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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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