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Chapter 14: Problem 56

For the elementary process $\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)$ the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are $154 \mathrm{~kJ} / \mathrm{mol}\( and \)136 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Short Answer

Expert verified
The energy profile for the given reaction can be sketched by drawing the reactants and products at different energy levels, and an arc representing the activation energy (Ea = 154 kJ/mol). The overall change in energy (ΔE) is labeled as 136 kJ/mol. To find the activation energy for the reverse reaction, we first calculate the energy difference between the peak and product energy levels (18 kJ/mol) and add it to the given ΔE. Thus, the activation energy for the reverse reaction is 154 kJ/mol.

Step by step solution

01

Determine the reactants and products

In the given reaction, the reactants are N2O5 (g) and the products are NO2 (g) and NO3 (g).
02

Draw the energy profile

Draw a horizontal axis labeled "Reaction Progress," and a vertical axis labeled "Energy." Start by drawing a horizontal line at a certain height representing the reactant's energy level, and another horizontal line at a higher height representing the product's energy level. Draw an arc above the reactant's energy level, indicating the transition state where the activation energy (Ea) is at its maximum.
03

Label the activation energy and overall change in energy

Label the distance between the reactant's energy level and the peak of the arc as the activation energy, "Ea = 154 kJ/mol". Label the difference in energy between the reactant and product energy levels as "ΔE = 136 kJ/mol". #b. Finding the activation energy for the reverse reaction#
04

Calculate the energy difference between the peak and product energy level

Determine the energy difference between the peak level (maximum energy) and the product energy level. Since we already know that Ea (forward) is 154 kJ/mol and ΔE is 136 kJ/mol, we can calculate this difference by subtracting ΔE from Ea. Energy difference = Ea - ∆E = 154 kJ/mol - 136 kJ/mol = 18 kJ/mol
05

Calculate the activation energy for the reverse reaction

Now, to find the activation energy for the reverse reaction (Ea_reverse), add the energy difference obtained in Step 1 to the overall change in energy (∆E). Ea_reverse = energy difference + ΔE = 18 kJ/mol + 136 kJ/mol = 154 kJ/mol Thus, the activation energy for the reverse reaction is 154 kJ/mol.

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Most popular questions from this chapter

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. (a) $E_{a}=75 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-20 \mathrm{~kJ} / \mathrm{mol}$ (b) $E_{a}=100 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=+30 \mathrm{~kJ} / \mathrm{mol}$ (c) $E_{a}=85 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-50 \mathrm{~kJ} / \mathrm{mol}$

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is $6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\(, what is the reaction rate when \)[\mathrm{NO}]=0.035 \mathrm{M}\( and \)\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})$ What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step $1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)$ (fast) Step $2: \quad \mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)$ (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.
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