Equilibrium II

Today I handed back exams and we continued to talk about equilibrium. The average on the exam was around 115/150.

In case you misplaced your problem set (or just want another one...), here's it is:

Chem 210 – Summer 2009 – Problem Set #4

1. Oxygen gas and chlorine gas react in the atmosphere to form chlorine dioxide, one of the substances responsible for ozone destruction. In a laboratory experiment, 18.237g of oxygen gas and 31.294g of chlorine gas are combined in a 8.00L vessel and allowed to reach equilibrium. At equilibrium, the chlorine concentration is found to be 48.9mM. What is the equilibrium constant for this reaction? Is this reaction product-favored?

2. Carbon dioxide reacts with ammonia (NH₃) in the gas phase to produce formamide (HCONH₂) and oxygen. If 15.215g of CO₂(g) and 4.139g of ammonia are combined in a 3.50L vessel and allowed to react equilibrium, the formamide concentration is found to be 24.8mM. What is the equilibrium constant for this reaction? Is this reaction product-favored or reactant-favored?

3. Graphite (solid carbon) reacts with oxygen gas to form carbon dioxide gas. You have sealed 23.84g of graphite and 53.97g of oxygen in a 12.00L vessel and allowed the system to react equilibrium at 73.91°C. If the equilibrium constant value is 294.7 at this temperature, what are the equilibrium concentrations of all reactants and products?

4. At 149.79°C, methane, CH₄(g), reacts with oxygen to form carbon dioxide and water with an equilibrium constant value of 6.43x10⁶. If you combine 6.85g of methane with 29.99g of oxygen in a 18.00L vessel, raise the temperature to 149.79°C, and allow the mixture to react equilibrium, what are the concentrations of all reactants and products?

Questions...

A couple questions came in from Exam 1, Summer 2007:

12. You have prepared a solution by diluting 18.64mL of 1.16M potassium iodate solution to50.00mL with water. What is the concentration of iodate ions in the resulting solution?
a. 3.11 M
b. 0.216 M
c. 0.432 M
d. 0.242 M
e. 2.31 M
For dilutions, you can use the formula C1V1 = C2V2. C1 = 1.16M (the stock concentration); V1 = 18.64mL (the stock volume); C2 = the "new" concentration; V2 = 50.00mL (the diluted volume). Every potassium iodate unit contains 1 iodate ion, so the concentration of potassium iodate is the same as the concentration of iodate.
(1.16M)(18.64mL) = C2(50.00mL)
C2 = 0.432M

13. You have prepared a solution by dissolving 1.38 mols of sugar (C6H12O6) in 500.0g of water. What is the boiling point of the resulting solution?
a. 1.4ºC
b. 105.13ºC
c. 101.4ºC
d. 98.6ºC
e. 5.13ºC
Boiling point elevation uses molality, so:
1.38mols sugar / 0.5000kg solvent = 2.76m
{delta}T = (0.52)(2.76)(1)
NOTE: I haven't included units because they would be hard to type. 0.52 deg.C/m is the boiling point elevation constant for water, it will be given on the front of the exam. Sugar is a molecular solute, so the van't Hoff factor is 1. Solving, the boiling point will CHANGE by 1.4 deg.C, so the boiling point of the solution is 100.0+1.4 = 1.104 deg.C

14. A reaction is found to be second order with respect to reactant A and zero order with respect to reactant B. If [A]o = 0.942M, [B]o = 0.613M and k = 2.49x10-2 M-1min-1, what is the initial rate of the reaction?
a. 1.44x10-2 M/min
b. 4.31x10-2 M/min
c. 2.21x10-2 M/min
d. 9.36x10-3 M/min
e. 2.81x10-2 M/min
You don't need a time for this one, the problem is basically giving you all the parts of the rate law equation except the rate itself. If the reaction is second order w.r.t. A and zero order w.r.t. B, the rate law expression is:Rate = k [A]^2Plugging in the given values for k and [A], should give you the correct answer.

15. A reaction is found to be second order with respect to ammonium ion, a reactant. If [NH4+]o = 3.34M and k = 1.28 M-1sec-1, what will the concentration of ammonium ion be after 2.16minutes have passed?
a. 0.326 M
b. 3.02x10-72 M
c. 0.575 M
d. 6.02x10-3 M
e. 1.18 M

Use the second order integrated rate law. If you got the wrong answer on this one, check your time units, k is given in seconds and the time is given in minutes. Convert t to seconds before you start.
1/[NH4+]t = (1.28 M-1sec-1)(129.6sec) + 1/3.34 = 166.2 M-1
[NH4+]t = 6.02x10-3 M

Other questions, let me know...

Kinetics...

I've posted a few more old exams on my web page, the full set from Summer 2008 and the first couple exams from Winter 2009. If you have questions, email me and I'll post answers to the blog. Good luck in your preparation for the exam.

Day 1...

Today we looked at states of matter, solutions, and colligative properties.

Your lab experiments will be available online, please visit my web page to print them out. (www.mnstate.edu/bodwin, click on the "Chem 210L" link in the left panel, then click on the experiment name) The class syllabus is also posted on the Chem 210L page. Be sure to print and read the experiment before class.

I will be updating many things on my web page throughout the next day or two...

Summer 2009

Welcome to the blog for summer 2009. This is where I will post answers to questions people ask and make class announcements, so if you have a question about class, check here first.