Problem 52 - Chemistry: An Atoms-Focused Approach

Question

Predict the order with respect to NO for the reaction \(\mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightarrow \mathrm{NOBr}_{2}(g)\) under each of the following conditions: a. The rate doubles when \([\mathrm{NO}]\) is doubled and \(\left[\mathrm{Br}_{2}\right]\) remains constant. b. The rate increases by 1.56 times when \([\mathrm{NO}]\) is increased 1.25 times and \(\left[\mathrm{Br}_{2}\right]\) remains constant. c. The rate is halved when \([\mathrm{NO}]\) is doubled and \(\left[\mathrm{Br}_{2}\right]\) remains constant.

Step-by-Step Solution

Answer

a) The rate doubles when the concentration of NO is doubled. b) The rate increases by 1.56 times when the concentration of NO is increased by 1.25 times. c) The rate is halved when the concentration of NO is doubled. Answer: a) The order with respect to NO is 1. b) The order with respect to NO is approximately 2. c) The order with respect to NO is -1.

Step 1: Part a - Rate doubles with double [NO]

We are given that the rate doubles when the concentration of NO is doubled while keeping the concentration of Br2 constant. Let's write the rate equation using initial concentrations and then doubled concentrations of NO: Rate1 = k[NO]^m [Br2]^n Rate2 = k(2[NO])^m [Br2]^n Since the Rate2 is double the Rate1: Rate2 = 2 x Rate1 2(k[NO]^m [Br2]^n) = k(2[NO])^m [Br2]^n We can cancel out the constants "k" and [Br2]^n from both sides: 2[NO]^m = (2[NO])^m This simplifies to: 2 = 2^m Therefore, m=1. The order with respect to NO in this case is 1.

Step 2: Part b - Rate increases by 1.56 with 1.25 times [NO]

Rate1 = k[NO]^m [Br2]^n Rate2 = k(1.25[NO])^m [Br2]^n Since the Rate2 is 1.56 times the Rate1: Rate2 = 1.56 x Rate1 1.56(k[NO]^m [Br2]^n) = k(1.25[NO])^m [Br2]^n We can cancel out the constants "k" and [Br2]^n from both sides: 1.56[NO]^m = (1.25[NO])^m Divide both sides by [NO]^m: 1.56 = 1.25^m Using logarithms, we can find the value of m: m = log(1.56)/log(1.25) m ≈ 2.07. The order with respect to NO in this case is approximately 2.

Step 3: Part c - Rate halved with double [NO]

Rate1 = k[NO]^m [Br2]^n Rate2 = k(2[NO])^m [Br2]^n Since the Rate2 is half of the Rate1: Rate2 = 0.5 x Rate1 0.5(k[NO]^m [Br2]^n) = k(2[NO])^m [Br2]^n We can cancel out the constants "k" and [Br2]^n from both sides: 0.5[NO]^m = (2[NO])^m This simplifies to: 0.5 = 2^m m = log(0.5)/log(2) m = -1. The order with respect to NO in this case is -1.

Key Concepts

Rate Law

In the study of reaction kinetics, the rate law is a fundamental concept that defines how the concentration of reactants affects the reaction rate. Understanding rate laws is essential as it allows chemists to predict how the reaction speed will change under different conditions. Rate laws are expressed in the form:

\[ \text{Rate} = k[A]^m[B]^n \]where:

• \( k \) is the rate constant specific to the reaction at a given temperature.
• \([A]\) and \([B]\) are the concentrations of the reactants.
• \( m \) and \( n \) are the reaction orders with respect to each reactant.

For example, in the reaction of \( \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightarrow \mathrm{NOBr}_2(g) \), the rate law helps determine how changes in the concentration of \( \mathrm{NO} \) affect the reaction speed while holding \( \mathrm{Br}_2 \) constant. A critical part of this process is determining the reaction order for each reactant to understand their specific influence on the rate.

Reaction Order

The reaction order is a component of the rate law that indicates the power to which the concentration of a reactant is raised in the rate equation. It gives insight into how sensitive the reaction rate is to concentration changes. The reaction order can take values that are integers, fractions, or even negative numbers, providing a flexible framework to describe diverse reactions:

- In Part a of the exercise, doubling the concentration of \( \mathrm{NO} \) results in the reaction rate doubling, which means the reaction order with respect to \( \mathrm{NO} \) is 1 since \( 2^1 = 2 \).- In Part b, increasing \([\mathrm{NO}]\) by 1.25 times led to a 1.56 times increase in rate, indicating a reaction order of approximately 2. This suggests the rate is highly influenced by changes in \( \mathrm{NO} \) concentration.- Part c demonstrates a negative reaction order, where doubling \([\mathrm{NO}]\) results in halving the rate. Here, the reaction order is -1, implying an inverse relationship between \( \mathrm{NO} \) concentration and reaction rate.

Understanding reaction orders is crucial for devising strategies to control reaction rates, which can be especially important in industrial applications where precise reaction control is critical.

Rate Equation

The rate equation is an algebraic depiction of the reaction rate as a function of the concentration of reactants. It combines the rate constant and the reaction order to provide a complete picture of the reaction kinetics. This is key in predicting how the rate of a chemical reaction will change depending on varying reactant concentrations.

In our exercise scenario, the rate equation for each situation aids in solving problems related to reaction dynamics. By knowing:

• The initial and final rates of reaction changes.
• The changes in reactant concentration.

For instance, by substituting known values into the rate equation, you can find unknown reaction orders, as demonstrated in the exercise. In Part a, \(2[\mathrm{NO}]^m = (2[\mathrm{NO}])^m\) simplifies to \(m = 1\). This straightforward substitution technique can be applied similarly in various complex scenarios to determine both the rate constant and reaction order, shaping the entire kinetic profile of a reaction.