SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 10

Chemical Equilibrium

Practice Exercises

1. For the reaction A + B  C + D, the equilibrium constant can be expressed as:

(A) 

(B) 

(C) 

(D) 

2. The concentrations in an expression of the equilibrium constant are given in

(A) mol/mL

(B) g/L

(C) gram-equivalents/L

(D) mol/L

3. In the equilibrium expression for the reaction

   BaSO4(s)  Ba2+(aq) + SO42−(aq)

   Ksp is equal to

(A)  [Ba2+][SO42−]

(B) 

(C) 

(D) 

4. The Kw of water at 298 K is equal to

(A) 1 × 10−7

(B) 1 × 10−17

(C) 1 × 10−14

(D) 1 × 10−1

5. The pH of a solution that has a hydrogen ion concentration of 1 × 10−4 mole/liter is

(A) 4

(B) -4

(C) 10

(D) -10

6. The pH of a solution that has a hydroxide ion concentration of 1 × 10−4 mole/liter is

(A) 4

(B) -4

(C) 10

(D) -10

7. A small value for K, the equilibrium constant, indicates that

(A) the concentration of the un-ionized molecules must be relatively small compared with the ion concentrations

(B) the concentration of the ionized molecules must be larger than the ion concentrations

(C) the substance ionizes to a large degree

(D) the concentration of the un-ionized molecules must be relatively large compared with the ion concentrations

8. In the Haber process for making ammonia, an increase in pressure favors

(A) the forward reaction

(B) the reverse reaction

(C) neither reaction

9. A change in which of these conditions will change the of an equilibrium given as a starting point?

(A) Temperature

(B) Pressure

(C) Concentration of reactants

(D) Concentration of products

10. If Ca(OH)2 is dissolved in a solution of NaOH, its solubility, compared with that in pure water, is

(A) increased

(B) decreased

(C) unaffected

The following questions are in the format that is used on the SAT Subject Area Test in Chemistry. If you are not familiar with these types of questions, study before doing the remainder of the review questions.

Directions: The following set of lettered choices refers to the numbered questions immediately below it. For each numbered item, choose the one lettered choice that fits it best. Every choice in the set may be used once, more than once, or not at all.

Questions 11–15

(A) the free energy is always negative

(B) the free energy is negative at lower temperatures

(C) the free energy is negative at high temperatures

(D) the free energy is never negative

(E) the system is at equilibrium, and there is no net reaction

Complete the sentence with the appropriate choice.

11. When enthalpy is negative and entropy is positive,

12. When enthalpy is positive and entropy is positive,

13. When enthalpy is negative and entropy is negative,

14. When enthalpy is positive and entropy is negative,

15. When ΔG, free energy, is zero,

Answers and Explanations

1. (C) The correct setup of Keq is the product of the concentration of the products over the products of the reactants.

2. (D) The concentrations in an expression of the equilibrium constant are given in moles/liter (mol/L).

3. (A) This reaction is the equilibrium between a precipitate and its ions in solution. Because BaSO4 is a solid, it does not appear in the solubility product expression. Therefore, Ksp= [Ba+2][SO42−].

4. (C) This is the expression of water.

5. (A) Because the pH is defined as the negative of the log of the H+ concentration, it is -(log of 10−4), which is -(-4) or 4.

6. (C) pH + pOH = 14. In this problem, the pOH = -log [OH] = -log [10−4] = -(-4) = 4. Placing this value in the equation, you have pH + 4 = 14. Solving for pH, you get 10.

7. (D) For Keq to be a small value, the numerator of the expression must be small compared with the denominator. Because the numerator is the product of the concentrations of the ion concentrations and the denominator is the product of the un-ionized molecules, the concentration of the un-ionized molecules must be relatively large compared with the ion concentrations.

8. (A) In the reaction for the formation of ammonia, N2 + 3H2 ↔ 2NH3 + heat (at equilibrium). An increase in pressure will cause the forward reaction to be favored, because the equation shows that four molecules of reactants are forming two molecules of products. This effect tends to reduce this increase in pressure by the formation of more ammonia.

9. (A) Only the changing of the temperature of the equilibrium reaction will change the of the equilibrium given at the starting point.

10. (B) Because there already is a concentration of (OH)- ions from the NaOH in solution, this common ion effect will decrease the solubility of the Ca(OH)2.

11. (A) When enthalpy is negative and entropy is positive, the free energy is always negative.

12. (C) When enthalpy is positive and entropy is positive, the free energy is negative at high temperatures.

13. (B) When enthalpy is negative and entropy is negative, the free energy is negative at lower temperatures.

14. (D) When enthalpy is positive and entropy is negative, the free energy is never negative.

15. (E) When ΔG, free energy, is zero, the system is at equilibrium and there is no net reaction.