Achieve #19

[Kaethe Zappacosta 2L]

A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.44 kJ of heat in 0.600 L of solution in the calorimeter (q=−1.44 kJ) , resulting in a temperature rise of 2.22 ∘C . In a subsequent experiment, 300.0 mL of 0.20 M HClO2(aq) and 300.0 mL of 0.20 M NaOH(aq) were mixed in the same calorimeter and the temperature rose by 4.94 ∘C . What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I don't really know what to do in terms of work when we are given a reaction. Do we write a balanced equation and do something with it?

[Jacqueline Wheeler 3I]

First you need to find the heat capacity of the calorimeter(C cal) which is equal to the heat gained by the calorimeter divided by the change in temperature. So C cal = 1.44 kJ / 2.22 degrees celsius. Since the system (the calorimeter) is closed, we assume the heat lost by the reaction is equal to the heat gained by the calorimeter. So q of the reaction is equal to the negative value of q cal. We can then use the equation q cal = C cal times the change in temperature. Using the value you calculated for C cal, multiply that by 4.94 degrees celsius. Then multiply this value by negative 1 to get the value for q of the reaction. The change in U is equal to the heat released by the reaction because no other processes are taking place.