Monday, February 14, 2011

January 25, 2011

It was the joint decision of the Period 2 class to postpone the blog for second semester. The Period 3 class has decided to maintain the daily procedure of a scribe capturing the essence of the lesson. The period 3 blog can be found here.

Tuesday, January 11, 2011

Wednesday, January 11

Today we started off class by reviewing the webassign reading sheet. It is located on page 25 in your unit packet.

Here are the answer to the Chapter 8.2 reading sheet
1. c
2. c
3. d
4. a, a
5. c
6. a.-520
b. -258.8
c. 0
7. d
8. -6534.8
9. incorrect
The delta H values must be multiplied by the coefficients of the equation.
10. incorrect
The proper method involves subtracting delta H reactants from products.

Next we did some problems on pages 8, 11 and 12 in our packets.

On page 8 we did problem 5 together.

2940kg HNO3 * 1000g/kg * 1 mol HNO3/63.01g HNO3 * 69.0 kJ/2 mol HNO3= 1.61*10^6kJ

On page 11 we did problem 2

C(s)+O2(g)---->CO2 Delta H= -394kJ
2H2(g)+O2---->2H2O Delta H=-572kJ
CO2(g)+2H2O---->CH4(g)+2O2(g) Delta H=890
------------------------------------------ -------------------
C+2H2---->CH4 Delta H=-76

On page 12 we did problem 4

2Ca+4C---> CaC2 Delta H= -125.6
2CO2----> 2O2+2C Delta H= 788
5O2+2CaC2----> 2CaCO3+2CO2 Delta H= -3076
--------------------------------------- ------------------
2Ca+2C+3O2---->CaCO3 Delta H= -2414

Lastly we got our lab notebooks out and finished our calculations for Lab TC7.
These were my calculations:
Qsurroundings=100g * 4.18J/g*C *39.1 C
moles Ca= 2/40.078=.05 moles Ca
.05 moles*16343.8J= -326.876kJ/mol

Delta H reaction= Delta (heat of formation of Ca(OH)2)-2(-286kJ)
-326.876= x+572
Delta heat of formation of Calcium hydroxide is -898.876kJ.

Theoretical Answer is 1003kJ

Monday, January 10

We started off class with Mr. Henderson returning our lab notebooks. We then went over the previous nights weba ssign which was reading assignment 8.3-4.

We then did problem #2 on page 7 in our packets. The problem gave us 2 CH3OH + 3 O2 ---> 2 CO2 + 4 H2O deltaT= -1204kJ. Part A of the question asked how much energy was released by 1 mole of CH3OH. This is a stoichiometry problem so you multiply 1.0 mol CH3OH x 1199kJ/2 mol CH3OH. This comes out to be 599.5 kJ produced and that is the answer.
Part B asked for the amount of heat 2.0 grams of CH3OH. This is similar to the first question but you nedd to find out how many moles 2.0 grams equals. Divide 2.0 by the molar mass which is 32.04 and then do the same multiplication as the problem before. The answer is 37.42 kJ.

We then did problems 1 and 3 on page 11. This page deals with Hess's Law. These problems are like a puzzle and you need to get the two equations to equal the original equation. Since C is in the reactants for both the first given equation and the equation you are trying to get you need to multiply the whole first equation by 2 to make the coefficients match. This means you also have to multiply the deltaT by 2 also. You then need to flip the second equation to get 2 CO into the product side. Then add the two equations together and you should get the original equation which is 2 C + O2 ---> 2 CO. Then add the two deltaT's together to figure out the total energy used and the answer is -221kJ.

We finished by starting the Heat of Formation Lab. We just did the measurements for the lab. We measured the volume of H2O, Mass of calcium, Initial Temperature of H2O and final Temperature of H2O. the purpose of the lab is to find the heat of formation of calcium hydroxide.

Thursday, January 6, 2011

Thursday, January 6

Today in class, Mr. Henderson discussed more on thermochemistry and introduced us to a new term called enthalpy. Basically, enthalpy is a form of chemical energy which is related to heat.

