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### Topic: Calculate Delta(H)rxn for the follwing reaction.  (Read 37188 times)

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#### santiano204

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• Mole Snacks: +0/-0 ##### Calculate Delta(H)rxn for the follwing reaction.
« on: April 11, 2010, 11:17:21 PM »
Hi, I have a question regarding enthalpy.

The question is as follows..

The addition of HCl to AgNo3 precipitates silver chloride according to the following reaction.

AgNO3 + HCl ---> AgCl + HNO3

When 50 ml of .100 AgNO3 is combined with 50 ml of .100 HCl in a coffecup calorimeter. The temperature changes from 24.21 celcius to 23.40 celcius.  Calculate delta(h)rxn for the reaction as written. Use 1.0g/ml density of the solution and C = 4.18J/g*c as specific heat capacity.

Now I basically used q=mc(delta)T.  So I did q=(100g)(4.18J/g*c)(.81) and ended up with qsoln = 338.58

Then to get qrxn I just added a negative. -338.58.  Now this is where I'm not sure exactly how this should turn out.  Do I divide by the mol's of HCl or mol's of AgNO3?  Looking at the reaction, they're both 1:1, with the same added volume, so then I realized I can't divide using either the AgNO3 or HCl because it would yield DIFFERENT results.  I'm a bit lost, sorry, any help would be appreciated.

#### UG

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« Reply #1 on: April 11, 2010, 11:35:39 PM »
Are you sure you have your signs correct? Is this an exothermic or endothermic reaction?
So you have your 'q' value, and exactly 0.005 mole of silver nitrate reacted with exactly 0.005 mole HCl, so now how do you find the change in enthalpy for one mole of the reaction?

#### santiano204

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« Reply #2 on: April 11, 2010, 11:37:24 PM »
Well I used the molar mass of HCl and divided it by the used amount(50g).  I then used that and divided it into the 338

#### santiano204

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« Reply #3 on: April 11, 2010, 11:39:14 PM »
Also, I have trouble understanding when to switch signs, sorry, I read over the book and I can't seem to understand it.  I always thought that by finding the q, you actually get the qsoln, and since delta H is in qrxn, you just switch signs and divide by the mole's of whatever is reacting.

#### UG

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« Reply #4 on: April 11, 2010, 11:42:26 PM »
No where does it say you have 50g of HCl, you have a 50 mL solution of HCl and most of it is water, in order to work out the number of mole, you use the conversion: n(mol) = concentration(mol L-1) x volume (L). Same thing for the silver nitrate.

And for the signs, it's actually a lot simplier than the books make it out to be. If the temperature of the surrounding (i.e. the solution) increases, it is an exothermic reaction and q IS NEGATIVE. If the temperature of the surrounding decreases, it is an endothermic reaction and q IS POSITIVE.

#### santiano204

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« Reply #5 on: April 11, 2010, 11:52:28 PM »
Oh, well it says use density 1g/ml for solution so I just converted them into grams. And for the changing of signs, isnt the temperature already an indication if it's endothermic or exothermic? and I ONLY change signs when finding delta H because the Q absorbed by the solution, is the Q released by the reaction, therefore when finding deltaHrxn, I need to switch signs?

#### UG

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« Reply #6 on: April 12, 2010, 12:01:27 AM »
Oh, well it says use density 1g/ml for solution so I just converted them into grams. And for the changing of signs, isnt the temperature already an indication if it's endothermic or exothermic? and I ONLY change signs when finding delta H because the Q absorbed by the solution, is the Q released by the reaction, therefore when finding deltaHrxn, I need to switch signs?
The density given is basically ignoring the weight of the HCl in the solution, if you had 50 mL of liquid HCl, and they told you to use density 1g/mL then you can just find the number of moles like you did before.
Temperature of the surroundings is an indicator of whether it is exothermic or endothermic, correct.
The 'proper' way you should have done your calculation is to use  :delta:T as being Tfinal - Tinitial which would have given you -0.81 oC, then when you put this into your calculations you get a negative q.
The way q and H are related is that  :delta:H = -q/n
So in the end two negatives gives you a positive and so the change in enthalpy is positive, meaning it is an endothermic reaction, but you could have deduced that from the change in temperature.
You have q, now you just gotta find 'n'

#### santiano204

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« Reply #7 on: April 12, 2010, 12:14:23 AM »
Oh I'm sorry actually the tfinal is the "24.21".  And this would mean that the solution had gained the q, which makes more sense therefore leaving that the reaction must be exothermic.  I had also found the answer now by using the moles of HCl or AgNO3 found using the concentrations and reacted volumes, sorry I didn't even notice that because in previous questions where instead of adding two liquids, it would instead say 1 g of magnesium is added to an HCl solution, and etcetera.  I had to divide the mass by molar mass, and in this question, it had hinted about the density so I assumed that the 1ml=1g and I could find the reacted amount of moles with their respective molar masses.

#### UG

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« Reply #8 on: April 12, 2010, 12:16:26 AM »
Well I'm glad you've sorted it out 