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| | From:  Abalo3 (Original Message) | Sent: 1/3/2008 9:15 PM |
I am stuck and need some help.
Please
Thank you
Questions
(f) calculate the calorimeter's heat absorption as
Qcal = Ccal * delta T (Joules):
(g) record the true heat of reaction as
Qreaction = Qobserved + Qcal (Joules):
2. Convert the total heat of dissolution to a molar enthalpy:
(a) moles of NH4Cl in 5g (MW = 53.492)
(b) enthalpy of the reaction, per mole of NH4Cl (J/mol):
3. Write out the reaction NH4Cl(s) -> NH3(g) + HCl(g) as a series of steps which include the reactions observed in Procedures 2 and 3.
Use the following known (previously determined) enthalpies of reaction as additional information:
Change of state of NH3(g) -> NH3(aq) (delta H = -34640 J/mol)
Change of state of HCl(g) -> HCl(aq) (delta H = -75140 J/mol)
Combine the above data and the experimental data values you determined in this lab as needed to allow calculation of the enthalpy change of the desired final reaction.
Be sure to show how the reaction steps must proceed so that "delta H" for the desired reaction can be calculated. And be careful to use the correct positive or negative enthalpy values depending on the DIRECTION of the reactions that you add together.
In your final answer specify whether the enthalpy change occurring in the combined final reaction causes heat to be released or absorbed.
Answers
1.
a. volume of water in the calorimeter was 25.00 ml
b. Mass of water in the calorimeter (assuming a density of 1.0 g/mL for water) (g):
=Volume * density
= 25ml* 1.0 g/ml
= 25 g
c. Total mass in the calorimeter at the end of the reaction:
= 25g + 5g
=30 grams
d. Temperature change from the reaction
=21.00 degrees C - 10.09 degrees C
= 10.91 degrees C
By adding Ammonium Chloride to the water , the temperature started increasing from 10.09 degrees C to 21.00 degrees C. Such reaction in temperature is proved to be endothermic because the change of heat is positive meaning no heat flows.
e. The observed heat of reaction in joules
=30g * 4.184 j/g-Deg C * 10.91Deg C
=1369.42 joules
f. The calorimeter's heat absorption
=
f. Calorimeter's heat absorption
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Reply
| | | From: Abalo3 | Sent: 1/7/2008 2:47 PM |
3. Write out the reaction NH4Cl(s) -> NH3(g) + HCl(g) as a series of steps which include the reactions observed in Procedures 2 and 3.
Qreaction in Procedure 2= 2235.93j
Qreaction in Procedure 3 = 1434.88j
1) NH4Cl(s)---------> NH3 (g) + HCl(g) Procedure 3
2) HCl(g) + NH3(g)---------> NH4Cl (s) Procedure 2
I am lost and need some suggestions.
Thank you
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Reply
| | (1 recommendation so far) | Message 8 of 21 in Discussion |
| | From: ·Steve· | Sent: 1/7/2008 3:39 PM |
Here are the two reactions you did (see message 2 of this thread to see where they fit in). For the time being I'm not specifying the signs of Qreaction.
HCl (aq) + NH3 (aq) ––�?gt; NH4Cl (aq) Qreaction = 44718.6 J (message 8 of "Enthalpy problem")
NH4Cl (s) ––�?gt; NH4Cl (aq) Qreaction = 15428.82 J (message 5 of this thread)
Note that the Qreaction values are the per one mole values. That corresponds to the coefficients in the reactions: 1 in front of HCl and NH3, and 1 in front of NH4Cl (s) (the ones are not actually written which is normal). In chemical calculations, the coefficients in the balanced reactions are the moles of reactants and products. That is why you divided the heats of reaction by the moles of HCl (aq), NH3 (aq), or NH4Cl (s), so that the values correspond to the two reactions as they are written.
Now, instead of saying "Qreaction", we will say "ΔH". They're the same numbers, but you have to be careful about the signs. ΔH is positive for endothermic processes and negative for exothermic processes. Once you have filled in the ΔH values for these two reactions correctly, you will be ready for the Hess's law part. 
