Today in chemistry we started by looking at the answers for the 6.5 reading sheet. Some tips that Mr. Henderson left for us were that you should remember to pay attention so see whether H2O is in a liquid state or a gas because the energy for each one is different. He also reminded us to use coefficients of balanced chemical equations as multipliers and also that for elements in standard states ΔHf=0
On the board were notes on using ΔHf values to calculate heats of reaction. They were the following:ΔH RXN = Sum of ΔH°f products – Sum of ΔH°f reactants
What the above basically means is that the heat of reaction from any reaction is equal to sum of the heats formation values for reactants subtracted from those of the product.
Next we moved to page 9 in our packets and we started on number 2. The problem was the following:
The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide can be used as an oxidizing agent. The balanced equation for the reaction is:
5N2H4(l) + 4N2H3CH3(l) -----------------> 3N2(g) +4H2O(g),
Determine the heat of the reaction.
What we did first was move the product side in the reaction over. Next we looked at the chart and we found the ΔH°f values of the compounds and in this N2 was 0 because it was an element in its natural state. For H2O in a GAS state the chart said it was -242 kJ/mol. Then we moved on to the reactant part of the equation. We found the ΔH°f values for the reactants and we multiplied them by the coefficients attached to each one of the compounds. So for this problem we had to multiply -20 kJ/mole by 5 for the 5N2H4(l) and that is -100 kJ/mole. Next we found that the heat formation values of N2H3CH3(l) were 54 kJ/mole, and because the compound was accompanied with a molar number of four, you have to multiply 54 by 4 to get 216 kJ/mole. When you add the two heat formation numbers you get -46 kJ/mole. Finally to finish the problem you subtract the first sum from the second one. So it would be -242 - -46 and you get -196 kJ/mole and that would be your answer.
Then we also did #3 following and if you weren’t here that one would be good practice. Next we looked at Wills blog and Mr. Henderson briefly went over its content.
We then went back to our packets and looked at page 11 and started with #2. The problem was the following:

Calculate the ΔH for: C(s) + 2 H2(g) → CH4 (g)
Given: C(s) + O2(g) → CO2 (g) ΔH = -394 kJ
2 H2(g) +O2(g) → 2 H2O (l) ΔH = -572 kJ
CH4(g) + 2O2 (g) → CO2 (g) + 2 H2O ΔH = -890 kJ

To solve the problem you basically have match up variables that aren’t in the model equation (the one on top) and cancel out the ones that aren’t by either switching around the product and reactants or by multiplying the whole equation by a number to put coefficients to the numbers could match up and cancel out. Then by adding up the ΔH of the equations to finally end up with the model one you wanted and that will be the answer to the problem which in this case is ΔH = -76 kJ. We finished the class by working on our labs TC4.