Achieve 5&6 #19

[andrea guzman 1F]

A constant volume calorimeter was calibrated by carrying out a reaction to release 1.75 kJ of heat in 0.500L of solution in the calorimeter (q= -1.75 kJ), resulting in a temperature rise of 3.43 C. In a subsequent experiment, 250.0 ml of of 0.10 HClO₂ (aq) and 250.0 ml of 0.10 NaOH (aq) were mixed in the same calorimeter and the temperature rose by 6.32 C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

Hi I'm a confused as to how to start this problem and the steps to solve it.

[George_Furey_1C]

start by finding the Ccal using qcal=Ccal*T. then once you have found the Ccal you can multiply that by the second change in temperature to get the qcal for the second part. the change in internal energy is equal to the opposite of this value because qcal=-qreaction

[Maleeha Amir 2E]

First, you would calculate the specific heat with q = c times delta T. Once you have the specific heat, you can input the temperature change and the specific heat of the new reaction to get q.

[906013563]

1. Use the calibration data to determine the calorimeter's heat capacity
2. use the temperature rise from the neutralization reaction to find the heat of the reaction
3. calculate the heat capacity of the calorimeter,
4. use the calorimeter's heat capacity to calculate the heat of the neutralization reaction
5. Since qrxn is the heat of the reaction for 0.500 L, you would need to adjust for the actual volume used in the neutralization reaction, which is 0.250 L of each solution (totaling 0.500 L), to find ΔU.
Hope this helps, if you would like me to provide equations at each step let me know