4A.13

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Fdonovan 3D
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

4A.13

Postby Fdonovan 3D » Sat Feb 08, 2020 2:47 pm

A constant volume calorimeter was calibrated by carrying out a reaction known to release 3.50kJ heat in a .200 L of solution in the calorimeter resulting in a temperature rise of 7.32 degrees C. In another experiment 100.0 mL of 0.200 M HBr and 100.0 mL of 0.200 M KOH mixed in the calorimeter and the temperature rose by 2.49 degrees C. What is the change in internal energy of the reaction?

In the solution book, none of the information aside from the temperature from the second reaction are used. Why don’t we need to use any of that info??

Charisse Vu 1H
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

Re: 4A.13

Postby Charisse Vu 1H » Sat Feb 08, 2020 2:54 pm

Since the calorimeter contains the same volume of liquid in both cases (0.200 L and 100.0 mL+100.0 mL), there is no change in volume and the final volume is constant at 0.100 L. That is why the volumes are not considered since there is no change in volume.

Michelle N - 2C
Posts: 117
Joined: Wed Sep 18, 2019 12:19 am

Re: 4A.13

Postby Michelle N - 2C » Wed Feb 12, 2020 11:49 am

Fdonovan 3D wrote:A constant volume calorimeter was calibrated by carrying out a reaction known to release 3.50kJ heat in a .200 L of solution in the calorimeter resulting in a temperature rise of 7.32 degrees C. In another experiment 100.0 mL of 0.200 M HBr and 100.0 mL of 0.200 M KOH mixed in the calorimeter and the temperature rose by 2.49 degrees C. What is the change in internal energy of the reaction?

In the solution book, none of the information aside from the temperature from the second reaction is used. Why don’t we need to use any of that info??


I am actually terribly lost in this question. I understand that the volumes are not included since it's constant, but how would you go on with solving this problem? Thank you in advance!


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