Homework Question 8.25

Moderators: Chem_Mod, Chem_Admin

aristotelis1H
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Homework Question 8.25

Postby aristotelis1H » Thu Jan 21, 2016 8:34 pm

"8.25 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 7.23oC. 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 2.49oC. What is the change in the internal energy of the neutralization reaction?"

I understand that in solving,
∆U = ∆H - P∆V
∆V = 0 because of constant volume
therefore ∆U = q
and
qsys + qsurr = 0
therefore qsys = - qsurr

In the solutions manual, whenever the given value of q is used in a calculation, the positive value is used. However, the question directly states that heat was released (meaning a negative q value should be used in calculations); however, this would result in a positive answer? Why is 3.5 kJ used in calculations and not -3.5 kJ??

Thank you :)

Chem_Mod
Posts: 17475
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 391 times

Re: Homework Question 8.25

Postby Chem_Mod » Fri Jan 22, 2016 2:15 pm

The heat released by running this reaction has to be released into the environment (in this case all the heat is absorbed by the calorimeter). Therefore you are correct in stating that qsys = - qsurr. Heat is being added to the calorimeter.

To answer your question, we are initially calculating the heat delivered to the calorimeter from the heat released by the initial reaction and this value is positive because we saw a rise in the temperature as well as the heat (the heat is going from the system to the calorimeter). This value gives us the specific heat equation of the calorimeter that we use to find the q for the neutralization reaction in part (ii).

Shirley Wong 2E
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Question 8.25

Postby Shirley Wong 2E » Sat Jan 23, 2016 2:27 pm

I had a question regarding the temperature in this problem. In the solutions manual, it just used the temperatures they gave us (7.32C and 2.49C) and changed the unit to K. I was just wondering why we wouldn't have to convert the temperature to K and why we can just change the unit like that?

aristotelis1H
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Question 8.25

Postby aristotelis1H » Mon Jan 25, 2016 9:17 pm

Because a change in temperature, whether in Kelvin or in Celsius, will still result in the same difference (since changing to Kelvin you just add 273 to both temperature values, it will make no difference in your answer).


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest