Choose the correct answer and explain why

a) We observe 2H2O2 2H2O and O2, but not the reverse. This is because

• O2 and H2O are too rare

• H of the reverse reaction is too negative

• The temperature is too high

• The Eact of the reverse reaction is too great

b) The system reaches equilibrium. Which reactions are still happening?

• None

• Both forward and reverse

• Only the forward

• Only the reverse

c) Which must be true at equilibrium?

• [reactant] = [product]

• [reactant] = 0

• k forward = k reverse

• forward reaction rate = reverse reaction rate

d) A B, Keq = 1. In the beginning [B] = 2, [A] = 1. So...

• [B] will increase

• [A] will increase

• [A] and [B] will remain unchanged

e) A + B 2C. Keq = 1. So at equilibrium...

• [A] = [B] = [C]

• 2[C] = [A] + [B]

• [C]2 = [A][B]

• (2[C])2 = [A][B]

f) Which of the following affects Keq?

• catalyst

• activation energy

• temperature

• initial concentration

g) In the reaction 2A B, the forward reaction rate = 0.1[A]2, the reverse reaction rate = 0.2[B]1. Keq = ?

• 2

• 1

• 0.5

• 0.2

h) A 2B. What is Keq in terms of [A] and [B] at equilibrium?

• [B] / [A]

• 2 [B] / [A]

• [B] x [B] / [A]

• 2[B] x 2[B] / [A]

I) In the reaction 2A B, at EQ, [A] = 0.33M = [B]. What is Keq?

• 3

• 1

• 0

• 0.33

j) N2O4 2 NO2, Keq = 2. In the beginning there was no NO2, and [N2O4] = 1 mol/L. What is [N2O4] at equilibrium?

• 2 mol/L

• 1.5 mol/L

• 1 mol/L

• 0.5 mol/L