Problem 30 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(a) The rate law is \( Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] \). (b) The reaction rate is \( 8.82 \times 10^{-1} M~s^{-1} \). (c) The reaction rate is \( 6.0 M~s^{-1} \). (d) The reaction rate is \( 1.8 \times 10^{-2} M~s^{-1} \).

Step 1: Write the Rate Law for the Reaction

Given that the reaction is first order in H₂ and second order in NO, we can express the rate law as follows: \( Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] \), where k is the rate constant.

Step 2: Find the reaction rate at given concentrations (part b)

The rate constant at 1000 K is given as \(6.0 \times 10^4 M^{-2}s^{-1}\). We are given the concentrations of NO and H₂ as 0.035 M and 0.015 M, respectively. Using the rate law derived in Step 1, the reaction rate can be calculated: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1})(0.035 M)^2 (0.015 M) \] Calculate the rate to find the answer.

Step 3: Find the reaction rate with new NO concentration (part c)

We are given a new concentration for NO (0.10 M) with the H₂ concentration remaining at 0.010 M. Use the same rate law equation as before: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1}) (0.10 M)^2 (0.010 M) \] Calculate the rate for this new set of concentrations.

Step 4: Find the reaction rate with new concentrations for both reactants (part d)

The concentrations of NO and H₂ are now given as 0.010 M and 0.030 M, respectively. Use the rate law equation once again: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1}) (0.010 M)^2 (0.030 M) \] Calculate the rate for this last set of reactant concentrations.

Key Concepts

Rate Law

In chemical kinetics, the rate law is a mathematical equation that describes how the rate of a reaction depends on the concentration of reactants. It is an expression that helps you understand the relationship between the rate of reaction and the concentration of reactants involved. The general form of the rate law can be indicated as:

• \( ext{Rate} = k[A]^m[B]^n \)

where

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

In the provided reaction, the rate law is first order in \( \mathrm{H}_2 \) and second order in NO, which makes the rate law:

• \( ext{Rate} = k[ ext{NO}]^2[ ext{H}_2] \)

This means that the concentration of \( \mathrm{H}_2 \) impacts the rate linearly, while the concentration of NO affects the rate quadratically.
Understanding this relationship helps in predicting how changes in concentration will affect the speed of the reaction.

Reaction Rate

The reaction rate is the speed at which reactants are converted into products. It tells you how fast a reaction occurs. Although we often express it as the change in concentration of a reactant or product with respect to time, it is determined experimentally and explained through the rate law.

• For our reaction with a rate constant \( k = 6.0 \times 10^4 \text{ M}^{-2}\text{s}^{-1} \) at 1000 K, the reaction rate is directly calculated using the concentrations plugged into the rate law formula.

In part (b) of the exercise, substituting the values \( [\text{NO}] = 0.035 \text{ M} \) and \( [\text{H}_2] = 0.015 \text{ M} \) into the rate law:

• \( \text{Rate} = (6.0 \times 10^4) \times (0.035)^2 \times (0.015) \)

In part (c) and (d), you can similarly substitute the new concentrations into the formula. Changes in concentration lead to different reaction rates due to their direct influence on the formula. This quantitative analysis via the formula allows for precise prediction of reactions under varying conditions.

Order of Reaction

The order of reaction refers to the sum of the powers to which the concentration terms are raised in the rate law. It provides insights into the mechanism of the reaction and shows how the rate is affected by the concentration of reactants. In our case:

• The order of reaction with respect to NO is 2, indicating that its concentration is squared in the rate law, and it has a significant impact on the reaction rate.
• The order with respect to \( \mathrm{H}_2 \) is 1, showing a linear relationship with the reaction rate.

The overall order of the reaction is simply the sum of these individual orders, which in this exercise is 3 (2 for NO and 1 for \( \mathrm{H}_2 \)). Knowing the order helps to understand how doubling the concentration of each reactant will affect the overall reaction rate. For NO, doubling the concentration squares its impact, whereas doubling \( \mathrm{H}_2 \) will double the rate. Understanding these concepts gives key insights into how chemical reactions proceed and are influenced by concentration changes.