HW Ch.8 question 8.25

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Eman_Burney_1D
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Joined: Wed Sep 21, 2016 2:57 pm

HW Ch.8 question 8.25

Postby Eman_Burney_1D » Sat Jan 28, 2017 11:57 pm

I was looking back at questions I had done, and for 8.25 I am still unsure as to why the internal energy result is negative. Here is the question:

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q=-3.50 kJ), resulting in a temperature rise of 7.32 degrees celsius. In a subsequent experiment, 100.0 mL of 0.200 M HBr (aq) and 100.0 mL of 0.200 M KOH (aq) were mixed in the same calorimeter and the temperature rose by 2.49 degrees celsius. What is the change in the internal energy of the neutralization reaction?

So, to be more specific, is the answer negative (the answer is -1.19 kJ) because the problems mentions a "release of heat," implying that it is an exothermic reaction, and hence delta U has to be negative (since q=delta H, and the equation involved is delta U=q, rather than delta U= q+w, since there is no work being done)? Thanks!

Alex Dib 4H
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Joined: Wed Sep 21, 2016 2:57 pm

Re: HW Ch.8 question 8.25

Postby Alex Dib 4H » Sun Jan 29, 2017 12:56 am

Hey, I don't have the problem in front of me, but from what I understand, you're correct. You can see this mathematically when you set q(system)+q(calorimeter)=0, which is rearranged as q(calorimeter)=-q(system). What we're given, we are only finding information of the calorimeter. We know that the calorimeter absorbs the heat, so all of that heat must have come from the system's reaction. Therefore, when we find that the calorimeter's change in heat was q(cal)=+1.19kJ, this must mean that the change in heat of the system was q(sys)=-1.19kJ. Because it's a bomb calorimeter with constant V, deltaU=q; w=0.
Good luck!

Eric_pierce_3E
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Joined: Wed Sep 21, 2016 2:55 pm

Re: HW Ch.8 question 8.25

Postby Eric_pierce_3E » Sun Jan 29, 2017 4:57 pm

Nice post!

Alex Uy 2D
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Joined: Sat Jul 09, 2016 3:00 am

Re: HW Ch.8 question 8.25

Postby Alex Uy 2D » Tue Feb 14, 2017 9:40 pm

In the solutions manual to this, they say that q reaction + q calorimeter = 0. Is that something you just have to know? or where in the problem does it suggest that the change in internal energy is 0?

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

Re: HW Ch.8 question 8.25

Postby Zoe Robertson 2H » Sun Mar 12, 2017 4:05 pm

Alex Uy 2D wrote:In the solutions manual to this, they say that q reaction + q calorimeter = 0. Is that something you just have to know? or where in the problem does it suggest that the change in internal energy is 0?


also wondering about this!


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