HW problem 4A. 13

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Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

HW problem 4A. 13

Postby Gisela F Ramirez 2H » Tue Feb 12, 2019 11:16 pm

Say youre given a reaction in a constant volume calorimeter, and you have to calculate the change in internal energy.
This question gives us the q of a first reaction, q= -3.50 kJ, resulting temp rise of 7.32 degrees celsius

Then a second experiment, 100.0 mL of 0.200 M HBr(aq) and 100.0 mL of 0.200 M KOH(aq) mixed in the same calorimeter, and the temp rose by 2.49 degrees celsius.

Since volume is constant, the work is equal to zero. Would I find the q of the second reaction and add it to the first? And if so how would I do it with molarity, and would the temp change then be 7.32+2.49?


Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

Re: HW problem 4A. 13

Postby Vincent Li 4L » Tue Feb 12, 2019 11:31 pm

The first reaction is intended to give you sufficient information to solve for the calorimeter constant using q = C*deltaT. Then, because the reaction parameters are similar in the subsequent chemical reaction (same amount of solution in the same calorimeter), we can use that same calorimeter constant C to solve for q using the same equation, same C value, but different deltaT. Because it's constant volume, there is no work done, and therefore internal energy can only change by the absorbing or releasing of heat. Hope this helps.

Summer de Vera 2C
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

Re: HW problem 4A. 13

Postby Summer de Vera 2C » Tue Mar 12, 2019 11:36 pm

if you use q=nCdeltaT, then do you simply convert the reaction substances into mols and add them together to get the total mols?

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