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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

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\) \(\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\), is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M}\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\), is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C}\), \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g)\). If the reac- tion is first order with a half-life of \(56.3\) min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature (K) & Rate Constant (s \(^{-1}\) ) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\)

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1} .\) Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .\) (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C} ?\)

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