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Author Topic: ΔHrxn for a reaction...  (Read 4175 times)

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methic

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ΔHrxn for a reaction...
« on: October 13, 2010, 01:40:11 PM »

I'm not quite sure what this question was asking or why I got it wrong.
Hoping you guys could tell me exactly how I read the question.

Quote
Ammonium Nitrate dissolves according to the following reaction:

NH4NO3 → NH4+NO3-

In order to measure the enthalpy change for this reaction, 1.25g of NH4NO3 is dissolved in enough water to make 25.00mL of solution. The initial temperature is 25.8 C and the final temp is 21.9 C.

Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g/mL as the density of the solution and 4.18 J/gC as the specific heat capacity.)

This was my process:

qsol = (25g)(4.18J/gC)(-3.9) = -407.55 J
qrxn=407.55

ΔHrxn = 407.55 / 0.0156 moles NH4NO3 = 26kJ

Now, I got this wrong. The correct answer is supposed to be .41kJ.
I see where that number is (first step in finding finding qsol), but
was I wrong in understanding exactly what the question was asking?
What did I end up finding? I mean, if it was only asking for qrxn,
then why did it bother giving grams of NH4NO3?
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Borek

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Re: ΔHrxn for a reaction...
« Reply #1 on: October 13, 2010, 09:40:44 PM »

You are not asked to calculate MOLAR enthalpy, just enthalpy of the reaction as it happened.

Additional information... it is up to you to decide what information is necessary to solve the question. But I agree that wording of the question can be a little bit misleading.
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methic

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Re: ΔHrxn for a reaction...
« Reply #2 on: October 14, 2010, 05:31:55 AM »

You are not asked to calculate MOLAR enthalpy, just enthalpy of the reaction as it happened.

Additional information... it is up to you to decide what information is necessary to solve the question. But I agree that wording of the question can be a little bit misleading.

Man, that's whack. So how exactly does ΔH differ from q?
q is heat, right? Are they exactly the same when its
exposed to the atmosphere? I was under the impression
that there was some conversioning I had to do between
q and ΔH

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Jorriss

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Re: ΔHrxn for a reaction...
« Reply #3 on: October 14, 2010, 06:35:28 AM »

If a system is kept at constant pressure without working being done, enthalpy equals q.
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sjb

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Re: ΔHrxn for a reaction...
« Reply #4 on: October 15, 2010, 10:42:44 AM »

ΔHrxn = 407.55 / 0.0156 moles NH4NO3 = 26kJ

You might want to check your units, as well, here.
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methic

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Re: ΔHrxn for a reaction...
« Reply #5 on: October 15, 2010, 11:21:22 AM »

ΔHrxn = 407.55 / 0.0156 moles NH4NO3 = 26kJ

You might want to check your units, as well, here.

Okay, so now I'm confused again. What did I do wrong
with my units?

407.55 J / 0.0156 moles = 26125 J = 26.125 kJ if there
were an entire mole of the stuff.

No?

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Coastie17

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Re: ΔHrxn for a reaction...
« Reply #6 on: October 15, 2010, 02:39:08 PM »

If you have 407.55 J/ 0.0156 mols your answer would be in J/mol as the mols wouldn't cancel out, not just J as you have.
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methic

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Re: ΔHrxn for a reaction...
« Reply #7 on: October 15, 2010, 02:52:06 PM »

If you have 407.55 J/ 0.0156 mols your answer would be in J/mol as the mols wouldn't cancel out, not just J as you have.

Ah, I get what you're saying. I'm really bad with units. That's probably something I should start being more careful with.

Anywho, since I'm here. I may as well ask another related question.

In this problem:

Quote
A calorimeter contains 23.0mL of water at 12.5 degrees C. When 1.70g of X (with a molar mass of 80.0 g/mol) is added, it dissolves via the reaction

X + H2O → X (aq)

and the temperature of the solution increases to 28.5 degrees C. Calculate ΔH per mol of X

So, I did this problem; and I got it right after a failed attempt. Originally, I did this:

(23g)(4.18 J /gC)(28.5°C-12.5°C).

and got q = 1538.24 J

I got that part marked wrong because it said I needed the total mass of the reactants and the water. I'm just wondering why I needed to do add them together. In the previous question - as well as examples in the book - it only figures in the mass of the water. I don't see the difference between the two.

It was supposed to be

(24.7g)(4.18 J/gC)(28.5°C-12.5°C) ... and then other stuff I did correctly (divided by moles of X, etc).
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