Revised Q 5 Practice, Chemistry 111, Lycoming College, Feb. 23, 2005, Dr. Mahler

 

Fair Game for Friday’s Quiz: All previous material (Ch. 11, 12, 13). New Material –Chapter 14 Sections 3, 4 and 5. Relevant Homework: Ch. 14 problems - all assigned (keys on web: http://www.lycoming.edu/chem/spring2005/111/keys.htm)

 

Practice problems:

 

1) Possible essay / short answer type questions:

·         Chapter 14 Review Questions 1-9 (page 628)

 

2) For the reaction N2 (g) + 3 H2 (g)  ó 2 NH3 (g), Kp is 41 at 400 K. a) Calculate the value of Kc at 400 K. b) Say that the initial pressure of each gas is 0.234 atm at 400 K. What is the value of Qp ? c) Which way will the reaction in b) shift to reach equilibrium?

d) If you start with only 0.234 atm NH3, what will be the equilibrium pressures of each gas? Show all work and reasoning.

 

3) For the reaction 4 NH3 (g) + 5 O2 (g) ó 4 NO (g) + 6 H2O (g) at 298 K, ∆H = -905 kJ/mol. Starting at equilibrium in each case, what would a separate change in the system have as an effect? Specifically, the table below gives a change in the first column, then asks you to consider the effect of this change on something else, given in the second column. Your answer can be increase, no change, or decrease (check one of these).

 

Change

Quantity

Increase

No change

Decrease

Add NO

 

Amount of H2O

 

 

 

Add NO

 

Amount of O2

 

 

 

Remove H2O

 

Amount of NO

 

 

 

Remove O2

 

Amount of NH3

 

 

 

Add NH3

 

Kc

 

 

 

Remove NO

 

Amount of NH3

 

 

 

Add NH3

 

Amount of O2

 

 

 

Raise  temp.

 

Amount of NO

 

 

 

Raise pressure

 

Amount of H2O

 

 

 

Add catalyst

 

Amount of NH3

 

 

 

Lower temp.

 

Amount of H2O

 

 

 

 

 

 

 

Key for 3) The basic reasoning here is to use le Chatelier's principle. Look at what is added or removed or changed in the first column. Then see what the shift caused by the change is, and see how that shift changes the quantity asked for in the second column.

 

As an example, for the first one, adding NO is adding a product. Adding a product causes the reaction to shift to the left as written, towards reactants (it makes more reactants). This causes the amount of all products (including H2O) to decrease.

 

Adding reactant shifts the reaction to the right (make more products). Removing reactant shifts the reaction to the left (make more reactants). Finally, removing product shifts the reaction to the right (make more products).

 

Exothermic reactions have a negative ∆H and heat is essentially a product. Endothermic reactions have a positive ∆H and heat is essentially a reactant. So increasing the temperature is like adding product if exothermic and adding reactant if endothermic. The reaction given is exothermic, so raising the temperature is like adding product and shifts the reaction to the left (make reactants), decreasing the amount of NO. Lowering the temperature is like removing product, shift the reaction right (make products) and increasing the amount of products, including H2O.

Pressure increases cause the reaction to shift to the side of the equilibrium with fewer moles of gas, here the reactants (with 9 moles of gas vs. 10 moles for the products). So raising the pressure causes the amount of all products (including H2O) to decrease.

 

Finally, Kp (or Kc) is a constant and is not changed by adding or removing chemicals (their amounts shift to get back to the value of Kp or Kc). Catalysts will speed up the reaction but not change Kp or Kp or the concentrations of reactants or products at equilibrium.