[earthnation112]

The dilemma I am having with the following question is that I have the right answer for both part (a) and (b), but I don’t know how to calculate part (a). Even though part (b) can’t be answered without knowing part (a), I was able to calculate it as I was told by a friend what part (a) is. The question is as follows:

Q4. (a) Given that the enthalpy of formation of CH3Cl (g) is -80.83 kJ mol-1 estimate the average bond dissociation energy of a C-Cl bond. (b) Estimate the enthalpy change for the reaction:
                                    H2C=CH2 + HCl ------------- CH3CH2Cl

Bond         Ave. Bond Dissociation energy / kJ mol-1    Element   H°f / kJ mol-1
C-H             414                            C (g)      716.7
C-C             343                            H (g)      218.0
C=C             611                            Cl (g)      121.3   
H-Cl             431               

I will illustrate below my attempt for part (a):

First I wrote the word equation to the reaction which is:

                             H2C=CH2 + HCl ------------- CH3CH2Cl
Then the values given for enthalpy of formation were used to calculate the change in enthalpy:
H2C=CH2    = 2(716.7) = 1433.4
HCl             = 218.0 + 121.3 = 339.3
CH3CH2Cl    = -80.83 + 716.7 = 635.17
Enthalpy of formation for H2 equals 0 so it was excluded from the calculations above.

Enthalpy of change = 635.17 – 1433.4 – 339.3
                             =  -1377.53 KJ mol  ^-1 (energy to break C=C and Cl-H bonds).

Then I tried working out the dissociation energy of (C-Cl) using the following method:


-1377.53 = D_have(C=C) + D_have(Cl-H)
-1377.53 - D_have(C=C) = D_have(Cl-H)
-1377.53 - 611 = D_have(Cl-H)
-1377.53 – 611 = -1988.53

The above was my attempt at working out the bond dissociation energy of (C-Cl), I was totally off as the real bond dissociation energy is 330.8 kj mol^-1 and not what I calculated which was -1988.53.

Where did I go wrong? How do I calculate this correctly? I know how to calculate the part (b) but I just can’t seem to get my head around on how to calculate the bond dissociation energy of (C-Cl).
Any help would be greatly appreciated, Thank You.

[mikasaur]

You're making this hard on yourself.

Part a) is asking for the bond dissociation energy of a C-Cl bond. You don't need the reaction H₂C=CH₂ + HCl  CH₃CH₂Cl.

Use the table you were given and the fact that the enthalpy of formation of CH₃Cl is -80.83 kJ/mol.

[earthnation112]

Cheers for the reply, just don't seem to be getting it.

I followed your advice and came to the following:

enthalpy of formation of CH3Cl = -83.83
-83.83 = 3D_have(C-H)+ D_have(C-Cl)
-83.83 -3D_have(C-H) =D_have(C-Cl)
-83.83 -3(414) =D_have(C-Cl)
-1325.83 = D_have(C-Cl)

Where do I keep going wrong?

[mikasaur]

Unfortunately I'm not sure what D_have means. Regardless, let's try to help you learn how to solve these sorts of problems here.

The problem is that the Enthalpy of Formation is not the sum of the average bond enthalpies. That's why you're not getting the right answer.

How do you define the Enthalphy of Formation? It's the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. (link)

Have you heard of Hess's Law? It states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. (link)

This means that you should be able to think of a series of "mini" reactions that get you to the formation of CH₃Cl from its constituent pure elements.

See if you can answer these questions:

1. What are the pure elements (and their states) that make up the compound CH₃Cl?
2. What must happen to them in order for them to form that compound?

The information you're provided, e.g. ΔH°_f of C(g), H(g), and Cl(g) and bond dissociation energies, are good hints for thinking about the second question.

[Enthalpy]

Not using the equation from question (b) is indeed better. However, you make the same mistake(s) as previously.

The most important is to realize that a heat of formation is computed starting from the elements in their normal state. This is a convention - a very sensible, meaningful and useful one. We could have given ΔHf starting from individual atoms, but that would be very inconvenient, and it's not the convention.

As an illustration, to make 2HCl from the elements, H₂ and Cl₂ must be broken. This (half) enthalpy to break H-H and Cl-Cl counts in the enthalpy of formation of (one) HCl, which is not the bond dissociation energy of H-Cl consequently.

Have you noticed the factor of two? Mind it in dissociation energies! If a table gives ΔHf for Cl° it's to create one mole of atoms starting from the normal state.

The risk to forget is worse with solid elements. We write just C or Fe but they are giant molecules with many chemical bonds between the atoms. So if you need a C atom to compute bond energies, don't forget to atomize it!

Signs... An enthalpy of formation is negative when the synthesis has released enthalpy. A bond dissociation energy is always positive.

You can try to learn many conventions about atoms vs molecules and about signs. My advice is rather: keep the possible sources of mistake in mind, and think.

I don't understand D_have. Is that a ΔH? Check the nice list of symbols over the typing window.

Maybe I haven't seen all error sources in your attempts. Try again now maybe?

[earthnation112]

(DHave) = average bond dissociation energy, this is what was meant by the term.

