180 165 180 180 OUESTIONS 1. What is the largest source of error in the experime

Question

180 165 180 180 OUESTIONS 1. What is the largest source of error in the experiment? 2. IHow should the two heats of reaction for the neutralization of NaOH and the two acids compare? Why? 3. The experimental procedure has you wash your thermometer and dry it after you measure the temperature of the NaOH solution and before you measure the temperature of the HCl solution. Why? 4. A 50.0 mL sample of a 1.00 M solution of CusO, is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 °C before mixing and 26.3 °C after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data, calculate AH for the process CuSO4(1 M) + 2KOH(2 M)- Cu(OH)2 (s) + K2SO4 (0,5 M) Assume that the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are additive

Explanation / Answer

1. The largest source of error in the experiment stems from the use of non-calibrated glassware like measuring cylinders. Measuring cylinders were used throughout for transferring 50.0 mL of NaOH and acid solutions. However, measuring cylinders are not as precise as a burette or pipette and introduce an uncertainty in transferring sample volumes quantitatively. It is highly likely that a slight excess or deficit of solutions were transferred. Since the heat of the reaction and hence the heat of neutralization depends on the mass of the solution transferred, supplying an erroneous mass of NaOH or the acid will introduce a large error in the calculation of heat of neutralization.

2. The two heats of reaction for the neutralization of NaOH with the two acids should be comparable to each other. We define the heat of neutralization as heat of reaction per mole of water generated. The same number of moles of NaOH were used in both the cases and hence the same moles of water were generated and hence the two heats of reaction should be comparable. Offcourse, the initial and final temperatures will be different but the change in temperature should be the same.

3. The thermometer must be thoroughly cleaned and dried before inserting in the HCl solution. If the thermometer is not thoroughly washed, small drops of NaOH will stick to the thermometer which will react with HCl. The moles of HCl neutralized will be small definitely, however, it will introduce a small error in the calculation of the temperature and lead to erroneous results overall.

4. Total volume of the solution = (50.0 + 50.0) mL = 100.0 mL.

Density of the solution is the same as that of water, i.e, 1 g/mL. Therefore, mass of the solution = (100.0 mL)*(1 g/1 mL) = 100.0 g.

Specific heat of the solution is the same as that of water, i.e, the specific heat of the solution is 4.18 J/g.C.

Temperature change of the solution = (26.3 – 20.2)C = 6.1C.

Heat absorbed by the solution = (mass of solution)*(specific heat of solution)*(temperature change of solution) = (100.0 g)*(4.18 J/g.C)*(6.1 C) = 2549.8 J.

Heat absorbed by the calorimeter = (Heat capacity of the calorimeter)*(change in temperature of the solution) = (12.1 J/K)*(6.1 C) = (12.1 J/K).(6.1 K) = 73.81 J (change in temperature in K = change in temperature in C).

Heat of reaction, H = (2549.8 + 73.81) J = 2623.61 J = 2.62361 kJ = 2.624 kJ (ans)

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