Kaplan MCAT General Chemistry Review

Adding sodium acetate increases the number of acetate ions present. According to Le Châtelier’s principle, this change will push this reaction to the left, resulting in a decrease in the number of free H⁺ ions. Because pH is determined by the hydrogen ion concentration, a decrease in the number of free protons will increase the pH. This problem can also be solved with the K_a equation: K_a = [CH ₃COO^-] [H⁺]/[CH₃COOH]. An acid’s K_a will remain constant under a given temperature and pressure, eliminating (C) and (D). For K_a to remain unchanged while [CH₃COO^-] increases, [H⁺] must decrease or [CH₃COOH] must increase. A decrease in products would require an increase in reactants, and vice versa, so the final effect would be both an increase in [CH₃COOH] and a decrease in [H⁺]. Again, removing hydrogen ions will increase the pH of the solution.

7. A

If the sum of the exponents (orders) on the concentration of each species in the rate law is equal to 2, then the reaction is second-order. That’s the case in this situation, so option I is correct. Option II is incorrect because the exponents in the rate law are unrelated to stoichiometric coefficients, so NO₂ and Br₂ could be present in any ratio in the original reaction and still be first-order. Option III is incorrect because the rate can be affected by a wide variety of compounds. A catalyst, for example, could increase the rate, but it wouldn’t be included in the rate law. Any compound that would preferentially react with NO₂ or Br₂ (including strong acids/bases and strong oxidizing/reducing agents) would decrease the concentration of reactants and decrease the rate. Only I is valid; therefore, (A) is the correct answer.

8. C

In the first two trials, the concentration of XH₄ is held constant, while the concentration of O₂ is multiplied by 4. Because the rate of the reaction is also increased by a factor of approximately 4, oxygen must be a first-order reactant. Analyze the other reactant in the last two trials, XH₄. When you double its concentration, the rate of the reaction quadruples. That means XH₄ is a second-order reactant.

Until it becomes intuitive, use math to get to the right answer.

Write the following equations:

Trial 1: rate₁ = 12.4 = k[XH₄]^x [O₂]^y = k(1.00)^x (0.6)^y

Trial 2: rate₂ = 49.9 = k[XH₄]^x [O₂]^y = k(1.00)^x (2.4)^y

Plug in the values given in the question stem. When you divide Trial 2 by Trial 1, it simplifies to 4.02 = 4^y, so y is approximately equal to 1. Next, a similar procedure allows you to calculate the order of [XH₄] using trials 2 and 3.

Trial 2: rate₂ = 49.9 = k[XH₄]^x [O₂]^y = k(0.6)^x (2.4)¹

Trial 3: rate₃ = 198.3 = k[XH₄]^x [O₂]^y = k(1.2)^x (2.4)¹

Inserting the values you know, you find that the x exponent must be 2. Based on this, conclude that the experimental rate law is = k [XF₄]² [O₂].

9. B

The equilibrium of a reaction can be changed by several factors. If you recall Le Châtelier’s principle, you can rule out any answer choice that would shift the direction of an equilibrium. Adding or subtracting heat would shift the equilibrium based on the enthalpy change of the reaction. Adding reactant concentration would shift the equilibrium in the direction of the product, and the opposite would occur if concentrations were decreased. Changing the volume of a reactant would change its concentration, so that would have the same effect. Only (B), adding or removing a catalyst, would change the reaction rate without changing where the equilibrium lies.

10. A

The first step to answering this question is to write out the balanced equation for the reaction of H₂ and N₂ to produce NH₃ [N₂ + 3H₂ 2NH₃]. This means that K_eq is equal to [NH₃]²/([H₂]³[N₂]). Because the volume is 1 L, the amount of each gas (in moles) is equal to the value of the concentration of each gas (in M). We can plug these concentrations back into the K_eq expression to get K_eq = (0.05)²/([3]³[1]), which is equal to (0.0025)/(27). This is approximately equal to 0.0001, and approximations are appropriate for the MCAT.

11. D

A system is exothermic if energy is released by the reaction. For exothermic reactions, the net energy change is negative, and the potential energy stored in the final products is lower than the potential energy stored in the initial reactants. Point E, which represents the energy of the final products, is lower on the energy diagram than point A, which represents the energy of initial reactants. Thus, energy must have been given off. While point A is useful for determining the energy of the overall reaction, point B represents the activation energy of the first transition state, and point C suggests an intermediate. (C) references the difference between points D and E, which only indicates the change in energy from the transition state of the second reaction step to the final products.

12. B