See if you can answer these questions:

1. What are the pure elements (and their states) that make up the compound CH3Cl?

The answer I got was 2C(s) + 3H_2(g) + Cl_2 (g) ---------> 2CH_3Cl(g)

2. What must happen to them in order for them to form that compound?
They must add together?

Since the heat of formation for CH3Cl is -80.83 does this mean that the heat of formation for 2CH_3Cl is double -80.83?

How do I use this information to calculate the average bond dissociation energy of a C-Cl bond? Thanks

[mikasaur]

1. What are the pure elements (and their states) that make up the compound CH3Cl?

The answer I got was 2C(s) + 3H_2(g) + Cl_2 (g) ---------> 2CH_3Cl(g)

2. What must happen to them in order for them to form that compound?
They must add together?

Since the heat of formation for CH3Cl is -80.83 does this mean that the heat of formation for 2CH_3Cl is double -80.83?

Yes, it's double.

So our starting point and ending point is captured by the equation:

2C(s) + 3H₂(g) + Cl₂(g)    2CH₃Cl(g)

You're right that we must add them together, but what must happen before we can do that? We must atomize the starting reagents to form gaseous (individual) atoms! Then we can combine them together.

Keep in mind Hess's Law. If we know the ΔH to atomize our starting materials, and then the ΔH to combine them together (to form CH₃Cl) we can calculate the heat of formation. We know the ΔH of atomizing our starting material, we know the ΔH°_f of the ending compound. We know a lot about the energy involved in combining them together, but do we know everything? Hmmm...

Can you take another shot of writing an equation that helps you to solve for the dissociation energy of C-Cl? It might help you to break up our current chemical equation into one with another step:

2C(s) + 3H₂(g) + Cl₂(g)  some stuff   2CH₃Cl(g)

[earthnation112]

.You're right that we must add them together, but what must happen before we can do that? We must atomize the starting reagents to form gaseous (individual) atoms! Then we can combine them together

The starting reagents are 2C(s) + 3H_2(g) + Cl_2

To break them up and atomise them to form gaseous individual  atoms would it be the following:

2C(g) + 6H(g) + 2Cl(g)

[mikasaur]

To break them up and atomise them to form gaseous individual  atoms would it be the following:

2C(g) + 6H(g) + 2Cl(g)

Right!

So now we have:

2C(s) + 3H₂(g) + Cl₂(g)  2C(g) + 6H(g) + 2Cl(g)    2CH₃Cl(g)

C(s) + [itex]\frac{3}{2}[/itex]H₂(g) + [itex]\frac{1}{2}[/itex]Cl₂(g)  C(g) + 3H(g) + Cl(g)    CH₃Cl(g)

We know the energy change involved in the first step (our heats of formation!). We almost know the energy change involved in the second step (we have bond energies for all but one bond, the bond we're trying to solve for!). And we know the energy involved in the whole reaction (the heat of formation of the final compound!).

Can you set up a mathematical equation that solves for the bond energy in C-Cl?

::EDIT:: to make the chemical equation a little simpler

[earthnation112]

To break them up and atomise them to form gaseous individual  atoms would it be the following:

2C(g) + 6H(g) + 2Cl(g)

Right!

So now we have:

2C(s) + 3H₂(g) + Cl₂(g)  2C(g) + 6H(g) + 2Cl(g)    2CH₃Cl(g)

C(s) + [itex]\frac{3}{2}[/itex]H₂(g) + [itex]\frac{1}{2}[/itex]Cl₂(g)  C(g) + 3H(g) + Cl(g)    CH₃Cl(g)

We know the energy change involved in the first step (our heats of formation!). We almost know the energy change involved in the second step (we have bond energies for all but one bond, the bond we're trying to solve for!). And we know the energy involved in the whole reaction (the heat of formation of the final compound!).

Can you set up a mathematical equation that solves for the bond energy in C-Cl?

::EDIT:: to make the chemical equation a little simpler

So the enthalpy change for the first part is equal to the heat formation which is:

C(s) + 3/2H2(g) + 1/2Cl2(g) is the
716.7 + 3/2(0) + 121.3 = 838

So the enthalpy change for the second part is:
C(g) + 3H(g) + Cl(g) is:
            3(414) = 1242

Since were trying to work out the C-Cl I have not added them above is that correct?

Then what I did was the enthalpy change of the whole reaction was -80.83, so

-80.83 = 838 – 1242 – (C-Cl)
-80.83 = -404 – (C-Cl)
-80.83 + 404 = -(C – Cl)
-323.17 = -(C – Cl)
323.17  = C – Cl

Have I calculated this correctly? The actual answer for this revision question for C-Cl is 330.8. Thanks

[mikasaur]

Very close.

So the enthalpy change for the first part is equal to the heat formation which is:

C(s) + 3/2H2(g) + 1/2Cl2(g) is the
716.7 + 3/2(0) + 121.3 = 838

Why do you say that the heat of formation of 3H(g) is 0?

[earthnation112]

Very close.

