Problem 52 - Chemistry and Chemical Reactivity

Question

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. Step 1 \(\quad\) Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2 Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is \(-\Delta\left[\mathrm{CH}_{3} \mathrm{OH}\right] / \Delta t=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

Step-by-Step Solution

Answer

(a) Overall: \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \). (b) Diagram: \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \) peak. (c) Rate law: \( k[\mathrm{CH_{3}OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \)."}

Step 1: Combine the Elementary Steps

To find the overall reaction equation, we combine the two steps of the mechanism: \[ \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \] and \[ \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} + \mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \] The \(\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}\) species is an intermediate and does not appear in the overall reaction equation. Add up reactants and products from both steps and cancel out the intermediate: Overall Reaction: \[ \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \]

Step 2: Reaction Coordinate Diagram

The reaction coordinate diagram shows the energy changes during a chemical reaction. - Start with reactants \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \) at normal energy level. - The first step is fast and endothermic, meaning a small or moderate uphill climb to the intermediate \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \). - The second step is slow and exothermic: a steep climb followed by a steep drop as \( \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \) forms, ending lower than the original reactants.

Step 3: Determine the Rate Law

In a multi-step reaction, the overall rate is determined by the slowest step, which is often called the rate-determining step. The rate law for the second step is based on its reactants: \[ \text{Rate} = k_2[\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}][\mathrm{Br}^{-}] \] However, \([\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}]\) is derived from Step 1's equilibrium expression: \[ [\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}] = K_1[\mathrm{CH}_{3}\mathrm{OH}][\mathrm{H}^{+}] \] Substitute this back into the rate law: \[ \text{Rate} = k_2K_1[\mathrm{CH}_{3}\mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \] - Define \( k = k_2K_1 \). Thus, the rate law for the overall reaction is \[ -\Delta[\mathrm{CH_{3}OH}] / \Delta t = k[\mathrm{CH_{3}OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \]

Key Concepts

Elementary Steps

In a chemical reaction mechanism, elementary steps are the simplest stages through which a reaction progresses. Each step represents a simple reaction leading from reactants to products or intermediates. Think of them as building blocks of the overall reaction. They're crucial because they show the path the reaction takes and how different molecules interact.

• The first step often involves a quick and simple transformation or combination of molecules.
• The second step might be slower and more complex.
• Intermediates like \(\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}\) in our example are created and consumed during these steps.

During our reaction, the first step quickly forms the intermediate \(\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}\). It's fast and endothermic, but this intermediate doesn't appear in the final reaction equation. Instead, it facilitates the transformation to the final products: this is a hallmark of elementary steps and their importance in understanding reactions.

Reaction Coordinate Diagram

A reaction coordinate diagram represents the energy changes during a chemical reaction. It displays how the energy of a system varies as the reaction proceeds from reactants to products. Visual learners often find these diagrams helpful because they make intangible concepts like energy changes easier to grasp.
A typical reaction coordinate diagram has:

• The x-axis as the reaction progress or "reaction coordinate."
• The y-axis shows energy levels.
• Hills and valleys corresponding to energy barriers and intermediate states.

For our reaction, start with the reactants at a moderate energy level. The diagram will then show a rise for the endothermic first step, indicating an energy absorption as \(\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}\) is formed. The second step, being slower and exothermic, will be a steep energy climb followed by a dramatic drop, illustrating energy release when \(\mathrm{CH}_{3}\mathrm{Br}\) and \(\mathrm{H}_{2}\mathrm{O}\) are formed. The ending energy level is lower than where we began, highlighting the whole reaction's exothermic nature.

Rate Law

The rate law explains how the rate of a chemical reaction is affected by the concentration of its reactants. For chemical reactions occurring in multiple steps, the slowest step, also known as the "rate-determining step," dictates the overall reaction rate. This concept aligns with the classic bottleneck analogy — when flow is slowest in one area, it restricts flow everywhere.
In our reaction example, the rate-determining step is the slow, exothermic conversion of \(\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}\) with \(\mathrm{Br}^{-}\). The rate law for this step is:

• Rate = \( k_{2}[\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}][\mathrm{Br}^{-}] \)

However, because \(\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}\) is an intermediate formed in equilibrium, we use its expression from the first step to derive the final rate equation:

• \([\mathrm{CH}_{3}\mathrm{OH}_{2}^{+}] = K_{1}[\mathrm{CH}_{3}\mathrm{OH}][\mathrm{H}^{+}] \)
• Combined to give the overall rate as: \( k[\mathrm{CH}_{3}\mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \)

Understanding the rate law is key to controlling reaction rates in industrial and laboratory settings.

Exothermic Reactions

Exothermic reactions are processes that release energy in the form of heat or light. They are common and essential to various natural and industrial processes. The term "exothermic" comes from the Greek "exo", meaning "outside," and "thermikòs," related to heat.

• Characteristic energy release leads to a decrease in the potential energy of the system.
• The surroundings experience a temperature increase.
• Reactions often proceed spontaneously.
• Think of combustion in engines or the heat from a campfire as everyday examples.

In the reaction between \(\mathrm{CH}_{3}\mathrm{OH}\) and \(\mathrm{HBr}\), while the initial step may be endothermic (requiring an input of energy), the sum of changes in both steps is exothermic overall. This means, by the time \(\mathrm{CH}_{3}\mathrm{Br}\) and \(\mathrm{H}_{2}\mathrm{O}\) are formed, more energy has been released than absorbed. Thus, the reaction tends to be self-sustaining after the initial energy input.