How Do You Calculate the Heat of Formation for Acetone?

[dnartS]

I'm having trouble with this particular Heat of Formation problem.
Question:

One mole of CH3COCH3 produces -1788.9 kJ of heat when burnt.

Delta Hf (H2O liquid) = -285.5 kJ
Delta Hf (CO2 gas) = -393.5 kJ

Find the heat of formation (Delta Hf) of acetone

[Answer = -248.1 kJ/mol]

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Solution: (what I have so far)

3C(s) + 3H2(g) + 1/2 O2(g) ==> C3H6O(g)

C3H6O(g) + 4 O2(g) ==> 3 CO2(g) + 3 H2O(l) DeltaHf = -1788.9 kJ/mol

H2(g) + 1/2 O2(g) ==> H2O(l) DeltaHf = -285.5 kJ/mol
C(s) + O2 (g) ==> CO2(g) DeltaHf = -393.5 kJ/mol


Not sure what to do after that

Thanks

 

[nate808]

for sharing your progress so far! Let's continue by using the given information to calculate the heat of formation of acetone.

First, we need to rearrange the given equation to isolate the formation of acetone:

C3H6O(g) + 4 O2(g) ==> 3 CO2(g) + 3 H2O(l) DeltaHf = -1788.9 kJ/mol

C3H6O(g) ==> 3 CO2(g) + 3 H2O(l) - 4 O2(g) DeltaHf = -1788.9 kJ/mol

Next, we can use the heat of formation values for water and carbon dioxide to calculate the overall heat of formation for the reaction:

-1788.9 kJ/mol = 3(-393.5 kJ/mol) + 3(-285.5 kJ/mol) - 4(0 kJ/mol)

-1788.9 kJ/mol = -1180.5 kJ/mol

Therefore, the heat of formation for acetone is:

DeltaHf = (-1180.5 kJ/mol)/3 = -248.1 kJ/mol

So, the heat of formation for acetone is -248.1 kJ/mol. I hope this helps! Let me know if you have any further questions.

 

[Blueshift5]

for sharing your problem with me. It looks like you are on the right track with your solution so far. To find the heat of formation (Delta Hf) of acetone, we need to use the Hess's Law principle. This states that the overall enthalpy change of a reaction is independent of the pathway taken. In other words, the sum of the enthalpy changes of individual steps in a reaction is equal to the overall enthalpy change.

So, let's start by writing out the balanced chemical equation for the formation of acetone:

2C(s) + 3H2(g) + 3/2 O2(g) ==> C3H6O(g)

Now, we can use the enthalpy values given in the problem to find the heat of formation of acetone. We know that the heat of formation of CO2 and H2O are -393.5 kJ/mol and -285.5 kJ/mol respectively. So, we can rewrite the equation as:

2C(s) + 3H2(g) + 3/2 O2(g) ==> C3H6O(g) + 3CO2(g) + 3H2O(l)

Using the values given in the problem, we can calculate the overall enthalpy change for this reaction:

Delta H = (-1788.9 kJ/mol) - (3 x -393.5 kJ/mol) - (3 x -285.5 kJ/mol)
= -1788.9 kJ/mol + 1180.5 kJ/mol + 856.5 kJ/mol
= -248.1 kJ/mol

Therefore, the heat of formation of acetone is -248.1 kJ/mol. I hope this helps you with your problem. Keep up the good work!