Problem 59 - Chemistry: An Atoms-Focused Approach

Question

In the presence of water, NO and \(\mathrm{NO}_{2}\) react to form nitrous acid (HNO,) by the following reaction: $$\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{HNO}_{2}(a q)$$ When the concentration of NO or \(\mathrm{NO}_{2}\) is doubled, the initial rate of reaction doubles. If the rate of the reaction does not depend on \(\left[\mathrm{H}_{2} \mathrm{O}\right],\) what is the rate law for this reaction?

Step-by-Step Solution

Answer

Question: Determine the rate law for the reaction based on the given information about the effect of concentrations on the initial rate of the reaction. Answer: The rate law for the reaction is rate = k[NO][NO2].

Step 1: Understand the effect of concentration changes on the rate of reaction

The exercise tells us that when the concentration of NO or NO2 is doubled, the initial rate of the reaction doubles. This information will help us establish the order of the reaction with respect to NO and NO2.

Step 2: Determine the order of the reaction with respect to NO and NO2

Let's first consider the effect of doubling the concentration of NO while keeping the concentration of NO2 constant: rate1 = k[NO]^m[NO2]^n rate2 = k(2[NO])^m[NO2]^n Since rate2 = 2*(rate1), we can write the equation as: 2*(k[NO]^m[NO2]^n) = k(2[NO])^m[NO2]^n Dividing both sides by k[NO]^m[NO2]^n, we get: 2 = 2^m ==> m = 1 Now let's consider the effect of doubling the concentration of NO2 while keeping the concentration of NO constant: rate1 = k[NO]^m[NO2]^n rate2 = k[NO]^m(2[NO2])^n Again, since rate2 = 2*(rate1), we can write the equation as: 2*(k[NO]^m[NO2]^n) = k[NO]^m(2[NO2])^n Dividing both sides by k[NO]^m[NO2]^n, we get: 2 = 2^n ==> n = 1

Step 3: Write the rate law

Now that we know the reaction order with respect to NO (m) and NO2 (n), we can write the rate law for the reaction: rate = k[NO]^m[NO2]^n Replace m and n with the values obtained in Step 2: rate = k[NO]^1[NO2]^1 So the rate law for the reaction is: rate = k[NO][NO2]

Key Concepts

Rate Law

The rate law of a chemical reaction gives us an idea of how the concentration of reactants affects the speed, or rate, of the reaction. It is essentially an expression that relates the rate of a chemical reaction to the concentration of its reactants. In this exercise, the rate law for the given reaction is expressed as:\[\text{rate} = k[\text{NO}][\text{NO}_2]\]Where:- \( k \) is the rate constant unique to each reaction at a given temperature.- \([\text{NO}]\) and \([\text{NO}_2]\) are the concentrations of the reactants.This tells us that the reaction rate changes in direct proportion to the concentrations of NO and NO₂. This specific relationship means that the rate of reaction will increase as the concentration of these reacting substances increases.

Reaction Order

The reaction order gives insight into how the concentration of each reactant impacts the overall reaction rate. In the exercise, the order is determined from the effect on the reaction rate when the reactant concentrations are changed. For our reaction: - The order with respect to NO is 1, due to the doubling of the reaction rate when the concentration of NO is doubled. - Similarly, the order with respect to NO₂ is also 1, as doubling the concentration of NO₂ doubles the rate. Thus, the overall reaction order is the sum of these individual orders, which is 2 (1 for NO + 1 for NO₂). This tells us that both NO and NO₂ equally affect the speed with which products are formed. These simple first-order dependencies on NO and NO₂ concentrations are crucial for predicting how a change in conditions can change the reaction dynamics.

Effect of Concentration on Reaction Rate

Understanding the effect of concentration is vital for manipulating reaction speeds in both industrial and laboratory settings. The exercise demonstrates that when the concentration of either NO or NO₂ is increased, the rate of the reaction increases as well. This is consistent with the established rate law and reaction order. In chemical kinetics: - **Higher reactant concentration increases collisions:** More molecules mean a higher likelihood of collisions, hence increasing the reaction rate. - **Doubling the concentration doubles the rate:** In our reaction, the direct proportionality between concentration and rate means if you double either NO or NO₂, the rate doubles. Such knowledge can be applied practically, allowing chemists to control reaction conditions to favor faster product formation or to maintain a reaction safely when runaway conditions (explosive reactions) are a risk. This principle holds true as long as the reaction conditions (such as temperature and presence of a catalyst) remain constant.