Topic 4A Exercises: 4A. 13

[Sean Phen]

4A-13: "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=−3.50kJ), 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 0.200M 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 reaction mixture as a result of the neutralization reaction?"

I am absolutely confused on what formula to use. I looked at the solution manual, but it did not make sense to me. Can someone elaborate on how to do this?

[Ivan Chen 2H]

The question wants you to find the change in internal energy for the neutralization reaction, so we need to find the amount of heat and work done in the reaction because deltaU = q + w. Since the calorimeter is at constant volume, and work can be calculated with w= -PdeltaV, that means that no work is done since deltaV is zero, which means we only need to see how much heat is released in the reaction. By comparing the amount of heat released to the temperature raised in the first reaction, you can figure out the heat capacity of the calorimeter with formula c= q/deltaT. Then you can apply the heat capacity to the temperature of the second reaction to figure out how much heat is released in the neutralization reaction.

[Asia Yamada 2B]

First, you find the heat capacity of the calorimeter with the formula q = C•deltaT (you use 7.32 for delta T and 3.50 for q). Then, you use that heat capacity and the same formula to find the heat of surroundings of the second experiment. The heat of the system is the opposite of the heat of surroundings and since delta U = q + w and there was no work done, then delta U = -1.19 kJ.