DOC PREVIEW
UT CH 302 - CH302 Worksheet 4 Answer Key
Type Miscellaneous
Pages 2

This preview shows page 1 out of 2 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Spring 2009 CH302 Worksheet 4 Answer Key: Chemical Equilibria and an Introduction to Water Chemistry Please note that all calculations below can be worked without the aid of a calculator. 1. Write a mass action quotient for the following chemical equation: CaCO3(s) Ù CaO(s) + CO2(g) K = [CO2] 2. Write a mass action quotient for the following chemical equation: H2O(l) Ù H+(aq) + OH-(aq) K = [H+]·[OH-] 3. Based on your answers to numbers 1 & 2, what types of reactants and products always appear in mass action quotients? What types never appear in mass action quotients? Why? Gases and dissolved species [e.g. CO2(g), H+(aq) & OH-(aq)] always appear and solids and liquids [e.g. CaCO3(s), CaO(s) & H2O(l)] never appear. This is because the former have a changeable activity/concentration and the latter have a constant activity. 4. What are the units of all equilibrium constants? Equilibrium constants have no units. 5. Complete the RICE diagram below. Express unknown quantities in terms of X. Reaction CO(g) + NO2(g) Ù CO2(g) + NO(g) Initial 5 M 5 M 0 M 8-X M Change -X M -X M +X M +X M Equilibrium 5-X M 5-X M X M 8 M 6. Assuming K = 0.5, what was the initial concentration of NO(g)? 0.5 = 8X / (5-X)2 0.5(5-X)2 = 8X 0.5X2 - 5X + 12.5 = 8X 0 = 0.5X2 - 13X + 12.5 X = {13 ± [169 - 4(0.5)(12.5)]0.5}/2(0.5) = 13 ± 12 = 1 The initial concentration of NO(g) was therefore 7 M. 7. How would the chemical system in numbers 5 & 6 respond to a decrease in volume? What about addition of CO(g)? What about removal of NO(g)? Since the number of moles of gas is the same on both sides of the equation, decreasing the volume of the system will not affect Q or K, so nothing will occur. Addition of CO(g) will push the reaction to the right. Removal of NO(g) will pull the reaction to the right. 8. The Clausius-Clapeyron equation and the van't Hoff equation are very similar in appearance. What is the main difference between the two equations and why are they so similar? The main difference is that the Clausius-Clapeyron equation relates vapor pressures to temperatures and the van't Hoff equation relates equilibrium constants. They are so similar because vaporization is an equilibrium process and since liquids have a constant activity, the pressure of the gas phase is directly proportional to the equilibrium constant. 9. Based on the van't Hoff equation, how will an exothermic reactions equilibrium constant respond to changes in temperature? What about an endothermic reaction? For an exothermic reaction, K should be inversely proportional to T. For an endothermic reaction, K should be directly proportional to T.10. In a 1 liter container you initially have one mole of each species below. 4Fe(s) + 3O2(g) Ù 2Fe2O3(s) K = 1012 What happens as this system approaches equilibrium? Since Q is substantially less than K initially (103 vs 1012), as the system approaches equilibrium the pressure of oxygen will decrease dramatically and most of the metallic iron will become rust. 11. List the 7 strong acids from memory. Hydrochloric (HCl), Hydrobromic (HBr), Hydroiodic (HI), Sulfuric (H2SO4), Nitric (HNO3), Chloric (HClO3) and Perchloric (HClO4) 12. List the 8 strong bases from memory. Lithium Hydroxide (LiOH), Sodium Hydroxide (NaOH), Potassium Hydroxide (KOH), Rubidium Hydroxide (RbOH), Cesium Hydroxide (CsOH), Calcium Hydroxide [Ca(OH)2], Strontium Hydroxide [Sr(OH)2], Barium Hydroxide [Ba(OH)2], 13. What would be the pH of 1 liter of a 10 M solution of nitric acid? pH = -1 14. How many grams of barium hydroxide would be needed to neutralize the solution in number 13 10 M × 1 L HNO3 = 10 mol HNO3, which would be 10 mol H+ So we need 10 mol OH-, which would require 5 mol Ba(OH)2 171 g·mol-1 × 5 mol = 855 g (OH)2 15. Rank the following solutions from lowest pH to highest pH: 0.5 M HI, 0.1 M Sr(OH)2, 2 M HClO4, 0.5 M H2SO4, 0.1 M NaOH, 2 M KOH. 2 M HClO4 < 0.5 M H2SO4 < 0.5 M HI < 0.1 M NaOH < 0.1 M Sr(OH)2 < 2 M KOH 16. What is meant by the term "autoprotolysis of water?" What chemical equation describes this? Autoprotolysis of water refers to water's tendency to spontaneously split into a hydroxide ion and a proton. H2O(l) Ù H+(aq) + OH-(aq) 17. What is Kw? What is its value to at room temperature? How does temperature influence Kw? Kw = [H+]·[OH-] = 10-14 at room temperature. Since autoprotolysis is endothermic, Kw is directly proportional to temperature. 18. What would be the pOH of a 0.5 M solution of a weak acid with a Ka = 2 × 10-4? [H+] = (Ka·Ca)0.5 = (2 × 10-4·0.5)0.5 = (10-4)0.5 = 10-2 pOH = 14 - pH = 14 - 2 = 12 19. Complete the RICE diagram for the reaction of a general weak base (:B) with water. Use general terms (e.g. Cb for initial concentration of base). Reaction :B(aq) + H20(l) Ù BH+(aq) + OH-(aq) Initial Cb 0 10-7 Change -X +X +X Equilibrium Cb - X X 10-7 + X 20. Write K for the reaction in number 19 and then substitute the values from the RICE diagram. K = [BH+]·[OH-] / [:B] = (X)·(10-7 + X) / (Cb -


View Full Document

UT CH 302 - CH302 Worksheet 4 Answer Key

Documents in this Course
Exam 2

Exam 2

6 pages

Exam 3

Exam 3

8 pages

Acids

Acids

21 pages

Exam 3

Exam 3

7 pages

SYLLABUS

SYLLABUS

15 pages

ex1s08

ex1s08

11 pages

Load more
Download CH302 Worksheet 4 Answer Key
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view CH302 Worksheet 4 Answer Key and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view CH302 Worksheet 4 Answer Key 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?