[JohnTravolski]

Using the balanced equation for the heat of formation of water and the following thermonuclear equation, calculate the amount of heat released when 27.9 grams of C(s) is incompletely combusted to give CO(g).

C(s) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 131.3 kj/mol

I think that I know how to do this problem, but I just need confirmation:

The following information was obtained from charts in the book:
H2O(l) = -285.8 kj/mol
H2O(g) = -241.8 kj/mol
CO(g) = -110.5 kj/mol
C(s) = 0 kj/mol
H2(s) = 0 kj/mol

Because H2O is a gas and in order for heat (enthalpy) of formation to occur properly, it needs to be in the standard state of a liquid. So I know that H2O(g) → H2O(l) = -44 kj/mol.

So the change in enthalpy should be:
Products - Reactants
therefore
[-110.5 + 0] - [-44 + 0] = -66.5 kj/mol.

Then, with this number, I can find the amount of heat released given the grams.

27.9 g C(s) * [1 mol C(s)/12.01 g C(s)] * [-66.5 kj/1 mol C(s)] ≈ -154

which means that approximately 154 kj of heat are released in the reaction. I'm hoping that I did this correctly, but I'm not sure. My two major concerns are:

Is there supposed to be some kind of limiting reactant? That's what "incompletely" suggests to me. I can't imagine how to find it if there is one.

Did I calculate the change in enthalpy correctly? Is -66.5 the number that I should be using?

The problem isn't hard, I just need to make sure that I'm not missing something somewhere. Maybe I put a negative sign in the wrong place, I don't know, but I need your help to make sure I'm on track here.

[thetada]

Hi, this is a weird problem. First the incomplete combustion bit is nothing to do with limiting reagents. It means that CO rather than CO2 is produced because of a shortage of oxygen. That's the weird part, the equation is 2C + O2 --> 2CO, which is basically just the formation of carbon monoxide. That value has been given as -110.5kJ/mol so if I was doing the question, that is the enthalpy change i'd use. That would mean the other equation is unnecessary but that's not unheard of in questions.

[Zyklonb]

Ok, this is actually really easy, you're over thinking it.
It gives the heat of reaction for one mole of carbon, all you had to do is find out what it is for 27.9 grams of carbon.
So, one mole is 12 grams and reacts using 131.3 kJ, so one gram is 131.3 ÷ 12 =10.9416667, × 27.9 =305.272501
Thus the reaction can proceed if it is given 305.3 kJ, that's the answer.
As for the above post, it contains so much nonsense it's hard to know where to begin...
In each sentence the poster says either an outright fallacy, or shows his/her fundamental misunderstanding of the laws of thermodynamics, equation balencing and this entire problem.

First the incomplete combustion bit is nothing to do with limiting reagents. It means that CO rather than CO2 is produced because of a shortage of oxygen.

That's a bit like saying "I'm not broke, I just ran out of money".

[thetada]

First the incomplete combustion bit is nothing to do with limiting reagents. It means that CO rather than CO2 is produced because of a shortage of oxygen.

That's a bit like saying "I'm not broke, I just ran out of money".

That's fair, my phrasing was clumsy. I should have said that it is not necessary to identify a limiting reagent for the purposes of determining the number of moles of carbon reacted.

Using the balanced equation for the heat of formation of water and the following thermonuclear equation, calculate the amount of heat released when 27.9 grams of C(s) is incompletely combusted to give CO(g).

C(s) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 131.3 kj/mol

Maybe the question is badly written but I cannot see how the quoted equation could refer to a combustion reaction. Oxford dictionary has the definition of combustion as: "Rapid chemical combination of a substance with oxygen, involving the production of heat and light." Collins dictionary supplements this with: "A chemical process in which two compounds, such as sodium and chlorine, react together to produce heat and light."

In the course I teach (UK A-level), enthalpy of combustion is defined as the energy released when one mole of a substance is completely reacted with excess oxygen. The carbon does not react with oxygen in the quoted equation. Even by the Collins definition, in which they suggest that it is any reaction that gives out heat and light, I cannot see how this reaction could qualify, since it is endothermic.

Again, from the question:

Using the balanced equation for the heat of formation of water and the following thermonuclear equation, calculate the amount of heat released when 27.9 grams of C(s) is incompletely combusted to give CO(g).

Which contradicts Zyklonb's answer:

So, one mole is 12 grams and reacts using 131.3 kJ, so one gram is 131.3 ÷ 12 =10.9416667, × 27.9 =305.272501
Thus the reaction can proceed if it is given 305.3 kJ, that's the answer.

The question specifically requests the amount of heat released, but Zynlonb's answer quotes an amount of energy required for the reaction to proceed.

In spite of my clumsy phrasing, I'd still go with my solution, but please come back and tell us the answer when you find it out.

[Zyklonb]

The reaction is well known. Its used to create a mixture of flammible gasses (CO and H₂) called browns gas I believe. I looked up the enthalpy for one mole and it's indeed 131.3 kJ/mol. Thus scaled up it will still be endothermic proprtionally.

