[vdemas]

Here's another one :

When 0.950g of sigma-toluic acid ( C₈H⁸O₂ ) udergoes combustion in a constant volume bomb calorimeter, the temperature rises. This temperature rise can be duplicated by applying 27.02kJ of electrical energy, which is a measure of the change in internal energy. Calculate the standard enthalpy of formation of sigma-toluic acid, given that :

Molar mass (toluic acid) = 136.15 g/mol
Delta H_f [H₂O(l) ] = -285.8kJ/mol
Delta H_f [ CO₂(g) ] = -393.5kJ/mol


This is what I have so far:

The balanced equation ror the combustion reaction is :

C₈H₈O₂ + 9O₂ = 4H₂O + 8CO₂


Because it is a constant volume calorimeter, the energy supplied as heat = q = delta U.
This value is 27.02kJ.


Working out the no. of moles of toluic acid :

0.950g x 1 mol / 136.15g = 0.006978 moles toluic acid


But, delta U = q_v / n_{(C8H8O2 )}
= 27.02kJ / 0.006978 mol
= 3.872 X 10³ kJ/mol

Now I know we are looking for delta H_f ( C₈H₈O₂ ) and to find that we need delta H for the entire combustion process. But we only have delta U.
How do I calculate delta H from delta U if I don't know the pressure or the volume or the temperature.
This is very confusing.
Please help.
Thanx.

[Yggdrasil]

Since you're calculating the standard enthalpy of formation, you need to calculate the change in enthalpy at 1 atm and 273K.  Recall that enthalpy is defined as:

ΔH = ΔU + Δ(PV)

Since P is fixed, all you need to figure out is ΔV.  To figure out ΔV, compare the moles of gas on the reactants side to the moles of gas on the products side.  Assuming the solids and liquids in the reaction take up no space, how does the total volume of the gases change after the reaction is complete?

[vdemas]

So, the initial volume would be :

V_i = n_i R T / P_i
= (9moles) (0.082057L.atm/K.mol) (273.15K) / (1)
=201.725L

And the final volume would be :
V_f = n_f R T / P_f
= (12) (0.082057L.atm/K.mol) (273.15) /(1)
=268.966L


So, delta PV = P_f V_f - P_i V_i
= (1) (201.725) - (1) (268.966)
=67.24 atm.L
= 67.24 J
= 0.06724 kJ


Thus,
deltaH = deltaU + delta (PV)
= 3.872 x 10³ kJ/mol + 0.06724 kJ
= 3872.06 kJ

Now, what value do I use for delta U ? Do I use the given value ( 27.02kJ) or the one I calculated per mole of C₈H₈O₂ ( 3.872 x 10³ kJ/mol)?

I think I should use the given one, but I'm not entirely sure because why would they give you the molar mass and mass of sigma-toluic acid if you don't have to use it. Or do I still have to divide the 0.06724kJ by the no. of moles of sigma-toliuc acid?

In the latter case,the value for delta H would be 3881.633 kJ/mol

Thanx for everything.

[Yggdrasil]

When you're calculating the Δ(PV), you're calculating Δ(PV) per mole of sigma-toluic acid reacted (for example, if two moles of sigma-toluic acid were reacted, you would have n_i = 18 instead of n_i = 9), so you would want to use the ΔU in kJ/mol.

Another small point, since you're given H_f for liquid water, water should not count as a gas and n_f = 8, not 12.  However, if you had used H_f for gaseous water, then you would be correct in using n_f = 12.  This is why it is important to include the phase subscripts when specifying a reaction since:

C₈H₈O_{2 (s)} + 9O_{2 (g)} --> 4H₂O _(l) + 8CO_{2 (g)}

and

C₈H₈O_{2 (s)} + 9O_{2 (g)} --> 4H₂O _(g) + 8CO_{2 (g)}

will have different heats of reaction.

[vdemas]

Thank You so much. I really get clumsy when working with these enthalpy values.

I see now that I made a mistake in n_f because the product is liquid water and not gaseous water.
Thus, n_f is 8 and not 12.

So, can you please confirm for me that the answer for delta_f H⁰ (C₈H₈O₂ ) is -8.16 x 10³ kJ/mol.


Thanx.

[Hunt]

Since the volume is constant, dH  = dU + VdP

You can assume that the gases are ideal so PV = nRT
It's required to find the enthalpy at T = 273/298 K. Keep T, V constant and write : dP = (RT/ V) dn

Substitute to get dH = dU + RT dn

ΔH = ΔU + RT Δn

This relation is also true if P = constant while V varies.

Actually here RT Δn is negligible << ΔU . this is why it doesnt matter if you write delta n correctly or even totally neglect the 2nd part. But still this is a general equation that serves as a good approximation for many gas-phase reactions.

Btw Yggdrasil 's point is really imp. If you were required to find the enthalpy of formation at T > 373 K , water would become vapour and delta n changes.


Quote
deltaf H0 (C8H8O2 ) is -8.16 x 103 kJ/mol.


an 8 or a 4 ?

[vdemas]

  What do you mean by 8 or 4 ?
This is how I worked delta_f H⁰ out :

Let x be delta_f H⁰ of toluic acid.


Reaction enthalpy = (sum of enthalpy of formation of products) - (sum of enthalpy of formation of reactants)

 3872 = [ (4) ( -285.32) + ( (-393.51) ] - [ (9) (0) + (1) (x) ]
 3872 = [ (-1141.28) + (-3148.08) ] - x
     -x = 8161.36
      x = -8161.36 kJ/mol
                         

[vdemas]

The face with the glasses on is suppose to be an 8 in brackets. I tried to correct it, but it doesn't change.

[Hunt]

I thought you made a typing mistake. Your final ans is wrong because you considered delta H combustion to be positive while it should be negative.

- 3872 = [ (-1141.28) + (-3148.08) ] - x

x is about - 417 kJ/mol

[vdemas]

Is delta H (combustion) a negative value because it is a combustion reaction and the reaction releases heat instead of using it?
 That is, because the question says there was a rise in temperature, I have to assume that the reaction enthalpy has a negative value?

[Hunt]

Yes vdemas, the combustion reaction releases heat. It's an exothermic reaction because the tempreature of the surrounding water increases.