When to add for Delta H vs formula

[MoooShuuu]

Can someone clarify for me when to add for the Delta H values (Hess's Law) vs. using the dH= sum(dh products)-(sum dH reactants)?

Going through my study guides I have found that sometimes they are added and sometimes the formula is used and I haven't been able to sort it out 😕

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[chessxwizard]

When you're given a reaction with heats of formation for products and reactants, just do the products-reactants.

However, when you're given many individual steps, you have to add them together to cancel out the intermediates and get the final reaction. That's when you add the heats for each reaction, which is Hess' law.

 

[IWantSmile]

The rules are practically as chessxwizard said.

Here is example,
Calculate deltaH for the process at 25 degree C of dissolving 1mol of KCl in a large excess of water:
KCl (s)-----> K+ + Cl-
delta H f(K+)= -60 kcal/mol
delta H f(Cl-)= -40 kcal/mol
delta H f(KCl)= -104 kcal/mol

then use the rules what chessxwizard said at first.
vs.

The heat released on neutralization of CsOH with all strong acids is 13.4kcal/mol. The heat released on neutralization of CsOH with HF (weak acid) is 16.4 kcal/mol. Calculate delta H of HF in water.

OH- + H+ ----> H2O deltaH= -13.4 kcal/mol
OH- + HF ----> H2O + F- deltaH= -16.4 kcal/mol

then use the rules what chessxwizard said at second.
Hope this help too. I just add some more information.
What do you think chessxwizard? did I complicate more? I hope not!

When you're given a reaction with heats of formation for products and reactants, just do the products-reactants.

However, when you're given many individual steps, you have to add them together to cancel out the intermediates and get the final reaction. That's when you add the heats for each reaction, which is Hess' law.

 

[chessxwizard]

Sounds good to me 👍

 

[waystinthyme]

When you're given a reaction with heats of formation for products and reactants, just do the products-reactants.

However, when you're given many individual steps, you have to add them together to cancel out the intermediates and get the final reaction. That's when you add the heats for each reaction, which is Hess' law.

exactly. maybe provide a problem or two that you're having trouble with, and we can walk you through it?

-waystinthyme

 

[MoooShuuu]

Thanks guys...okay here is one that has confused the heck out of me.

Given the following, what is the heat of formation in kcal/mole of methane?

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

H2(g) + 1/2O2(g)---> H2O (l) dH -68.3

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

the solution gives dHformation= -94.1+ 2(-68.3)+ 212.9 = -17.9 and uses Hess's law.

So I guess my issue here is that I want to rearrange the reactions so they match up to Eq 3. I would reverse Eq 1 so CO2 is a reactant (+94.1) and reverse and multiply EQ2 so that H20 is a reactant and 02 is a product.

Where am I going wrong? Thanks!!!

 

[waystinthyme]

Thanks guys...okay here is one that has confused the heck out of me.

Given the following, what is the heat of formation in kcal/mole of methane?

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

H2(g) + 1/2O2(g)---> H2O (l) dH -68.3

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

the solution gives dHformation= -94.1+ 2(-68.3)+ 212.9 = -17.9 and uses Hess's law.

So I guess my issue here is that I want to rearrange the reactions so they match up to Eq 3. I would reverse Eq 1 so CO2 is a reactant (+94.1) and reverse and multiply EQ2 so that H20 is a reactant and 02 is a product.

Where am I going wrong? Thanks!!!

the question asks for the dh of the heat of formation of methane, so you want to:

1) rearrange the equations so you are only left with the reaction for the heat of formation of methane, and all the other intermediates cancel out
2) use your manipulations of the original equations to calculate the dh of the heat of formation of methane.

let's look at the equations we're given.

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

H2(g) + 1/2O2(g)---> H2O (l) dH -68.3

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

1) first, we need to figure out the equation for the formation of methane

C(s) + 2H2 (g) --------> CH4 (g)

this is the equation that we want to end up with when all the intermediates cancel out.

