Kaplan MCAT General Chemistry Review

2) How does this rate law differ from the one that you might expect if this reaction were to be carried out in solution, instead of in the gas phase?

3) How would the key intermediates differ between this reaction in the gas phase and in solution?

Remember: Intermediates are assumed to exist for only a brief period of time because they are produced in one step and consumed in another. Therefore, their concentration cannot be measured, and they must be eliminated from the rate law.

3) Combine constants and simplify the rate law.

Combine all of the constants and concentrations.

Remember: A constant times a constant times a constant, and so on, is just another constant.

Things to Watch Out For

In this case, you may assume that the stoichiometric coefficients of each reactant are equal to the order. When you are presented with rate data, you may not make this assumption but must use the rate data to determine order.

Le Châtelier’s Principle

Key Concepts

Chapter 5

Le Châtelier’s principle

System stress

A chemistry student adds solid copper sulfate to water at room temperature. The resulting solution has an emerald blue color reminiscent of azulene. The student then adds a piece of aluminum foil to the solution and watches as a small hole develops in the foil. What could the student do to increase the rate at which the hole forms?

CuSO₄ (s) + 5 H₂O (l) CuSO₄ • 5 H₂O (aq) H < 0, G < 0

2 Al (s) + 3 Cu²⁺ (aq) 2 Al²⁺ (aq) + 3 Cu (s) H < 0, G < 0

1) Identify the direction of the desired reaction.

To the right of equation 2

The hole in the aluminum foil indicates that the solid is dissolving into solution. Looking at the reaction equations, solid aluminum reacts with Cu²⁺ to form Al²⁺, the desired end product.

Takeaways

Le Chatelier’s principle basically puts K_eq and Q into words. Chemical reactions attempt to reach equilibrium. Adding more reactant, for instance, makes the reaction move to the right: More product needs to be formed to balance out the addition.

2) List the different types of stress that can be applied to any system.

Pressure, temperature, concentration

On the MCAT, liquids and solids are incompressible, so altering the pressure at which the reaction occurs should have no effect. However, we can alter the temperature or the concentration of the reactants or products. To increase the rate at which the hole forms, we are looking to push the reaction to the right.

3) Consider the effect of various system stresses: heat.

Run the reactions at a lower temperature to increase the rate of hole formation.

We are told that both reactions are exothermic; thus, we can rewrite them in this generic format: A B + , where is heat. Heat is a product of this reaction. Increasing the heat will push the reaction to the left: It is as if we’ve added more of the “product.” Alternatively, if we drop the temperature at which the reactions are conducted (i.e., remove the heat product), we push the reaction to the right and favor the formation of the hole.

MCAT Pitfall: Increasing the temperature would favor formation of the original CuSO₄solid due to the exothermic nature of the two reactions.

Things to Watch Out For

Be particularly careful with ionic species and gases. Ionic species can dissociate in water, depending on their electrolytic strength, and may result in multiples of the original concentration of solid. An increase in pressure favors the side of the reaction with fewer molecules of gas.

4) Consider the effect of various system stresses: concentration.

Increase the concentration of the reactants or remove the product.

From the second reaction, it is clear that it is the Cu(II) ion that is reacting with the aluminum foil. How can we get more Cu(II) ion in solution? Add more solid copper sulfate. Alternatively, we could remove the final product. Removing Cu(II) after copper sulfate pentahydride dissociates will disable the second reaction. Removing SO₄^2-, however, will cause more hydrate to form. Better yet, we could simply remove the solid copper that plates out when the aluminum atoms are ionized.

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