Steve
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Reply
| | | From: Abalo3 | Sent: 1/7/2008 11:31 PM |
3.
HCl (aq) + NH3 (aq) ------>NH4Cl (aq) DeltaH deg=44718.6j
NH4Cl(s)-----> NH4Cl(aq) DeltaH Deg= 15428.82j
--------------------------------------------------------------------
After cancelling NHCl (aq) I have:
HCl(aq) +NH3(aq)-------->NH4Cl(s) DeltaH Deg=60147.42j
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| | | From: ·Steve· | Sent: 1/8/2008 12:54 AM |
You still have to fix the signs on the ΔH values for those two reactions. That was my excuse to review the meanings of endothermic and exothermic reactions with you, since these terms do apply here.  An endothermic reaction absorbs heat from the surrounding solvent in the calorimeter, and from the thermometer too. The final thermometer temperature will be less than the initial temperature in an endothermic reaction. An exothermic reaction produces heat which is absorbed by the surroundings and thermometer. The final temperature is higher than the initial temperature after the reaction. So what you need to do is go back and check the temperature changes for these two reactions from your data: If Tf < Ti, the reaction is endothermic and ΔH is positive. If Tf > Ti, the reaction is exothermic reaction and ΔH is negative. The sign of ΔH tell us whether the system, which is the reaction, gained energy in the form of heat or lost energy in the form of heat. The definition of ΔH is the change in heat energy in a reaction or process under constant pressure conditions, which is what we have in the open room under a constant atmospheric pressure. If the system loses heat energy (exothermic process) it's like we made a "withdrawal" (negative change) of heat from the system. If the system gains heat energy (endothermic) it's like we made a "deposit" (positive change) of heat energy to the system. If you reverse a reaction, change the sign of its ΔH. You will need to use all four reactions in message 2 to get the desired reaction (also in message 2). Substances have to be on opposite sides in order to cancel. In your last message, the NH4Cls do not cancel because they are on the same side. You will have to reverse one of these reactions before adding them together. So first, be sure you have the correct signs for the ΔH values for your two experiment reactions, and then you will be set to do this last part, using the four reactions with known ΔHs to determine ΔH of a new reaction. Steve P.S. Thermodynamics is a challenging subject! Do you have a textbook to use with this online course? Even with a textbook, learning thermodynamics can be a slow process. Keep at it. |
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| | | From: ·Steve· | Sent: 1/8/2008 2:11 AM |
I forgot that you said earlier that you have a textbook for the class. Who are the authors? I was just curious to know if it's one I'm familiar with. |
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| | | From: Abalo3 | Sent: 1/8/2008 4:19 PM |
| The authors names are McMurray and Fae Fourth Edition. |
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| | | From: ·Steve· | Sent: 1/8/2008 4:59 PM |
We should have a copy here. We are using the Chang text currently. Carefully read over the section about Hess's law (again!) and this last part should make sense! |
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| | | From: Abalo3 | Sent: 1/8/2008 5:06 PM |
You are not the problem. I am the one to blame for all these mistakes I am making.
NH3 (g) ––�? NH3 (aq) ΔH = �?4640 J/mol
HCl (g) ––�? HCl (aq) ΔH = �?5140 J/mol
NH4Cl (s) ––�? NH4Cl (aq) ΔH = -15428.82J/mol
NH3 (aq) + HCl (aq) �? NH4Cl (aq) ΔH= - 44718.6J/mol
--------------------------------------------------
Cancelling NH3(aq),HCl(aq),NH3(aq) and HCl(Aq) gives me:
NH3(g)+HCl(g)+NH4Cl(s)------->2[NH4Cl(aq)
Combination of Delta H=�?4640 J/mol +(�?5140 J/mol) + 2(-15428.8J/mol)+ 2(- 44718.6J/mol) =-199217.2J/mol
The answer is :
NH3(g)+HCl(g)+NH4Cl(s)------->2[NH4Cl(aq) DeltaH=-199217.2J/mo
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| | | From: ·Steve· | Sent: 1/8/2008 5:22 PM |
I think you are on the right track. But first, we have to get the signs of those two experimental reactions right. This is a case where the online approach is not good. When you do these reactions and measure the temperature changes in the lab, it's much easier to see and understand what's happening. Here are the four "given" reactions, with their corresponding ΔHs: NH3 (g) ––�?gt; NH3 (aq) ΔH = �?4640 J/mol
HCl (g) ––�?gt; HCl (aq) ΔH = �?5140 J/mol
NH4Cl (s) ––�?gt; NH4Cl (aq) ΔH = +15428.82 J/mol
NH3 (aq) + HCl (aq) ––�?gt; NH4Cl (aq) ΔH = �?4718.6 J/mol The "desired" reaction is
NH4Cl (s) ––�?gt; NH3 (g) + HCl (g) ΔH = ?