So the enthalpy change for the first part is equal to the heat formation which is:

C(s) + 3/2H2(g) + 1/2Cl2(g) is the
716.7 + 3/2(0) + 121.3 = 838

Why do you say that the heat of formation of 3H(g) is 0?

I read my notes and it said:
"The heat formation of elements in their normal chemical state at 1 atmos and 298 K is taken to be zero. i.e.   of C(s) = 0".

Just checked now and that only applies to H2(g) and not 3H(g)

So the enthalpy change for the first part is equal to the heat formation which is:

C(s) + 3/2H2(g) + 1/2Cl2(g) is the
716.7 + 3(218.0) + 121.3 = 1492

So the enthalpy change for the second part is:
C(g) + 3H(g) + Cl(g) is:
            3(414) = 1242

Using the data above I calculated the following:
-80.83 = 1492 – 1242 – (C-Cl)
-80.83 = 250 – (C-Cl)
-80.83 - 250 = -(C – Cl)
-330.83 = -(C – Cl)
330.83  = C – Cl

Yes I finally got the answer!!!!!
Really appreciate all the help, I wasn't getting it for some reason at all. Again thank you means alot.

One last question why is it that we are allowed to use the dissociation energy and not the heat formation energy when calculating the enthalpy change for the second part of the equation which was:
C(g) + 3H(g) + Cl(g) is:
            3(414) = 1242

Why did we use the 414 which is the dissociation energy for C-H and not the heat formation energy for 3H which would be 3(218.0)?

[mikasaur]

Glad we can assist!

So we're using the bond dissociation energy because that is what is going on in our second "step" of our multistep process:

C(g) + 3H(g) + Cl(g)    CH₃Cl(g)

We are starting with individual atoms and combining them into bonds. The energy associated with this is by definition the bond dissociation energy, i.e. ΔE = -ΣD₀ = 3(-414) - D_C-Cl.

The heat of formation for H(g) is the energy required to atomize diatomic Hydrogen. This is not what we're doing in the step above. This is what we're doing in the first step.

Make sense?

[earthnation112]

Yes makes perfect sense, the bond dissociation energy is used when you break bonds, and the enthalpy of formation is when you form bonds?

Any good links on how to atomize reagents to produce gaseous (individual) atoms? I've tried using Google and the book Chemistry, The Central Science(13th Edition - Brown, LeMay et al) with no real luck.

[mikasaur]

Yes makes perfect sense, the bond dissociation energy is used when you break bonds, and the enthalpy of formation is when you form bonds?

Mmm, only sorta, unfortunately. Bond dissociation energy is the energy needed to break a bond between two atoms to separate them into separate gaseous atoms. (link). This is why before we can use our second step and the bond energies, we had to atomize our H₂(g) and Cl₂(g) into individual H(g) and Cl(g) atoms. Only then could we use the sum of the bond energies as the enthalpy associated with the second reaction. The key here is that the energy assumes something going from a bond to separate gaseous atoms.

Enthalpy of formation is the energy involved when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. The key here is that the energy assumes something going from pure elements at a standard state to a new compound. CH₃Cl breaks up into the individual atoms of C(g), 3H(g), and Cl(g) but none of those atoms are elements in their standard state. Those are, respectively, C(s) H₂(g), and Cl₂(g).

We can use bond dissociation energy to find the enthalpy of either of the reactions below:

C(g) + 3H(g) + Cl(g)    CH₃Cl(g)
CH₃Cl(g)    C(g) + 3H(g) + Cl(g)

In the first instance, bonds are being formed and the enthalpy of the reaction is -ΣD₀. This is because we are forming bonds and so we use the negative of the dissociation energy (the energy required to break a bond).

In the second instance, bonds are being broken and the enthalpy of the reaction is ΣD₀ for the opposite reason. Note that in both reactions above we have a compound on one side (a collection of bonds) and individual gaseous atoms on the other. Also note that these calculations are just estimations because we use average bond energies.

We can use standard heat of formation to find the standard enthalpy of either of the reactions below:

2C(s) + 3H₂(g) + Cl₂(g)    2CH₃Cl(g)
2CH₃Cl(g)    2C(s) + 3H₂(g) + Cl₂(g)

In the first instance we are forming a compound from its pure elements in their standard state. The enthalpy of the reaction is simply the standard enthalpy of formation, ΔH°_f, for Chloromethane.

In the second instance we're breaking apart Chloromethane into its constituent elements in their standard state. The enthalpy of the reaction is -ΔH°_f of Chloromethane.

Any good links on how to atomize reagents to produce gaseous (individual) atoms? I've tried using Google and the book Chemistry, The Central Science(13th Edition - Brown, LeMay et al) with no real luck.

How? You mean experimentally? Or do you just want to understand the theory? Hmmm, I'm not sure exactly where to point you. Understanding the Born-Haber cycle may help. If you look here you can learn about it. In step 3 they talk about turning Cl₂ into 2Cl. I hope you don't get too confused, though. Because the heat of formation, ΔH°_f of 2Cl(g) is also the dissociation energy for Cl-Cl(g)! Can you see why?