2) next we need to make sure that our reactants/products, that we need in the final equation, are on the correct side.

from the first equation, we see that our C is on the reactant side.
from the second equation, we see that our H is on the reactant side.
from the third equation, we see that our CH4 is on the products side.

so all of the reactants/products that are in our final equation are already on the correct side.

note: if one of the reactants or products were not on the correct side, we would have to reverse the reaction, and therefore the sign of the HF would change (if the hf were (-), the hf for the reverse reaction would be (+) and visa versa)

3) we need to manipulate the equations so that we have the correct number of reactants/products and all other intermediates (elements that are not contained in our final equation) cancel out (meaning there are equal numbers of the intermediates on the product side of one equation and reactant side of another equation).

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

H2(g) + 1/2O2(g)---> H2O (l) dH -68.3

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

looking at our equation for methane we need: Reactants [1 C(s), 2 H2(g)] and Products [1 CH4(g)]

equation 1: we already have 1 C(s) on the reactant side, so this equation does not need to be manipulated

equation 2: we only have 1 H2(g) on the reactant side, so we must multiply this equation by 2, giving us...

2H2(g) + O2(g)---> 2H2O (l) dH 2(-68.3)

PLEASE REMEMBER: when we multiply the equation by 2, we must also multiply the DH by 2.

equation 3: we already have 1 CH4(g) on the product side, so we don't have to do anything to this equation.

4) make sure all of the intermediates cancel out, and we are only left with our equation for the formation of methane

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

2H2(g) + O2(g)---> 2H2O (l) dH 2(-68.3)

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

equation 1: C(s) is a reactant in the final equation, O2 and CO2 are intermediates

equation 2: H2 is a reactant in the final equation, O2 and H2O are intermediates

equation 3: CH4 is a product in the final equation, CO2, H2O and O2 are intermediates

Intermediates: O2, CO2, H2O

how do we cancel? intermediates cancel when there are equal amounts in the reactants of one equation and the products of another equation.

note: on paper, you may actually want to mark through the intermediates when they cancel out...so it's easier to see.

O2: there is 1 mole on the reactant side of equation 1, and 1 mole on the reactant side of equation two, giving us a total of 2 moles on the reactant side.

there is 2 moles on the product side of equation 3, so O2 cancels out.

CO2: there is 1 mole in the product side of equation 1, and 1 mole in the reactant side of equation 3, so CO2 cancels out.

H2O: there is 2 moles in the product side of equation 2, and 2 moles in the reactant side of equation 3, so H2O cancels out.

we are left with our equation for the formation of methane.

C(s) + 2H2(g) --------> CH4(g)

5) now we use our new equations to find our dh for the formation of methane, CH4

Just add the dh of the new equations, to find the dh for the formation of methane.

C(s) + O2 (g)---> CO2 (g) dH= -94.1 kcal/mol

2H2(g) + O2(g)---> 2H2O (l) dH 2(-68.3)

CO2(g) + 2H2O (l) ---> 2 O2(g) + CH4(g) dh 212.8

dH = -94.1 + 2(-68.3) + 212.8 = -17.9

i'm not sure how hess's law is used here, but that's how you solve this type of problem. it looks complicated, but with practice you should be able to do this in 90 seconds or less.

let me know if this was confusing, or if you need any more information. hope it helped a bit 🙄

-waystinthyme

 

[MoooShuuu]

Thank you so much. It was a perfect explaination. For some reason I had it stuck in my head that I was suppose to use that equation with the methane in it as my formation equation. So nothing added up or made sense to me. Thanks again.

 

[supraman]

Holy extensive response! Hey also remember even though its off topic, when you reverse an Keq (equilibrium constant) for a given rxn, the Keq is not negative like change of heat above would be, but is 1/Keq.

 

[IWantSmile]

😱 .... Wao~ it is sooo clear cut an awesome response !! 👍

 

[waystinthyme]

no problem...glad i could help.

-waystinthyme