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### Textbook question 4A.13

Posted: Wed Feb 05, 2020 11:48 pm
A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solu- tion in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. 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 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I keep getting +1.19kJ instead of -1.19kJ. Could someone please explain how to get -1.19kJ?

### Re: Textbook question 4A.13

Posted: Thu Feb 06, 2020 12:43 am
Qreaction=-Qcalorimeter and so that is why the answer is negative 1.19kJ!

### Re: Textbook question 4A.13

Posted: Thu Feb 06, 2020 7:45 am
The reaction is mixing with the products so heat will transfer to the insulated chamber of the calorimeter, causing the internal energy to decrease.

### Re: Textbook question 4A.13

Posted: Thu Feb 06, 2020 11:17 am
The reaction is transferring heat to its surroundings, so the system itself is losing heat. Therefore, it will have a negative value.

### Re: Textbook question 4A.13

Posted: Fri Feb 07, 2020 12:28 am
If a reaction is exothermic it loses heat to surroundings so the change must be negative. I think you are getting value for calorimeter not system.

### Re: Textbook question 4A.13

Posted: Fri Feb 07, 2020 1:17 pm
try to think of the - and + signs as how we express the energy change from one perspective. Like it was said before, the reaction is exothermic so it'll leave to it's surroundings, lessening the amount of heat available. From this perspective, heat is being taken away and moved somewhere else.