The following experiment was performed. The mass of an empty coffee cup calorime

Question

The following experiment was performed. The mass of an empty coffee cup calorimeter was 5.25 g. To the calorimeter was added 80.0 mL of 1.50 M hydrobromic acid; the mass of the calorimeter and solution was 85.15 g. The temperature of the solution in the calorimeter was 20.00°C. To the calorimeter was added 45.0 mL of a 2.00 M aqueous solution of sodium hydroxide, also at 20.00°C. After the two solutions had been mixed and the temperature of the resultant solution was no longer changing, the temperature of the solution in the calorimeter was 28.57°C, and the mass of the calorimeter and contents was 132.75 g. The specific heat of liquid water is 4.18 J/g·°C; the specific heat of the solution present in the calorimeter at the end of the experiment is 4.20 J/g·°C. The energy change per mole of hydrobromic acid that reacts in the reaction between hydrobromic acid and sodium hydroxide is -52.4 kJ/mol. Calculate the heat capacity of the calorimeter.

Explanation / Answer

Let heat capacity of the calorimeter be C J/K.

Moles of HBr(hydrobromic acid)= 0.080L*1.50Mol/L=0.12mol

Moles of NaOH added=0.045L*2Mol/L=0.090mol

So moles of HBr reacted =Moles of NaOH added(limiting reagent)=0.090mol

Energy released=0.090mol*52.4 kJ/mol=4.716KJ=4716J

M(HBr Solution)=85.15g-5.25g=79.90g

M(NaOH)=132.75 g-85.15g=47.6g

Now energy released in reaction=Energy taken by solutions and calorimeter.

4716=(C+4.2*127.5)*(28.57-20)=(C+4.2*127.5)*8.57=8.57C+4589.24

We Get C=14.8J/K which heat capacity of the calorimeter.

Comment in case of any doubt.

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