So, you need to add the four given reactions in such a way so that you get this reaction.
Hint: Notice, for example, that the desired reaction has 1 NH4Cl (s) on the left side. The third given reaction also has 1 NH4Cl (s) on the left side. None of the other given reactions have NH4Cl (s) in them. Therefore, the third reaction is OK as is, you do not have to reverse it or anything. Also, you will need to use each given reaction only once.
Steve
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| | | From: Abalo3 | Sent: 1/8/2008 5:44 PM |
NH3 (g) ––�? NH3 (aq) ΔH = �?4640 J/mol
HCl (g) ––�? HCl (aq) ΔH = �?5140 J/mol
NH4Cl (s) ––�? NH4Cl (aq) ΔH = +15428.82 J/mol
NH3 (aq) +HCl (aq)-->NH4Cl (aq)ΔH = �?4718.6 J/mol
------------------------------------------------------
There are 2 NH4Cl(aq) in the third reaction as well as the forth one.Therefore I cancelled one.
Cancelling NH3(aq),HCl(aq),NH3(aq) , NH4Cl(aq) and HCl(aq) gives me:
NH3(g)+HCl(g)+NH4Cl(s)------->NH4Cl(aq)
ΔH= -139069.78J/mol |
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| | | From: ·Steve· | Sent: 1/8/2008 5:56 PM |
You have to reverse one or more of the given reactions before adding them together. As written, the four reactions do not add up to give the desired reaction. |
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| | | From: Abalo3 | Sent: 1/8/2008 6:49 PM |
Calculating DeltaH= -34640-75140+15428.82=-94351.18J/mol
Reversing the reactions give:
NH3(g) + HCl(g)----------->NH4Cl(s) ΔH = -94351.18J/mol |
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| | | From: ·Steve· | Sent: 1/8/2008 7:53 PM |
Well, keep at it.... the reaction you should end up with is NH4Cl (s) ––�?gt; NH3 (g) + HCl (g) That's what you're calculating ΔH for. I'll let you keep chipping away at it! |
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| | | From: Abalo3 | Sent: 1/8/2008 9:55 PM |
After calculating ΔH I am still getting the same answer as before:
-34640-75140+15428.82=-94351.18J/mol
Therefore,
NH4Cl (s) ––�? NH3 (g) + HCl (g) ΔH= -94351.18J/mol which signals that we are dealing with an exothermic reaction.
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| | | From: ·Steve· | Sent: 1/9/2008 1:52 AM |
Well, not there yet! You should write the four reactions you are adding together, with their ΔHs, so we can both see how they add up to give the desired reaction. I think we may have a copy of you textbook at school, but I didn't have a chance to look for it. It should have a few examples of doing this under "Hess's law." Remember, if the same thing is on both sides, they "cancel." That's not the best way to word it, but here's an example: NH4Cl (s) ––�?gt; NH4Cl (aq) ΔH = +15428.82 J/mol
NH4Cl (aq) ––�?gt; NH3 (aq) + HCl (aq) ΔH = +44718.6 J/mol <–�?Note the sign change since this reaction was reversed. NH4Cl (s) ––�?gt; NH3 (aq) + HCl (aq) ΔH = +60147.42 J/mol See how the NH4Cl (aq)s "cancel"? When you add the reactions together, there is one NH4Cl (aq) on each side. Like an algebra equation with "5x" on each side. You can subtract 5x from each side and simplify the equation. We can do likewise with chemical equations. The yield sign is like the "equals" sign. |
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