HW #18

[505801516]

Im confused on this question on the homework
A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.51 kJ
of heat in 0.400 L
of solution in the calorimeter (q=−1.51 kJ)
, resulting in a temperature rise of 3.98 ∘C
. In a subsequent experiment, 200.0 mL
of 0.40 M HClO2(aq)
and 200.0 mL
of 0.40 M NaOH(aq)
were mixed in the same calorimeter and the temperature rose by 4.93 ∘C
. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

[Gregory_Kislik_2C]

The first step is to figure out the Calorimeter Constant:

q = -CcaldeltaT
-1.51 * 10^3 = -Ccal(3.98)
Ccal = 379.4 J/K

Then, we can use the same equation to figure out the q of the neutralization reaction:

q = - (379.4 J/K) (4.93 K)
q = -1870 J

Plug this into the internal energy equation, and since this is a constant volume calorimeter, and that this was happening in solution that took up the entire volume, we know that there is not work of expansion, so w = 0:

deltaU = q + w = q + 0 = -1870 J

I hope this helps and please let me know if there are any errors.

[105988375]

Hi! adding on to that, I don't think that there is any need to multiply 1.51 by 10^3 as its already in kJ, not just J. Other than that, the process looks like what I did!