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The activation energy for some reaction $$ \mathrm{X}_{2}(g)+\mathrm{Y}_{2}(g) \longrightarrow 2 \mathrm{XY}(g) $$ is 167 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is \(+28 \mathrm{kJ} / \mathrm{mol}\) . What is the activation energy for the decomposition of XY?

Short Answer

Expert verified
The activation energy for the decomposition of XY can be found using the formula Ea(Reverse) = Ea(Forward) + ΔE. Given Ea(Forward) = 167 kJ/mol and ΔE = +28 kJ/mol, we can calculate Ea(Reverse) = 167 kJ/mol + 28 kJ/mol, resulting in Ea(Reverse) = 195 kJ/mol. Therefore, the activation energy for the decomposition of XY is \(195 \, \mathrm{kJ/mol}\).

Step by step solution

01

Identify the given values and the unknown

We are given the activation energy Ea(Forward) = 167 kJ/mol, the change in energy ΔE = +28 kJ/mol, and we need to find the activation energy for the reverse reaction Ea(Reverse).
02

Understand the energy profile of a reaction

The activation energy for a reaction is the difference between the energy level of the transition state and the energy level of the reactants. ΔE is the difference between the energy levels of the products and the reactants. In the reverse reaction (decomposition), the reactants and products switch roles. Therefore, the activation energy for the reverse reaction is the difference between the energy level of the transition state and the energy level of the initial products (which are now the reactants in the reverse reaction). The energy profile can be visualized as a graph with energy on the vertical axis and reaction progress on the horizontal axis.
03

Calculate the activation energy for the reverse reaction

We know that Ea(Reverse) = Ea(Forward) + ΔE. Plug in the given values: Ea(Reverse) = 167 kJ/mol + 28 kJ/mol Ea(Reverse) = 195 kJ/mol So, the activation energy for the decomposition of XY is 195 kJ/mol.

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Key Concepts

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

Understanding the Energy Profile
An energy profile is a visual representation of a chemical reaction's energy changes. On the graph, you will see energy on the vertical axis and the progress of the reaction on the horizontal axis. You can think of it like a journey that starts at the energy level of the reactants and moves to the products.

In a typical energy profile, there is an initial climb to a peak known as the transition state, which indicates the energy barrier that the reactants must overcome to transform into products. After reaching the peak, the energy drops down to the level of the products.
  • The peak indicates the maximum potential energy the substances have during the reaction.
  • The height of this peak from the starting point is the activation energy (Ea) for the reaction.
For a clearer understanding, remember that the energy profile provides insight into the energetic demands of breaking and forming bonds during the reaction process.
Spotlight on Transition State
The transition state is a fascinating concept in chemistry. It represents a very brief moment in time during a reaction when reactants turn into products. You can think of it as the top of a hill or a tightrope — a precarious position where the molecules are neither reactants nor products.

The transition state is essential because it is the point at which the molecules have maximum energy and are in a highly unstable state. From here, they can either fall back and become reactants again or move forward to form products.
  • It's a fleeting arrangement not stable enough to be isolated or observed directly.
  • Being at the highest energy point, it requires activation energy to be attained.
Understanding the transition state helps chemists design catalysts that can lower the activation energy, making reactions faster and more efficient.
Exploring the Reverse Reaction
In chemistry, every reaction has a reverse reaction, which occurs when products convert back into reactants. The reverse reaction is important in equilibrium reactions, where both forward and reverse reactions occur simultaneously.

For the reverse reaction to happen, the concept of activation energy is crucial. Unlike the forward reaction, the activation energy for the reverse reaction is calculated from the products' energy level to the transition state.
  • In our reaction example, you can find the reverse activation energy using the equation: \( Ea_{(Reverse)} = Ea_{(Forward)} + \Delta E \).
  • The activation energy for the reverse reaction often differs from the forward reaction because \( \Delta E \), the net energy change between reactants and products, must be considered.
This concept is essential for understanding how reversible reactions can be controlled and manipulated in various chemical processes.

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Most popular questions from this chapter

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .^{\circ} \mathrm{C},\) respectively. What is the value of the activation energy?

Define stability from both a kinetic and thermodynamic perspective. Give examples to show the differences in these concepts.

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\) a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section 12.7 is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section 12.7\(?\)

Sulfuryl chloride undergoes first-order decomposition at \(320 .^{\circ} \mathrm{C}\) with a half-life of 8.75 \(\mathrm{h}\) . $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What is the value of the rate constant, \(k,\) in \(\mathrm{s}^{-1}\) ? If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25-\mathrm{L}\) container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after 12.5 \(\mathrm{h}\) ?

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