentration of bromine is \(0.030 \mathrm{M}\).

Step 2: 2. Substitute the values into the rate law equation

Insert the given values into the rate law formula: \(1.6 \times 10^{-8} \mathrm{~mol/L\cdot min} = k (0.020 \mathrm{M})^{2}(0.030 \mathrm{M})\)

Step 3: 3. Solve for k

Rearrange the equation to isolate k: \(k = \frac{1.6 \times 10^{-8} \mathrm{~mol/L\cdot min}}{(0.020 \mathrm{M})^{2} (0.030 \mathrm{M})}\) Calculate the value of k: \(k = 1.33 \times 10^{-3} \mathrm{~L^{2}\cdot mol^{-2}\cdot min^{-1}}\) The value of k is \(1.33 \times 10^{-3} \mathrm{~L^{2}\cdot mol^{-2}\cdot min^{-1}}\). #b) Concentration of bromine for the given rate and nitrogen oxide concentration#

Step 4: 4. Use the rate law and the given rate and nitrogen oxide concentration values

The rate law for this reaction is given as: \(rate = k[\mathrm{NO}]^{2}[\mathrm{Br}_{2}]\) We want to calculate \([\mathrm{Br}_{2}]\) when \(rate = 3.5 \times 10^{-7} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\) and \([\mathrm{NO}] = 0.043 \mathrm{M}\).

Step 5: 5. Substitute the values into the rate law equation

Insert the given values and the calculated k from part (a) into the rate law formula: \(3.5 \times 10^{-7} \mathrm{~mol/L\cdot min} = (1.33 \times 10^{-3} \mathrm{~L^{2}\cdot mol^{-2}\cdot min^{-1}}) (0.043 \mathrm{M})^{2}[\mathrm{Br}_{2}]\)

Step 6: 6. Solve for the concentration of bromine

Rearrange the equation to find \([\mathrm{Br}_{2}]\): \([\mathrm{Br}_{2}] = \frac{3.5 \times 10^{-7} \mathrm{~mol/L\cdot min}}{(1.33 \times 10^{-3} \mathrm{L^{2}\cdot mol^{-2}\cdot min^{-1}}) (0.043 \mathrm{M})^{2}}\) Calculate the value of \([\mathrm{Br}_{2}]\): \([\mathrm{Br}_{2}] = 0.033 \mathrm{M}\) The concentration of bromine required is \(0.033 \mathrm{M}\). #c) Concentration of nitrogen oxide for the given rate and bromine ratio#

Step 7: 7. Express the concentration of bromine in terms of nitrogen oxide concentration

Given that the bromine concentration is one-fourth the concentration of nitrogen oxide, we can write \([\mathrm{Br}_{2}] = \frac{1}{4}[\mathrm{NO}]\).

Step 8: 8. Substitute the rate law, rate value, and bromine expression into the equation

We want to calculate \([\mathrm{NO}]\) when \(rate = 2.0 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\), and \([\mathrm{Br}_{2}] = \frac{1}{4}[\mathrm{NO}]\). Use the rate law: \(2.0 \times 10^{-6} \mathrm{~mol/L\cdot min} = (1.33 \times 10^{-3} \mathrm{~L^{2}\cdot mol^{-2}\cdot min^{-1}}) [\mathrm{NO}]^{2} \left(\frac{1}{4}[\mathrm{NO}]\right)\)

Step 9: 9. Solve for the concentration of nitrogen oxide

Rearrange the equation to isolate \([\mathrm{NO}]\): \([\mathrm{NO}]^{3} = \frac{2.0 \times 10^{-6} \mathrm{~mol/L\cdot min}}{(1.33 \times 10^{-3} \mathrm{~L^{2}\cdot mol^{-2}\cdot min^{-1}})(\frac{1}{4})}\) Now, calculate the value of \([\mathrm{NO}]\): \([\mathrm{NO}] = 0.078 \mathrm{M}\) The concentration of nitrogen oxide required is \(0.078 \mathrm{M}\).

Key Concepts

Understanding Reaction Rates

In chemical kinetics, the reaction rate is a measure of how fast a chemical reaction occurs. It is usually expressed as the change in concentration of a reactant or product per unit time. For the reaction between nitrogen oxide and bromine (NO + 1/2 Br₂ → NOBr), the reaction rate can be determined by measuring how quickly the reactants are consumed or the products are formed. This rate can be impacted by factors such as temperature, pressure, the presence of a catalyst, and the concentrations of the reactants.

For the given reaction, a rate of 1.6 x 10^-8 mol/L·min was observed under specific conditions. Understanding the factors affecting reaction rates helps chemists control reactions, whether to accelerate them for industrial processes or to slow them down for safety purposes. The study of reaction rates is a fundamental aspect of chemical kinetics, providing insights into the steps involved in a reaction and helping to design better products and systems.

The Rate Law

A rate law expresses the relationship between the reaction rate and the concentrations of the reactants. For the reaction in our exercise, the rate law is given by the equation rate = k[NO]²[Br₂], where k is the rate constant, and the concentrations of nitrogen oxide and bromine are raised to the powers of their respective reaction orders.

To determine the rate constant, k, we plug in the known values for the concentrations and the reaction rate. With the rate constant known, the rate law can be used to calculate reaction rates for any combination of reactant concentrations, which is essential for predicting how a reaction will behave under different conditions. The concept of the rate law is a cornerstone in the study of chemical kinetics, permitting the calculation of rates from reactant concentrations and vice versa.

Determining Reaction Order

The reaction order is the power to which the concentration of a reactant is raised in the rate law and indicates how the rate is affected by the concentration of that reactant. In our example, the reaction is classified as second-order in nitrogen oxide because the rate law contains the term [NO]². Similarly, it is first-order in bromine as indicated by the single power in [Br₂]. The overall reaction order is the sum of the orders with respect to each reactant, which in this case would be three.

Understanding the reaction order is crucial for predicting how changes in concentrations affect the reaction rate. For instance, doubling the concentration of nitrogen oxide would quadruple the reaction rate, while doubling the concentration of bromine would only double the rate. Knowing the reaction order helps in controlling the speed of a reaction, which is especially important in industrial applications where precise control over the reaction rate can be the difference between success and failure.