Hess' Law

[whc235]

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Im really confused by the solution manual... so im wondering if any1 can help me

Given the average bond energies, N-H(389 kJ/mol), H-F(565 kJ/mol), and F-F(155 kJ/mol), calculate the average bond energy for N-F. NH3(g) + 3F2(g) à NF3(g) + 3HF(g) ∆H = -873.81 kJ




Thx!!

 

[Sublimation]

Im really confused by the solution manual... so im wondering if any1 can help me

Given the average bond energies, N-H(389 kJ/mol), H-F(565 kJ/mol), and F-F(155 kJ/mol), calculate the average bond energy for N-F. NH3(g) + 3F2(g) à NF3(g) + 3HF(g) ∆H = -873.81 kJ




Thx!!

Remember that Hess law is the diff in product and reactants. Wat you need to do is this.

N-H 3(389KJ/mol) Remember you have NH3 so you need to account for each N-H bond that you break.

F-F 3(155kJ/mol) Same reason for above.

Total is 2612kJ/mol. I think i did the calculations quickly, this is the energy required to break these bonds.

Then u calculate the energy released when u form the products. Now since the delta H for the reaction is -873. that means the total energy released by the products is 873+2612= 3485kJ/mol. This is the energy released by the products from the formation of the new bonds. Now u kno H-F is 565kJ/mol. Multiply this by 3 and subtract it from 3484 and then divide by 3.

3485-1695=1790 / 3 = 590. answer should be 590.

 

[Intrigue]

for the bond energy problems it is basically
reactant - products =dH
energy input - energy release = dH

 

[whc235]

Ooo am i confusing this with Hess' Law? Do i do reactants - products instead?

So i've been doing this...

∆H = [3∆H(N-F) + 3∆H(H-F)] - [3∆H(N-H) + 3∆H(F-F)]
-873.81 = [3∆H(N-F) + 3(565)] - [3(389) + 3(155)]

So my final answer was 312 ish


The solution manual says.. 270... cuz they did

∆H = -[3∆H(N-F) + 3∆H(H-F)] + [3∆H(N-H) + 3∆H(F-F)]
-873.81 = -[3∆H(N-F) + 3(565)] + [3(389) + 3(155)]
-3∆H(N-F) = -810.81
∆H(N-F) = 270.27 kJ/mol


The difference between the two is the order of products and reactants. So .. this is not Hess Law at all?

 

[bebebebebebe]

Yes it is somehow;the strategy for estimating reaction enthalpies is a straightforward application of Hess's law:

As we know Hess's law states that if a rxn is carried out in a series of steps, delta H for the reaction will equal the sum of the enthalpy changes for the individual steps.
And he stated that all chemical reactions can be broken down into formation reactions( Enthalpies of formation), so basically all chemical reactions can be broken down to steps that their enthalpies are enthalpies of formation.When we do(by using hess law), we obtain the general result that the standard heat of reaction is the sum of the standard heats of formation of the products minus the standard heats of formation of the reactants. ( I don wanna make this more complicated by explaining why you put minus in the formula)

Also for calculating the enthalpies of reactions by using bond enthalpies, we again used the Hess' law:
We can imagine that all reactions occur in 2 steps: 1) first step, breaking all bonds. 2) second step, formation of bond
So the enthalpy of the reaction is estimated as the sum of the bond enthalpies of the bonds broken(reactants), minus the sum of the bond enthalpies of the new bonds formed(products), and keep in mind we put minus because, formation of bond is an exothermic reaction.So here, it's important to put (-) before the bond enthalpies of products, because the new bonds only form in the products. And bonds only break in the reactants( endothermic). In your sln manual, they wrote it
-(bonds formed in products)+ (bonds broken in reactants) instead of (bonds broken)-(bonds formed), same thing.

 

[Intrigue]

Ooo am i confusing this with Hess' Law? Do i do reactants - products instead?

So i've been doing this...

∆H = [3∆H(N-F) + 3∆H(H-F)] - [3∆H(N-H) + 3∆H(F-F)]
-873.81 = [3∆H(N-F) + 3(565)] - [3(389) + 3(155)]

So my final answer was 312 ish


The solution manual says.. 270... cuz they did

∆H = -[3∆H(N-F) + 3∆H(H-F)] + [3∆H(N-H) + 3∆H(F-F)]
-873.81 = -[3∆H(N-F) + 3(565)] + [3(389) + 3(155)]
-3∆H(N-F) = -810.81
∆H(N-F) = 270.27 kJ/mol


The difference between the two is the order of products and reactants. So .. this is not Hess Law at all?

I got this answer when I did it. it is not hess law. it is called bond dissociation energy.