8.25

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Matthew Lee 3L
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Joined: Fri Sep 29, 2017 7:07 am

8.25

Postby Matthew Lee 3L » Thu Jan 18, 2018 5:42 pm

I'm having trouble on problem 8.25. I'm not sure what it's exactly asking and how to solve it. Thanks!

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 7.32 C. 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 C. What is the change in the internal energy of the neutralization reaction?

Joyce Lee 1C
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Re: 8.25

Postby Joyce Lee 1C » Thu Jan 18, 2018 6:36 pm

So the question is asking you to find ΔU using the information provided.
calibration of the calorimeter: q = -3.5- kJ, V = 0.200 L, ΔT = 7.32 degrees Celsius
subsequent experiment: 100.0 mL of 0.200 M KOH, ΔT = 2.49 degrees Celsius
-q = qcal and Ccal = qcal/ΔT
to find the C (specific heat) from the calorimeter we use the equations above.
Ccal = 3.50 kJ/7.32 K = 0.478 kJ/K
then, plug this specific heat into the equation q = -Ccal ΔT to find ΔU which is equal to q
Δ U = q = -1(0.478 kJ/K)(2.49 K) = -1.19 kJ

Gianna Apoderado 1B
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.25

Postby Gianna Apoderado 1B » Thu Jan 18, 2018 10:20 pm

You can also look at it as finding the q of the reaction, or qrxn. So, using the relationship of qrxn = -qcal, your calculated value of the specific heat of the calorimeter and the change in temperature, you can find the solution (much like how Joyce did!)


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