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The gas-phase reaction \(\mathrm{Cl}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\) \(\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Short Answer

Expert verified
The activation energy for the reverse reaction between Cl(g) and HBr(g) to form HCl(g) and Br(g) is 73 kJ.

Step by step solution

01

Sketch the energy profile

Begin by drawing a horizontal axis to represent the reaction progress and a vertical axis to represent the energy. On the left side of the graph, label a point as 'Reactants' and on the right side of the graph, label another point as 'Products'. This will help visualizing the starting and the ending point of the reaction.
02

Plot the activation energy and overall enthalpy change

We are given the activation energy (\(E_a\)) for the reaction as 7 kJ and the overall enthalpy change (\(\Delta E\)) as -66 kJ. Draw an arrow pointing upwards from the Reactants point, representing the activation energy. Label this arrow \(E_a = 7 \mathrm{~kJ}\). From the tip of this arrow, draw another arrow pointing downwards towards the 'Products' point, representing the overall enthalpy change. Label this arrow \(\Delta E = -66 \mathrm{~kJ}\).
03

Complete the energy profile

Connect the tips of the two arrows (activation energy and overall enthalpy change) with a peak that represents the transition state of the reaction. Draw a curve from the 'Reactants' point, passing through the transition state, and ending at the 'Products' point, creating the energy profile for the reaction. At this point, your energy profile sketch should show the activation energy arrow, the overall enthalpy change arrow, and a curve representing the energy changes during the reaction, from reactants to products.
04

Determine the activation energy for the reverse reaction

To find the activation energy for the reverse reaction, we need to determine the difference in energy between the transition state and the products. The reverse reaction would start at the products, go through the transition state and end at the reactants. Since the overall enthalpy change for the forward reaction is -66 kJ, the overall enthalpy change for the reverse reaction will be the opposite, 66 kJ. We can see from the energy profile that the activation energy for the reverse reaction is the sum of the activation energy for the forward reaction and the overall enthalpy change for the reverse reaction. Thus, the activation energy for the reverse reaction is: \(E_{a(reverse)} = E_{a(forward)} + \Delta E (reverse)\) \(E_{a(reverse)} = 7 \mathrm{~kJ} + 66 \mathrm{~kJ}\) \(E_{a(reverse)} = 73 \mathrm{~kJ}\) The activation energy for the reverse reaction is 73 kJ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change, symbolized as \( \Delta H \), is a measure of the heat content change in a system during a chemical reaction under constant pressure. It is an essential concept in thermodynamics and chemical kinetics, where a negative \( \Delta H \) value, such as \( -66 \mathrm{~kJ} \), indicates an exothermic reaction, meaning energy is released into the surroundings. Conversely, a positive \( \Delta H \) signifies an endothermic reaction, where the system absorbs energy.

To understand this with more clarity, envision the reaction as a journey from a higher energy level (reactants) to a lower one (products). The enthalpy change is like the decrease in elevation when descending a hill—it's the difference in height from top to bottom. In our case, the 'descent' releases \( 66 \mathrm{~kJ} \) of energy, making the surroundings warmer, which is typical for many combustion and bond-forming reactions.

Remember, enthalpy change doesn’t only tell us about the energy difference; it also provides insights into the stability of the products compared to the reactants and can hint at the spontaneity of a reaction.
Energy Profile Sketch
An energy profile sketch is a graphical representation that maps the energy changes a chemical system undergoes during a reaction. It is an excellent tool for visualizing the concept of activation energy and enthalpy change.

Creating an Energy Diagram

To create an energy profile sketch, imagine plotting a hiking trail on a graph. Start with marking 'Reactants' and 'Products' on either side of the 'path.' The vertical 'climb' from reactants represents the activation energy (\( E_a \) = \( 7 \mathrm{~kJ} \) for the reaction given), depicting the initial energy input required to start the reaction.

After reaching the peak, or the transition state, the 'trail' slopes downwards towards 'Products,' representing the energy release (\( \Delta E \) = \( -66 \mathrm{~kJ} \)). The depth of this 'descent' reflects the reaction’s exothermic nature. The curve you sketch from reactants to products shows the energy's ebb and flow, essentially the reaction's energy landscape.
Reverse Reaction
A reverse reaction travels the opposite 'path' of the forward reaction, converting products back into reactants. The most intriguing aspect of reverse reactions in terms of energy is that the activation energy and enthalpy change flip roles.

Consider the analogy of our trail: if the down-hill signifies the forward reaction, then the reverse reaction would mean trekking back up the hill. The activation energy for this 'upward journey' combines the initial climb (forward reaction's activation energy) and the descent you made (\( \Delta E \) of the forward reaction).

For our specific reaction, the activation energy for the reverse reaction (\( E_{a(reverse)} \) = \( 73 \mathrm{~kJ} \) is the sum of \( 7 \mathrm{~kJ} \) and \( 66 \mathrm{~kJ} \). This energy requirement is higher because you have to overcome the initial barrier and also input energy equivalent to what was released in the forward reaction. Hence, it is often more difficult for the reverse reaction to occur spontaneously, especially in exothermic reactions.

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Most popular questions from this chapter

Zinc metal dissolves in hydrochloric acid according to the reaction $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(a q)--\rightarrow \operatorname{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Suppose you are asked to study the kinetics of this reaction by monitoring the rate of production of \(\mathrm{H}_{2}(g)\). (a) By using a reaction flask, a manometer, and any other common laboratory equipment, design an experimental apparatus that would allow you to monitor the partial pressure of \(\mathrm{H}_{2}(g)\) produced as a function of time. (b) Explain how you would use the apparatus to determine the rate law of the reaction. (c) Explain how you would use the apparatus to determine the reaction order for \(\left[\mathrm{H}^{+}\right]\) for the reaction. (d) How could you use the apparatus to determine the activation energy of the reaction? (e) Explain how you would use the apparatus to determine the effects of changing the form of \(\mathrm{Zn}(s)\) from metal strips to granules.

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

(a) What part of the energy profile of a reaction is affected by a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst?

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) Write rate laws for each elementary reaction in the mechanism. (d) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

Urea (NH \(_{2} \mathrm{CONH}_{2}\) ) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M}\) \(\mathrm{HCl}\) occurs according to the reaction \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(2 \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} .\) (a) What is the value for the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?(\mathrm{c})\) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

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