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Offline Big-Daddy

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Heat Release Question
« on: November 05, 2012, 06:47:54 PM »
In a question I have, gaseous NH3 is burned with O2 in a container of fixed volume according to the equation given below.

4 NH3 (g) + 3 O2 (g) → 2 N2 (g) + 6 H2O (l)

The initial and final states are at 298.15 K. After combustion with 14.40g of O2, some NH3 remains unreacted.

Calculate the heat given out in the process.

Given: ΔHf[NH3 (g)]=-46.11 kJmol-1, ΔHf[H2O (l)]=-285.83 kJmol-1

First of all, I thought heat release always referred to ΔH and energy release to ΔE - I can work out ΔH for this reaction easily but I have no idea how to calculate ΔE or even what it is really.

Thinking the question was asking for ΔH (since that's what "heat change" means to me - what's the difference, please explain?), I solved:

ΔHr=((6*-285.83))-((4*-46.11))
ΔHr=-1530.54 kJmol-1
ΔHr per mole of O2 (since O2 is the one not in excess) = -1530.54/3 = -510.18 kJ(mol O2)-1
qe=-ΔHr*n[O2]
qe=-(-510.18)*(14.40/(16*2))
qe=229.581 kJ

The actual answer uses ΔE instead of ΔH, qv instead of q (the e is there just to indicate evolved heat), and indicates that ΔE=ΔH-Δng*R*T. I recognize the second half of this as from the ideal gas equation, but how is Δng worked out? This is probably the main difference between myself and the right answer calculation-wise, but understanding-wise there are more: I don't understand what ΔE is, what qv is, and possibly my understandings of ΔH and q are also incomplete if I thought this problem referred to them, so major holes of understanding need to be patched.

At any rate, the correct answer is -227.7 kJmol-1. My wrong sign comes from my idea that q should be heat evolved whereas presumably it should actually be heat absorbed (which is why my answer should have been negative but was not). The slight discrepancy in the figure comes from Δng*R*T, which I don't know how to calculate (T=298.15 obviously and R is the ideal gas constant but how do I work out Δng - what even is Δng?).

I use the arbitrary notation qe to indicate heat evolved, but I'm willing to change this if better standard notation is available. The answer uses the notation qv for what is essentially the same thing - what does this mean? Just another mystery. My best guess is that the "v" indicates constant volume, and q indicates heat taken in (so when heat is evolved more than absorbed then the sign on q is negative); this is surely the same q as in the formula ΔU=q+w - but what is the relationship between q and qv?

Offline Big-Daddy

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Re: Heat Release Question
« Reply #1 on: November 08, 2012, 12:34:16 PM »
Could I get some help on this please?

Offline fledarmus

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Re: Heat Release Question
« Reply #2 on: November 09, 2012, 06:41:01 AM »
Δng refers to a change (Δ) in the total number of moles (n) in the gas phase (g). Does that help?

Offline Big-Daddy

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Re: Heat Release Question
« Reply #3 on: November 09, 2012, 12:29:56 PM »
Δng refers to a change (Δ) in the total number of moles (n) in the gas phase (g). Does that help?

Ah that would make sense since ΔU=q+w - qv suddenly seems to be = ΔH (I don't know why; q as I know it refers to a total amount of heat gained rather than a heat per mole which is ΔH), and there's still the mystery of what the subscript v indicates on qv.

Meanwhile, w is work done so for a gas expanding this is -pΔV (where ΔV=V2-V1; since expansion requires work to be done by the system and is thus negative, w=-pΔV). And since pV=nRT, -pΔV can be replaced by -ΔnRT.

But calculating Δn for this reaction is not something I've understood yet. That would be a necessary step towards calculating ΔU which appears to be what they are referring to when they say "calculate heat released".

Offline fledarmus

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Re: Heat Release Question
« Reply #4 on: November 09, 2012, 06:25:21 PM »

But calculating Δn for this reaction is not something I've understood yet.

Go back to your original chemical equation. You are reacting 14.40 g of oxygen - how many moles is that? How many moles of ammonia will it react with? How many moles of products will be produced? In the ideal gas law, it doesn't matter what the identity of the gas molecules is, only how many total moles of gas molecules are present, so you are looking for the change in number of moles of your total gas mixture.

Offline Big-Daddy

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Re: Heat Release Question
« Reply #5 on: November 10, 2012, 06:51:16 AM »

But calculating Δn for this reaction is not something I've understood yet.

Go back to your original chemical equation. You are reacting 14.40 g of oxygen - how many moles is that? How many moles of ammonia will it react with? How many moles of products will be produced? In the ideal gas law, it doesn't matter what the identity of the gas molecules is, only how many total moles of gas molecules are present, so you are looking for the change in number of moles of your total gas mixture.

OK, but this approach does not seem to yield the right answer:

Moles of O2 (g) in Reactants=14.40/32=0.45mol
Moles of Gas in Reactants: ((4/3)+1)*0.45=1.05mol
Moles of Gas in Products: (2/3)*0.45=0.30mol
Δng=0.30-1.05=-0.75mol

Whereas according to the answer this should be Δng=ng=-1.25mol.

Also, leaving the calculations themselves aside for a second to focus on the definitions: ΔE is the change in internal energy of the system, ΔU, which =q+w, right? So why does this sum say qv=ΔE? Particularly when it is evident that w does not equal 0 since there is a decrease in the total volume of gases (decrease because this is at constant pressure and temperature and the number of moles is decreasing).

If we take ΔE=q+w, w here =-pΔV=-Δng*R*T (which makes sense, because if the number of moles of gas drops and the gas thus contracts, Δng will be negative so -Δng will be positive and represent a positive work done by the system, adding to its total energy, which is what we want and expect), so why does q now equal ΔH? They are fundamentally different I thought - q refers to the quantity of heat taken in, ΔH to the increase in heat per mole.

At the end, qv and ΔE seem to revert back to their meanings as we know them - amounts, i.e. previously they referred to the energy increase for 1 mole and now they are being multiplied out by the number of moles used. I don't understand any of this. Does it mean ΔE=ΔH-Δng*R*T applies only for 1 mole of the reaction? Then how is Δng found? And how can ΔE be defined to describe just 1 mole of a reaction? Please help me with this.

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