Tuesday November 27th

Hey Molly, we missed you today. We reviewed for the test, which included about 3 problems from the practice test and going over key topics to study in each chapter. After everyone left I finished problem 7. I'll try to write it out here but the subscripts won't work, so keep that in mind. Here it is in steps.
1. Looking at the final equation you need, the bottom two equations in the list need to be switched. If you do this, the equations are as follows
MnO2(s) + CO(s) --> MnO(s) + CO2(g)
3MnO(s) + CO2(g) --> Mn3O4(s) + CO(g)
2Mn3O4 (s) + CO2 --> 3Mn2O3(s) + CO(g)

2. In order to start crossing out spectator species, you need to perform some stoichiometric adjustments in the coefficients. For example, to cross out MnO, multiply all species in the top equation by 3 so that "3MnO" can be crossed out. But doing this is not enough. You also need to multiply the bottom equation by 2 in order to cross out 2Mn3O4. Then go back to equation one to make everything even and multiply all species by two. Doing this you end up with...
6MnO2(s) + 6CO(s) --> 6MnO(s) + 6CO2(g)
6MnO(s) + 2CO2(g) --> 2Mn3O4(s) + 2CO(g)
2Mn3O4 (s) + CO2 --> 3Mn2O3(s) + CO(g)

3. Now you are ready to cross out spectators. Look for repeats on both sides. What happens when you add them up?
6MnO2(s) + 6CO + 3CO2 --> 3Mn2O3(s) + 6CO2 + 3CO

4. You can simplify this...
2MnO2(s) + CO(g) --> Mn2O3(s) + CO2(g)

5. If you multiply the enthalpies by the same amount as you did in step 2 and add them up you get your answer: 150 + (120*2) + (-140*6) = 570 kJ But remember that this number is two times what the simplified reaction you are given to solve for. Dividing by two we get 285 kJ. The only thing close to this is -290 kJ. There's a reason for this, sig figs! Cool huh? Okay, maybe not "cool" in the iPod sense, but cool nonetheless.

Remember, we have a study session for those interested tomorrow after school.