Problem 136 - Chemistry: An Atoms-Focused Approach

Question

Ammonia reacts with nitrous acid to form an intermediate, ammonium nitrite (NH_NO \(_{2}\) ), which decomposes to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}:\) \(\mathrm{NH}_{3}(g)+\mathrm{HNO}_{2}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) a. The reaction is first order in ammonia and second order in nitrous acid. What is the rate law for the reaction? What are the units on the rate constant if concentrations are expressed in molarity and time in seconds? b. The rate law for the reaction has also been written as $$ \text { Rate }=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}-\right]\left[\mathrm{HNO}_{2}\right] $$ Is this expression equivalent to the one you wrote in part \((a) ?\) c. With the data in Appendix \(4,\) calculate the value of \(\Delta H_{\text {ren }}^{\circ}\) for the overall reaction \(\Delta H_{\mathrm{f}, \mathrm{HNO}_{2}(a q)}^{\circ}=\) \(-128.9 \mathrm{kJ} / \mathrm{mol}\) d. Draw a reaction-energy profile for the process with the assumption that \(E_{\mathrm{a}}\) of the first step is lower than \(E_{\mathrm{a}}\) of the second step.

Step-by-Step Solution

Answer

Based on the given information, the rate law for the reaction is Rate = k[NH\(_{3}\)][HNO\(_{2}\)]\(^{2}\), and the units for the rate constant k are M\(^{-2}\)s\(^{-1}\). The provided rate law expression is not equivalent to the one derived in part a. The value of \(\Delta H_{\text {ren }}^{\circ}\) for the overall reaction is calculated to be 456.6 kJ/mol. To draw a qualitative reaction-energy profile, start at the energy level of the reactants (NH\(_{3}\) and HNO\(_{2}\)), represent the activation energy of the first step with a hill, followed by the intermediate (NH\(_{4}\)NO\(_{2}\)), the activation energy of the second step with a higher hill, and finally the energy level of the products (N\(_2\) and 2H\(_{2}\)O). The difference in energy levels between the reactants and products represents the overall enthalpy change, \(\Delta H_{\text {ren }}^{\circ}\).

Step 1: Determine the rate law for the reaction

According to the given information, the reaction is first order in ammonia (NH\(_{3}\)) and second order in nitrous acid (HNO\(_{2}\)). Thus, the rate law can be written as the following equation: Rate = k[NH\(_{3}\)][HNO\(_{2}\)]\(^{2}\)

Step 2: Find the units of the rate constant

To find the units of the rate constant, we can use the equation: Rate = k[NH\(_{3}\)][HNO\(_{2}\)]\(^{2}\) The Rate has the units of M/s (Molarity per second), and the concentrations are expressed in M (Molarity). Therefore: M/s = k(M)(M)\(^{2}\) So, the units for the rate constant k are: k = M\(^{-2}\)s\(^{-1}\)

Step 3: Check if the given expression is equivalent to the rate law found in part a

The given expression of the rate law is: Rate = k[NH\(_{4}^{+}\)][NO\(_{2}^{-}\)][HNO\(_{2}\)] To check if this expression is equivalent to the one found in part a, we should examine its relationship with the original rate law: Rate = k[NH\(_{3}\)][HNO\(_{2}\)]\(^{2}\) Since NH\(_{4}^{+}\) is formed by NH\(_{3}\) and H\(_{+}\), and NO\(_{2}^{-}\) is formed by the reaction of HNO\(_{2}\) with H\(_{2}\)O, it can be inferred that: [NH\(_{4}^{+}\)] \(\propto\) [NH\(_{3}\)] [NO\(_{2}^{-}\)] \(\propto\) [HNO\(_{2}\)] Therefore, the given expression for the rate law is not equivalent to the one we found in part a.

Step 4: Calculate the value of \(\Delta H_{\text {ren }}^{\circ}\) for the overall reaction

To calculate \(\Delta H_{\text {ren }}^{\circ}\) for the overall reaction, we will use the following equation: \(\Delta H_{\text {ren }}^{\circ}\) = \(\sum \Delta H_{\mathrm{f}, \mathrm{products}}^{\circ}\) - \(\sum \Delta H_{\mathrm{f}, \mathrm{reactants}}^{\circ}\) First, we need to find the change in enthalpy for the reactants and products using the data from Appendix 4. For the reactants: \(\Delta H_{\mathrm{f}, \mathrm{NH}_{3}(g)}^{\circ}\) = -45.9 kJ/mol \(\Delta H_{\mathrm{f}, \mathrm{HNO}_{2}(a q)}^{\circ}\) = -128.9 kJ/mol For the products: \(\Delta H_{\mathrm{f}, \mathrm{N}_{2}(g)}^{\circ}\) = 0 kJ/mol \(\Delta H_{\mathrm{f}, 2\mathrm{H}_{2} \mathrm{O}(\ell)}^{\circ}\) = -2 \(\times\) (-285.8) kJ/mol Now we can calculate the value of \(\Delta H_{\text {ren }}^{\circ}\) using the equation: \(\Delta H_{\text {ren }}^{\circ}\) = (-2 \(\times\) -285.8 + 0 kJ/mol) - (-45.9 -128.9 kJ/mol) \(\Delta H_{\text {ren }}^{\circ}\) = 456.6 kJ/mol

