Chemistry 6.5 Hess' Law

Hess' Law continued

C2H4(g) + 6F2(g) ----> 2CF4(g) + 4HF(g) -Look at first reactant. Where does this reactant show up in the provided reactions? C2H4 shows up in last reaction on products side. You need it to be on reactants side so you have to flip this whole reaction backwards. (picture) -Therefore delta H will become -52.3kJ -Next reactant: F2 is in the reactants side of the first reaction AND the second reaction. If any of the species in your desired reaction show up in more than 1 of the provided reactions, skip it and come back to it at the end. -Next product is 2CF4. CF4 appears as a product in the second reaction. But we want 2 moles of it so multiply reaction by 2 (picture) -Therefore delta H will become (2)(-680)= -1360kJ -Last product is 4HF(g). 2HF appears in the products side of the first reaction. But we want 4 moles of it so multiply reaction by 2 (picture). -Therefore delta H will become (2)(-537)= - 1074 -In this case, we have used all three of the provided reactions.If you have done everything correctly and you have used all of the provided reactions you are usually done. -Go back to 6F2(g). If you look at the reactions you used, you have ended up with 6 moles of F2, which is exactly what we needed. Everything else better cancel out, so you are left with the reaction that you need. -Anything that shows up on both sides gets canceled out. -The three reactions added together give you the desired reaction. -Which means the delta Hs of the three reactions added together will give you the delta H of the desire reaction. Delta H of rxn= (-52.3) + (-1360) + (-1074) = (-50) - 1300 - 1000 = -1350 - 1000 = -2350 kJ

Third way of calculating delta H (Hess' Law)

Hess' Law -When you are looking for delta H of a certain reaction and you are given delta H's of a whole bunch of other reactions (picture) - Yo have to figure out how to add these three reactions in such a way that it adds up exactly to the one below. - You have an infinite number of delta Hs. If you make the coefficients of the first reaction 2, 2 and 4, delta H will be (2)(-537). -If you want the delta H of the reverse reaction it will be positive 537. C2H4(g) + 6F2(g) ----> 2CF4(g) + 4HF(g)

Hess' Law cont.

You don't have enough time on exam to do all of this work. - don't bother writing new equations. Just look at what you need and at the end if you have used every reaction, you're done. - If you want to double check your work, check the F2s. 2x2=4 and 2x2=2 2+4=6. SO you have 6 F2. -Then just add up all the delta Hs to the delta H of the desired reaction.