Kaplan MCAT General Chemistry Review

A. E _system + E _surroundings = E _universe.

B. S _system + S _surroundings = S _universe.

C. H_system + H_surroundings = H_universe.

D. S_universe = 0 at T = 0 K.

16. A reaction coordinate for a chemical reaction is displayed in the graph below.

Which of the following terms describes the energy of this reaction?

A. Endothermic

B. Exothermic

C. Endergonic

D. Exergonic

Small Group Questions

1. Why is energy required to break bonds? Why is it released when bonds are formed?

2. Why is H so much easier to determine than instantaneous H (enthalpy)? What standard methods have been developed to help scientists find H through calculations?

3. What is the difference between exergonic and exothermic? What is the difference between endergonic and endothermic?

Explanations to Practice Questions

1. A

The process is adiabatic. Adiabatic describes any thermodynamic transformation that does not involve a heat transfer (Q). An adiabatic process can be either reversible or irreversible, though for MCAT purposes, adiabatic expansions or contractions refer to volume changes in a closed system without experimentally significant losses or gains of heat. The internal energy of the system changes (U = - , where U = Q - W, W = , and Q = 0). (B) is incorrect because isobaric processes occur at constant pressures. (C) is incorrect because an isothermal process requires a heat transfer at constant temperature, and there is no heat transfer along an adiabatic curve. Once it becomes clear that the process is adiabatic, that rules out (D).

2. D

There is not enough information in the problem to determine whether or not the reaction is nonspontaneous. Start with G = H-. If the signs of enthalpy and entropy are the same, the reaction is temperature-dependent. If the signs of these terms are different, we can find the sign of G without using temperature. The most common thermochemistry questions on the MCAT test your ability to manipulate and interpret the G = H- equation. Memorize it.

3. B

Sodium oxidizes easily at standard conditions. No calculation is necessary here. There is enough information in the problem to predict the equilibrium constant, eliminating (D). If K_eq <1, the reverse reaction is favored, indicating that the forward reaction is nonspontaneous. If K_eq = 1, the reaction is at equilibrium. If K_eq >1, the forward reaction proceeds spontaneously. The question states that the sample spontaneously combusts at room temperature (i.e., 25°C). The answer is (B), K_eq >1.

4. C

Combustion involves the reaction of a hydrocarbon with oxygen to produce carbon dioxide and water. Longer hydrocarbon chains yield greater amounts of combustion products and release more heat in the process (i.e., the reaction is more exothermic). Isobutane combusts less easily than n-butane because of its branched structure.

5. A

At first glance, this might seem like a math-heavy problem, but it really doesn’t require any calculations at all. We just have to keep track of which bonds are broken and which bonds are formed. Remember, breaking bonds requires energy, while forming bonds releases energy.

Nucleophile: methanol’s oxygen CH₃O–H Electrophile: acetic acid’s carbonyl carbon CH₃COOH

Through nucleophilic attack and leaving group separation, substitution occurs so that CH₃O takes the place of acetic acid’s OH to create methyl acetate. A proton transfer occurs between methanol and acetic acid’s leaving group, OH.

Bonds broken:

1 O–H (-464 kJ/mol): Proton leaves methanol.

1 C–O (-360 kJ/mol): Between the carbonyl carbon and the hydroxyl oxygen in acetic acid

Bonds formed:

1 C–O (+360 kJ/mol): Between the carbonyl carbon and the attacking oxygen

1 O–H (+464 kJ/mol): Between the leaving group oxygen and the transferred proton

The sum of all our bonding events is 0 kJ. We can reason through this intuitively. If one O–H bond is broken and another is made, the two values will cancel each other out, and the net energy change must be 0 kJ. Similarly, if one C–O bond is broken and another is made, the net energy change will also be 0 kJ.

6. C

Standard temperature and pressure indicates 0°C and 1 atm. Gibbs free energy is temperature-dependent. If a reaction is at equilibrium, G = 0. (C) is the correct answer.

7. B

The correct answer is (B), using K_eq = e^-G°/RT from G° = -RT ln(K_eq). Use e = 2.7. R is the gas constant, 8.314 Jmol^-1 K^-1, and T = 298 K, because of the standard-state sign.

Convert -4.955 kJ to -4,955 J and then round -4,955 to -5,000. Let’s also round 298 K to 300 K. The exponent’s denominator will be 8.314 × 300, which we can estimate as 2,500.

So we’re left with e^{(-5,000/-2,500)} = K_eq, or e² = K_eq.