Step 5: Draw a reaction-energy profile

To draw the reaction-energy profile, we will use the information given that \(E_{\mathrm{a}}\) of the first step is lower than \(E_{\mathrm{a}}\) of the second step. But without knowing the actual values of activation energies, we can only draw a qualitative energy profile. The energy profile should start at the energy level of the reactants, NH\(_{3}\) and HNO\(_{2}\). Next, draw a hill representing the activation energy of the first step, and follow it with a valley that corresponds to the intermediate, NH\(_{4}\)NO\(_{2}\). Then, draw a higher hill representing the activation energy of the second step, and finally a line representing the energy level of the products, N\(_{2}\) and 2H\(_{2}\)O. The difference in energy levels between the reactants and products represents the overall enthalpy change, \(\Delta H_{\text {ren }}^{\circ}\). Please note that the drawing will not be quantitative, so the heights of the hills and valleys should be visually representative but do not have precise values without additional information.

Key Concepts

Rate Law

In chemical reactions, the rate law describes how the concentration of reactants affects the rate of the reaction. For this reaction, it's given as first order in ammonia (\(\mathrm{NH}_{3}\)) and second order in nitrous acid (\(\mathrm{HNO}_{2}\)). This means:

• The rate is directly proportional to the concentration of ammonia.
• The rate is proportional to the square of nitrous acid's concentration.

This rate law is expressed as:\[\text{Rate} = k[\mathrm{NH}_{3}][\mathrm{HNO}_{2}]^{2}\]where \(k\) is the rate constant. If concentrations are in molarity (M) and time in seconds (s), the units of \(k\) are \(M^{-2}s^{-1}\). This ensures the rate is in \(\frac{M}{s}\). Understanding how each reactant affects the rate helps control and predict the reaction speeds effectively.

Enthalpy Change

Enthalpy change, \(\Delta H\), measures the heat change in a reaction at constant pressure. For the reaction, we calculate \(\Delta H_{\text{ren}}^{\circ}\) using known enthalpies of formation.
First, find enthalpies for reactants and products:

• Reactants: \(\mathrm{NH}_{3}(g)\) is \(-45.9\ \mathrm{kJ/mol}\), \(\mathrm{HNO}_{2}(aq)\) is \(-128.9\ \mathrm{kJ/mol}\).
• Products: \(\mathrm{N}_{2}(g)\) is \(0\ \mathrm{kJ/mol}\), and \(2\ \mathrm{H}_{2} \mathrm{O}(\ell)\) is \(-2 \times (-285.8)\ \mathrm{kJ/mol}\).

Then the enthalpy change is:\[\Delta H_{\text{ren}}^{\circ} = [(0) + (571.6)] - [(-45.9) + (-128.9)] = 456.6\ \mathrm{kJ/mol}\]This positive value indicates the reaction absorbs energy, making it endothermic.

Reaction Mechanism

The reaction mechanism describes the step-by-step sequence by which reactants transform into products. Here, the mechanism involves a two-step process:

• In the first step, ammonia reacts with nitrous acid to form the intermediate, ammonium nitrite (\(\mathrm{NH}_{4}\mathrm{NO}_{2}\)).
• In the second step, \(\mathrm{NH}_{4}\mathrm{NO}_{2}\) decomposes into \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\).

Each step involves its own transition state and activation energy. The overall rate is determined by the slowest step, known as the rate-determining step. Understanding the mechanism helps predict reaction paths and identify possible intermediates.

Activation Energy

Activation energy (\(E_{a}\)) is the minimum energy required for a reaction to occur. It's depicted as a peak in a reaction-energy profile. For this reaction:

• The first step has a lower \(E_{a}\) compared to the second step. This means the first step is easier to achieve energetically.
• Higher \(E_{a}\) for the second step means it requires more energy, thus is slower.

The reaction-energy profile shows energy levels for reactants, intermediate, and products, with hills representing activation energies. The height of each hill corresponds to the energy needed for each step. This concept explains why some reactions occur faster and others are slower, guiding chemists in altering conditions to improve reaction rates.