Achieve Question #19

[Naman Jain 3F]

A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.27 kJ of heat in 0.400 L of solution in the calorimeter (q=−1.27 kJ) , resulting in a temperature rise of 3.78 ∘C . In a subsequent experiment, 200.0 mL of 0.30 M HClO2(aq) and 200.0 mL of 0.30 M NaOH(aq) were mixed in the same calorimeter and the temperature rose by 5.58 ∘C . What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I understand how to get the answer, but I was wondering why we don't have to factor in the mass of the reactants of the neutralization reaction when finding Q of the reaction mixture.

[Rena Wu 3E]

Hi Naman, I believe that you don't need to factor in the components of the neutralization reaction because the subsequent reaction is occurring in the same constant-volume calorimeter. From using the information, you can calculate the heat capacity of the calorimeter (in kJ/C). Your q value would then be equal to the heat capacity constant for the calorimeter multiplied by the temperature change in the second reaction. A side note, this process is only possible because the two reactions use the SAME calorimeter. Heat capacity is an extensive property, so each calorimeter used would have a different heat capacity from the other.

I also think that it isn't necessary to find Q of the reaction mixture in this problem, since we are not dealing with concentrations of the products and reactants. If you were referring to q, the heat energy transferred in the reaction, then the approach above should work. Hope this helps!

[Naman Jain 3F]

Hi Naman, I believe that you don't need to factor in the components of the neutralization reaction because the subsequent reaction is occurring in the same constant-volume calorimeter. From using the information, you can calculate the heat capacity of the calorimeter (in kJ/C). Your q value would then be equal to the heat capacity constant for the calorimeter multiplied by the temperature change in the second reaction. A side note, this process is only possible because the two reactions use the SAME calorimeter. Heat capacity is an extensive property, so each calorimeter used would have a different heat capacity from the other.

I also think that it isn't necessary to find Q of the reaction mixture in this problem, since we are not dealing with concentrations of the products and reactants. If you were referring to q, the heat energy transferred in the reaction, then the approach above should work. Hope this helps!

Hi Rena, thank you for answering my question, your explanation cleared things up perfectly!