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Chapter 12: Problem 70

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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Most popular questions from this chapter

A certain reaction has the form $$ \mathrm{aA} \longrightarrow $$ At a particular temperature, concentration versus time data were collected. A plot of 1\(/[\mathrm{A}]\) versus time (in seconds) gave a straight line with a slope of \(6.90 \times 10^{-2} .\) What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If \([\mathrm{A}]_{0}\) for this reaction is \(0.100 M,\) what is the first half-life (in seconds)? If the original concentration (at \(t=0 )\) is \(0.100 M,\) what is the second half-life (in seconds)?

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 \(\mathrm{s}\) at \(21.0^{\circ} \mathrm{C}\) and 13.0 \(\mathrm{s}\) at \(27.8^{\circ} \mathrm{C}\) a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at \(30.0^{\circ} \mathrm{C} ?\) c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

Consider the hypothetical reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow\( \)2 \mathrm{AB}(g),$ where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is $2.45 \times 10^{-4} \mathrm{L} / \mathrm{mol}\( \)\mathrm{s},\( and at \)508^{\circ} \mathrm{C}\( the rate constant is 0.891 \)\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$ . What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C} ?\)

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, 0.0048 mole of \(\mathrm{PH}_{3}\) is consumed in a \(2.0-\mathrm{L}\) container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?
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