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Chapter 12: Problem 68

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
For each case, we draw the energy profiles as follows: Case a: \(\Delta E = +10 \mathrm{kJ/mol}, E_a = 25 \mathrm{kJ/mol}\), we plot an initial energy level at 0 kJ/mol, a transition state at 25 kJ/mol, and a final energy level at 10 kJ/mol, with a smooth curve connecting them. Case b: \(\Delta E = -10 \mathrm{kJ/mol}, E_a = 50 \mathrm{kJ/mol}\), we plot an initial energy level at 0 kJ/mol, a transition state at 50 kJ/mol, and a final energy level at -10 kJ/mol, with a smooth curve connecting them. Case c: \(\Delta E = -50 \mathrm{kJ/mol}, E_a = 50 \mathrm{kJ/mol}\), we plot an initial energy level at 0 kJ/mol, a transition state at 50 kJ/mol, and a final energy level at -50 kJ/mol, with a smooth curve connecting them.

Step by step solution

01

Case a:

Given: \(\Delta E = +10 \mathrm{kJ/mol}, E_a = 25 \mathrm{kJ/mol}\) 1. Draw the x-axis (Reaction coordinate) and y-axis (Energy). 2. The initial energy level (Reactants) is at 0 kJ/mol. 3. Since \(\Delta E\) is positive, the final energy level (Products) will be higher than the initial level by 10 kJ/mol. 4. The activation energy is given as 25 kJ/mol, so the energy barrier (Transition state) will be 25 kJ/mol above the initial energy level. 5. Connect the initial, transition state, and final energy levels with a smooth curve.
02

Case b:

Given: \(\Delta E = -10 \mathrm{kJ/mol}, E_a = 50 \mathrm{kJ/mol}\) 1. Draw the x-axis (Reaction coordinate) and y-axis (Energy). 2. The initial energy level (Reactants) is at 0 kJ/mol. 3. Since \(\Delta E\) is negative, the final energy level (Products) will be lower than the initial level by 10 kJ/mol. 4. The activation energy is given as 50 kJ/mol, so the energy barrier (Transition state) will be 50 kJ/mol above the initial energy level. 5. Connect the initial, transition state, and final energy levels with a smooth curve.
03

Case c:

Given: \(\Delta E = -50 \mathrm{kJ/mol}, E_a = 50 \mathrm{kJ/mol}\) 1. Draw the x-axis (Reaction coordinate) and y-axis (Energy). 2. The initial energy level (Reactants) is at 0 kJ/mol. 3. Since \(\Delta E\) is negative, the final energy level (Products) will be lower than the initial level by 50 kJ/mol. 4. The activation energy is given as 50 kJ/mol, so the energy barrier (Transition state) will be 50 kJ/mol above the initial energy level. 5. Connect the initial, transition state, and final energy levels with a smooth curve.

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Most popular questions from this chapter

A popular chemical demonstration is the "magic genie" procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncatalyzed) reaction is 70.0 \(\mathrm{kJ} / \mathrm{mol}\) . When the catalyst is added, the activation energy (at \(20 .^{\circ} \mathrm{C} )\) is 42.0 \(\mathrm{kJ} / \mathrm{mol}\) . Theoretically, to what temperature ( \((\mathrm{C})\) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at \(20 .^{\circ} \mathrm{C} ?\) Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.

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