5.3 Hess' law enthalpy of formation Flashcards
Delta H is a state function
* meaning its independent of path as long as you have the same inital and final state, it doesnt matter how you got there
Hess’s Law
We need to add up some combination of the 3 rxns he’s provided us to find the delta H of the rx were looking for
Start by just looking at the first species in your rxn. In our case its NO(g). Then look to see if its in the other 3 rxns. If it shows up in more than 1 of those 3 rxns you skip it
* in our case for NO it shows up in 2/3 of those equations so we skip it. Dont want to worry about balancing that right now
We then move to the next species which is N2O. The only palce this shows up is in the 2nd equation. So this is a good candate
* so now I need to edit this 2nd reaction where I can get N2O to match the one in my base rxn.
* So that means 1 need 1 mole of N2O (because thats how it appears in the base rxn) and I need it to be on the products side
* So its on the correct side, but its 2 moles. So we want to cut this rxn in half.
Next lets move onto the next species which is NO2. This happens to only show up in 1/3 of the rxns. Which means its a good candate to balance.
* So to match our base equation I was again want 1 mole of NO2 and I want it on the products side of the rxn
* So its on the wrong side, and we have two moles instead of one. So were going to need to reverse this equation, and cut it in half
Lastly to get the 3 moles of NO on the products side were going to have to reverse our third equation. One of our other 2 equations had 1 mol of NO already on the products side, so we only need to get 2 mols of NO on the product side instead of 3
if you have a rxn thats 113kj, well you can just reverse the rxn and then it would be -113 kj
you can also double everything as well
There are ways to work through these problems much quicker. However, should go back through and verify if time allows
* if you don’t use all 3 equations in this method, than defiently go back and verify
our first reactant is C2H4 gas.
* The only place this shows up is in the last rxn
* so i need 1 mole of C2H4 gas on the prodcts side
* Well I have 1 mole, but its on the wrong side, so i just need to reverse the entire thing.
* which means I can just stick a negative sign infront of that 52.3 kj (instead of writing the entire thing out)
Next, ill move onto F2.
* I need 6 moles of this and I need it on the reactants side.
* Its in the first and second rxns, so ill skip this and move on for now
Next, ill move onto CF4
* this only shows up in the middle equation
* I need this CF on the products side and I need 2 moles of it
* its already on the products side, but i need to multiply by 2 because i only have 1 mole of it
Finaly looking at HF, I need 4 moles of it on the products side.
* its onl found in the first equation
* its on the products side, but theres only 2 moles of it which means I need to double it
So we havent worried about the 6F2 yet, but we’ve used every rxn. And when we use every rxn, were actaully done
* can also briefly look at make sure you get the correct # of F2’s
* can look at see where it was douvled / flipped and make sure the correct # of F2’s end up where they’re supposed to be, matching the base equation
below im just attaching the quick method. I’m putting the fully worked out problem on the answer slide
This is the entire problem worked out
This is the cleaned up best version of this
Enthalpy of Formation
Were going to use it to calculate delta H
* the other way to calculate is hesses low, but this is easier
Delta H rxn = sum of delta H’s of formation of the products - the sum of delta H’s formation of the reactants
* the m and the n in the equation is multiplying by how many moles of the products and reactants you have
The delta H of formation for NO, N2O and NO2 were all provided
Its products - reactants
Enthalpy of formation
Formation reaction
1) Forms 1 mole of a single product
* When you have the enthalpy of formation of HCl then the formation rxn will form exactly 1 mole of HCl
2) Reactants are elements in their standard states.
* So you can’t have any compounds on the reactants side, only elements
* need to be in standard states (these need to be memorized, this means if their standard state is gas they need to be in gas form etc..)
* note the products dont have to be in the standard phase just the reactants
Enthalpy of formation is just the enthalpy of a formation rxn.
Second example
Remember, carbon in its most stable lowest E way is graphite not diamond, so must write carbon = solid graphite, not just sold, when doing formation rxns
Name all the elements in the standard state of gas
All of group 18
F, Cl, O, N, H
Name the two atoms that are liquid in standard state
Hg, Br2
Alotrope = when an element has more than 1 form in the same phase
* for carbon you have graphite and diamond
graphite is the stnadard form
Remember, delta H is how much E is either absorbed by the rxn, or excreted into the environment
* delta H = negative = system lost heat aka exothermic, the surroundings gained
* When a rxn breaks into multiple parts delta H is positive, meaning the system gained heat. This is because it took E to break apart the bonds. whenever bonds are broken, you’re putting energy into the system to overcome the attraction between atoms and break the bonds.
The answer being -1640kj makes since. The base equation went from 2 molecules to 1, meaning bonds were formed.
* When reactants come together to form bonds, they’re moving to a more stable, lower energy state. That stability is what releases E
* So if the net result is more stable products (strong bonds than what you started w/), the reaction releases heat and has a negative delta H, meaning its exothermic.
* Think about two magnets snapping together. At first, if they’re far apart, yes it takes effort to move them. But once they’re close eneough they naturally pull together and release E, when they snap into place. Thats like bond formation: its spontaenous once they’re in the right position and orientation
This is a really hard hess law problem. Work through this again
Didnt actually answer the kj part there but its -107kj
Dont have to use hess law because the delta H of the rxn is given (thank god)
Essentially all we do is sume the enthalpys of formation and multiply by them by the mole amounts. Hardest part is remembering that its products - reactants to find the total E used
O2 - any element in its elemental form has a delta H of 0, which is why its not on the table provided)
* elements in their standard form have a delta Hf = 0 by defintion we set this as their baseline and its 0. if its not in standard form (like O instead of O2) it does not have a zero delta Hf
Reason its products - reactants
* You’re essentially asking how much energy is released or absorbed when i form the products from their elements, and how much energy would have been needed to form the reactants from their elements. Whats their difference
* You’re basically comparing starting energy vs ending energy. Just like you would do final - initial.
* Think of it this way, you start w/ $50 (reactants) and you end w/ $30. You lost $20, so its a change in $20 (30-50 = -20)
Remember, stnadard enthalpy of formation of a compound is the amount of heat absorbed or released when 1 mole of the compound is formed from its elements in their standard states
* so for example IBr’s enthalpy of formation is 41kj. This means that hwen I and Br combine the system (in this case IBr) uses 41kj of energy to synthesize this compound. Meaning the system gains E, making it an endothermic rxn
