Kaplan Final gchem #16

[yakuza]

Advertisement - Members don't see this ad

Given the reactions and thermodynamic data below, calculate the

for C6H5OH in kcal/mol.
reaction ΔH°(kcal)
C6H5OH + 7 O2→ 6 CO2 + 3 H2O +729.8
C + O2→ CO2 +94.4
2 H2 + O2→ 2 H2O +136.8


Hess's law tells us that the standard change in enthalpy of a reaction, ΔH°, is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Therefore, for the first reaction in the table, we can write:

O2 is already in its standard state, and so its enthalpy of formation is zero. The last term in the equation thus vanishes. The enthalpy of formation of carbon dioxide is the enthalpy change of the second reaction in the table, i.e. 94.4 kcal/mol. The enthalpy of formation of one mole of H2O is one half the enthalpy change of the third reaction, i.e. 68.4 kcal/mol. (The reaction leads to the formation of two moles of H2O.) We can now substitute in these values and solve for the unknown, the enthalpy of formation of C6H5OH:

​

What the hell is going on here? Someone explain please

 

[Svart Aske]

You're given the ΔH of the combustion reaction, which is +729.8 kcal/mol. We know that:
ΔH = ΔHf of products - ΔHf of reactants
so
729.8 = ΔHf of products - ΔHf of reactants (reactant is obviously C6H5OH)

Since they give you the ΔHf of CO2 (94.4) and H2O (68.4), you can substitute them into the above equation as products.

729.8 = [6(94.4) + 3(68.4)] - ΔHf of reactants

So ΔHf of C6H5OH = 41.8 kcal/mol

They simplified the numbers at the end, which actually gives you an answer that's way off the mark.

 

[yakuza]

You're given the ΔH of the combustion reaction, which is +729.8 kcal/mol. We know that:
ΔH = ΔHf of products - ΔHf of reactants
so
729.8 = ΔHf of products - ΔHf of reactants (reactant is obviously C6H5OH)

Since they give you the ΔHf of CO2 (94.4) and H2O (68.4), you can substitute them into the above equation as products.

729.8 = [6(94.4) + 3(68.4)] - ΔHf of reactants

So ΔHf of C6H5OH = 41.8 kcal/mol

They simplified the numbers at the end, which actually gives you an answer that's way off the mark.

There's no way to balance the equation when doing the Hf calculation though.

Maybe I'm just thinking way too in depth, but it still doesn't make sense bc the equation doesn't balance out

 

[Svart Aske]

Hmm... not sure what you mean.

 

[tranv117]

what's the answer? I got -26.6. give or take some, maybe my math is off.

H came out to be --> -729.8 + 6(94.4) + 136.8

Moving the equations around gets you

6C + 2H2 + H2O -> C6H5OH

Notice how the reactants are all in its standard state form.

 

[supraman]

what's the answer? I got -26.6. give or take some, maybe my math is off.

H came out to be --> -729.8 + 6(94.4) + 136.8

Moving the equations around gets you

6C + 2H2 + H2O -> C6H5OH

Notice how the reactants are all in its standard state form.

I actually got 729.8 - 6(94.4) - 136.8, I'm probably wrong lol

 

[Svart Aske]

You have to halve the 136.8 because that's the heat of formation of 2 moles of H20, then multiply by 3.

 

[tranv117]

You have to halve the 136.8 because that's the heat of formation of 2 moles of H20, then multiply by 3.

you dont halve the 136.8 because if you look at the equations, in the first equation 3moles of H2O will cancel out the 2 moles of H2O leaving 1 mole of H2O.

 

[tranv117]

I actually got 729.8 - 6(94.4) - 136.8, I'm probably wrong lol

youre forming C6H5OH so the first equation has to be flipped, making the H of 729.8 to -729.8.

 

[Svart Aske]

you dont halve the 136.8 because if you look at the equations, in the first equation 3moles of H2O will cancel out the 2 moles of H2O leaving 1 mole of H2O.

I see your reasoning, but the OP's/my method also seems to makes sense. Can you see what's wrong?

 

[yakuza]

This question has to be a mistake, I just did it again.

You want to get

6 CO2 + 3 H2O --->>> C6H5OH + 7 O2
--------------------------- so we mix the equations up so it equals this


[CO2 ---------->>>> C + O2 ] x6 since we want 6CO2 on left
6O2 ---------- >>>> 6C + 6O2......this is what we get

[2 H2O ------->>>> O2 + H2 ] x3/2 since we want 2 H2O on left
3 H2O ------->>>> 3 O2 + 3 H2.........this is what we get


Now look, theres no freaking way to balance the left over C, O2, and H2

 

[Svart Aske]

Doesn't the Kaplan solution make sense to people? It gives you 41.8 kcal/mol.

 

[tranv117]

is kaplans answer on this 41.8?

 

[Svart Aske]

This is what the OP posted:

If you ignore the simplification it gets you 41.8