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Postby 704709603 » Mon Jan 16, 2017 4:32 pm

#25 says: A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50kJ of heat in 0.200L of solution in the calorimeter (q=-3.50kJ), resulting in a temperature change of 7.32C. In a subsequent experiment, 100.0mL of 0.200M HBr(aq) and 100.0mL of 0.200M KOH (aq) were mixed in the same calorimeter and the temperature rose by 2.49C. What's the change in the internal energy of the neutralization reaction?

So, why in the solution is q(reaction) + q(calorimeter) = 0? Why don't we use the heat capacity of water to and 0.04mol and 2.49C to calculate the enthalpy?

Thank you!

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Re: ch.8/#25

Postby mikezargari » Wed Jan 18, 2017 5:27 pm

I am pretty sure that q of the reaction + q of the calorimeter = 0 because the heat given off by the reaction is the same heat that is being absorbed by the calorimeter.

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Re: ch.8/#25

Postby Chem_Mod » Wed Jan 18, 2017 6:01 pm

The composition of the solution in experiment 1 is not specified, so you cannot figure out the heat capacity of the solution, nor can you figure out the mass of the solution. The reason that the answer does not use the other numbers is because you only need qcalorimeter to solve the problem, and the shortest way to solve this is to assume that the calorimeter will absorb any heat that the reaction gives off. To do this, you solve for C (calorimeter heat capacity) using the given q and delta T value. Since it is the same calorimeter, C remains the same for the second experiment and you simply plug in the second delta T value. qrxn is equal to -qcalorimeter because the heat of the reaction is negative since it is giving off heat, and qcalorimeter is positive because it is absorbing the heat of the reaction. I hope this helps.

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