Ch 8 Q 53

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Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

Ch 8 Q 53

Postby Ariana de Souza 4C » Sun Jan 15, 2017 3:38 pm

CO(g) + H20(g) --> CO2(g) + H2(g)

reaction of 1.4 g of CO in a bomb calorimeter causes the temp to go from 22.113 C to 22.799 C. The calorimeter assembly is known to have a total heat capacity of 3.00 kJ/C
b) calculate the internal energy change, delta U, for the reaction of 1 mol CO

So the solution manual does this
q (reaction) + q(calorimeter) = 0
q(reaction= -q(calorimeter)
why do we choose the calorimeter to be negative? Couldn't you just as easily make the reaction heat negative, and get the whole problem wrong? What am I missing here?

Cherry_Deng_1K
Posts: 12
Joined: Wed Sep 21, 2016 2:55 pm

Re: Ch 8 Q 53

Postby Cherry_Deng_1K » Sun Jan 15, 2017 4:57 pm

Hello,

The reaction is considered the system, while the calorimeter, where the reaction is taking place, is the surroundings. The heat of the system and the heat of the surroundings should add up to 0, meaning q(system) + q(surroundings)=0, or, in the case of #53, q(reaction) + q(calorimeter)=0. Any heat that is given off by the reaction is absorbed by the calorimeter, meaning the heats have opposite signs and add up to 0. If you subtract q(calorimeter) from both sides, you will get q(reaction) = -q(calorimeter).

To be honest, I don't think it matters whether q(reaction) or q(calorimeter) is subtracted to the other side. The signs of the heat depends on whether the heat was absorbed or released by the reaction, meaning the q(reaction) and q(calorimeter) will have opposite signs no matter what. You just have to read the question carefully to determine the sign of q(reaction) and q(calorimeter). The hint in this question is the direction of the temperature change. Hope this helps!
Last edited by Cherry_Deng_1K on Tue Jan 17, 2017 6:58 pm, edited 1 time in total.

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

Re: Ch 8 Q 53

Postby Ariana de Souza 4C » Mon Jan 16, 2017 3:28 pm

Yeah, it does! I appreciate the explanation. So when you're doing the math part, you're doing --

q(reaction) = -q(calorimeter)
q(Reaction)= -q*C*deltaT
so are you calculating the heat of the reaction then by saying it's the opposite of the heat of the calorimeter?

Cherry_Deng_1K
Posts: 12
Joined: Wed Sep 21, 2016 2:55 pm

Re: Ch 8 Q 53

Postby Cherry_Deng_1K » Tue Jan 17, 2017 7:38 pm

That's the first step! By finding q(reaction), we can find ΔU. If you look on page 10 of the course reader, it's mentioned that ΔU=qv, meaning the volume is constant. Since we know that a reaction taking place in a bomb calorimeter will not experience a change in volume, we can use ΔU=q (at constant volume). We can find q(calorimeter) by multiplying the calorimeter's heat capacity and temperature change together (qcal=Ccal x ΔT). Once you know q(calorimeter), we know q(reaction) since q(reaction)=-q(calorimeter). The q that we get is for 1.4 g of CO, but the question asks for 1.00 mole of CO, so you can make a ratio to get the final result. Hope this helps!

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

Re: Ch 8 Q 53

Postby Ariana de Souza 4C » Wed Jan 18, 2017 1:41 pm

yes, thanks so much!


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