  • Here are the answers to part 1 of page 5 which was completed in class.
a. EX g. EN
b. EN h. EX
c. EX i. EN
d. EN j. EX
e. EX k. EX
f. EN l. EX

  • Exothermic reactions will always release heat to the surroundings, and all combustion reactions will have heat. Therefore, heat is listed on the products' side.
CH4 + 2O2 -> CO2 + 2H2O + 185 kJ
reactants -> products + heat
ΔH = -185 kJ
The 185 kJ and the heat goes to the surroundings. The ΔH is the change in enthalpy. The reason why ΔH in this reaction is negative is because the energy in the system is lost to the surroundings, which gains it. Therefore, the enthalpy in exothermic reactions will always have a negative value.

  • Endothermic reactions will always gain energy from the surroundings. Heat is listed on the reactants' side.
6.0 kJ + H2O(s) -> H2O(l)
heat + reactants -> products
ΔH = +6.0 kJ/mol
Heat is needed in all endothermic reactions to make the products. This particular reaction is in the process of melting, also called fusion. So you would say the ΔH of fusion in this reaction equals 6. ΔH is positive in this reaction because the system gains energy while the surroundings lose it.

If A -> B + 100 J
then... 100 J + B -> A
then... 2A -> 2B + 200 J
then... 300 J + 3B -> 3A

Heat can be a stoichiometry term.

  • In class, we did page 4, problem 4 in the packet.
a. A 250.0 mL sample of liquid water (density = 1.0 g/mL) is cooled from 65.0°C to 38.5°C as 70.4 grams of ice is melted. Determine the amount of heat lost by the liquid water. PYSW

In order to solve this, we would have to use the formula q = m*c*Δt.

So we plug in our information:
(250.0 g) (4.18 J/g°C) (-26.5 °C)
The -26.5 comes from the 38.5 - 65.0, because the initial is subtracted from the final.
Answer = q(H2O) = -27692.5 J
The negative sign means that heat was lost from the water.

b. (This question was changed to:) Determine the moles of ice that melted. PYSW
70.4 g of ice * 1 mol/18.01 g of H2O = 3.911 mol of ice = Answer
(Easy stochiometry :D)

c. Determine the heat of fusion of ice in units of kJ/mol.
ΔH = 27692.5/3.911 mol * 1 kJ/1000 J
(Converting factor to make J to kJ)
Answer = 7.08 kJ/mol

We ended class by working on our previous labs until the bell rang.
Don't forget to do Webassigns!

Wednesday, January 5, 2011

Wensday, January 5th

to start class we went over the reading sheet on page 36. here are the answers.

next we worked on page 2 in out packet. all of page 2 deals with the consept of exothermic and endothermic processes. an exothermic processes is one where the heat leaves the system to heat up the sorroundings. endothermic is the opposite, in an endothermic process the system is colden and recives energy from the sorroundings. the answers are below.

After page 2 we did half of page 3. the backbone of page 3 is math, specificly the equation Q=M*C*Delta-T. for question 1 you must simply insert the values into there place in the equation. you get Q=100*4.18*6.6. the 100 is the mass of the H2O in grams, the 4.18 is the specific heat of H2O, and 6.6 is the inital temp-the final temp. if you solve that out you should get ~2760 if you are using 3 digit significant figures so the answer A) is 2760 J. B) is the same as A exept it is in KJ so the answer is 2.76 KJ. C is found almost the same way as A. you must plug in numbers to the equation but first you have to move the equation around. change Q=M*C*Delta-T to the equation C=Q/M*Delta-T. in this case it looks like this: C=-2760/6.52*-57.2. the 2760 is the Q from the last equation, the M is the mass of iron, and the -57.2 is the change in temp from the starting temp to the final temp. you can apply this method to finish our lab from monday.

after that we did a lab called heat of fusion of ice. to finish this lab you must use the formulia Q=M*C*Delta-T again. you must solve for Q, the M is 100 for the grams of H2O, the C is 4.18 for the specific heat of H2O and the final value is your groups temp change. you solve for Q my group got -9070 J. after you find the quantity of heat find the number of moles of H2O you had by deviding the mess in grams of your ice by 18. my group got 1.361 moles of H2O. ofter you find the moles you devide the quantity of heat by that number and get your final answer in J/M. for examplw my final answer way 6660 J/M