[JohnTravolski]

From what I understand, though, the heat of formation needs to use all of its constituent elements or compounds in its normal states, which is why I need to change the H2O(g) to H2O(l).  Here's another example from my book that's similar, but not quite the same, with all of the work showing the CORRECT answer.  I could check it in the back of the book so I know it is correct.



The negative, at the bottom, is correct, because that negative sign indicates the amount of heat released.  However, because all of the elements in that example are in their standard states, and this is enthapy of formation, I don't have to change anything.  The problem is that, no matter how hard that I look through my book, I cannot find what to do for enthalpy of formation whenever one of the constituent elements or compounds is NOT in its standard state, like water in my original problem.

[thetada]

The following information was obtained from charts in the book:
H2O(l) = -285.8 kj/mol
H2O(g) = -241.8 kj/mol
CO(g) = -110.5 kj/mol
C(s) = 0 kj/mol
H2(s) = 0 kj/mol

These quotes are enthalpies of formation, right? Can you not just use the value released when water is produced in the gaseous phase rather than the figure for the liquid phase formation?

Make a hess cycle with the original reaction between carbon and water forming the top two points. Then put the separate elements below, to form the third point of the triangle.

[Borek]

Off topic - you wrote "approximately equals" and gave the answer with ten digits. Do you understand what "approximately" means? Pi is approximately 3. Giving it as 3.141592654 (same 10 digits) is highly accurate.

[Big-Daddy]

Why do you need to do anything with heats of formation for these problems? You are given the enthalpy changes of the reaction, for every mole of the reaction that occurs. Just multiply that by the number of moles that will occur (which will be the number of moles of the limiting reagent divided by the stoichiometric coefficient of the limiting reagent in the reaction equation; limiting reagent being defined as the species with lowest value of number of moles divided by stoichiometric coefficient). That's all they are asking for.

In the OP problem, you don't use -44 or -66.5 or anything like that. Just use the data you are given directly. Incomplete combustion would be 2C(s) + O2(g) => 2CO(g), for which your formation enthalpies directly tell you the reaction enthalpy. That is, unless they expect you to just assume the incomplete combustion equation is the one they gave.

[JohnTravolski]

I know what approximately means; I just wrote that because I'm so used to rounding in my answers that I didn't think about it.  Plus, one could also assume that there isn't exactly 1.26 x 10^4 grams of Ammonia.  But that's not important.

But here's the problem; when I just used the exact data that's given to me, I would have this:
27.9 g C(s) * (1 mol C(s)/12.01 g C(s)) * (131.3 kj/1 mol C(s)) which gives me a positive answer of 305 kj; that would mean that it's endothermic and takes 305 kj, but the question clearly asks how much heat is released, so the answer would have to be negative.

I'm assuming that the "balanced equation for the heat of formation of water" (as the question originally asks) is this: H2O(g) → H2O(l) = -44 kj/mol.
So I'm assuming that I have to use that number somewhere.  When I do, I then get the original answer I posted, which is negative, which indicated the amount of heat released.

Also, here's a quote from my book:
"Remember that a ΔH°rxn is only a ΔH°f when there is just one product, just one mole is produced, and all of the reactants are elements in their standard states."

When I put all of this together, I'm just really confused.

[thetada]

No that's not the formation of water, that's the phase change of water. Formation of water is 2H2 + O2 --> 2H2O. Like I said, put the formation of steam figure that you quoted in your data table in a Hess triangle with the equation in the question. Basically the sum of the enthalpies of formation of water and the equation in the question gives you the enthalpy of formation of CO ( which you have in your data table anyway)

[JohnTravolski]

That means that I could just use the ΔH°rxn = 131.3 kj/mol that's originally given to me.
So that makes sense, but then when I do the following:

27.9 g C(s) * (1 mol C(s)/12.01 g C(s)) * (131.3 kj/1 mol C(s))

I get an answer of positive 305 kj, which would indicate the amount of heat needed for the reaction to take place, not the amount released, which is what the question asks for.

[thetada]

If you look at the attached image, it shows a Hess cycle for a similar reaction. Looking at the ΔH_B reaction, notice how it's basically the same as the enthalpy of formation of CO₂ because hydrogen is in its elemental form before and after and the enthalpy of formation for any element is defined as 0. Hess' Law states that the sum of the clockwise enthalpies is equal to the sum of the anticlockwise enthalpies, so in this case: ΔH_A + ΔH_B = ΔH_C. You basically need to apply the same thinking to your question.

[JohnTravolski]

.

[thetada]

Because you should be using the -110.5kJmol^-1 value, not the 131.3kJmol^-1 value. The enthalpy of formation of CO, which is equivalent to the enthalpy of incomplete combustion of C, is -110.5kJmol^-1. Work out the number of moles in 27.9g of carbon, multiply that figure by -110.5 and that's your answer.