Procedures:
Reaction of calcium with dilute hydrochloric acid
- The mass of a clean and dry polystyrene cup is weighted and recorded.
- 1g of calcium metal is weighted out approximately and the mass is recorded.
- A measuring cylinder is used to measure out 100 ml of approximately 1.0 M hydrochloric acid and it is placed in the polystyrene cup.
- The temperature of the acid is measured and it is recorded.
- The weighed portion of calcium is added and it is stirred thoroughly with thermometer until all the metal has reacted.
- The maximum temperature attained by the solution is recorded
- The experiment is repeated and the average temperature rise is determined if there is time.
Reaction of calcium carbonate with dilute hydrochloric acid
- The mass of a clean and dry polystyrene cup is weighted and recorded.
- 2 – 3 g of dry, powdered calcium carbonate is weighted out directly into the polystyrene cup.
- 100 ml of approximately 1.0 M hydrochloric acid is placed in a measuring cylinder.
- Temperature is recorded and the acid is poured on the carbonate in the polystyrene cup.
- It is stirred briskly with the thermometer and the maximum temperature is recorded reached by the solution.
- The experiment is repeated and the average temperature rise is determined if there is time.
Result:
Calculations:
The enthalpy change of formation of water: -286 kJ mol-1
The enthalpy change of formation of carbon dioxide: -393 kJ mol-1
Part A
The ionic equation for the reaction taken place
Ca(s) + 2H+(aq.) → Ca2+(aq.) + H2(g)
Heat evolved by one mole of calcium.
Heat change = (100/1000)(4.2)(28.5)
= 1197
= 11.97kJ
Number of moles of calcium = 1.01/40.1
= 0.025187 mol
Heat evolved by one mole of calcium
=11.97/0.25187
= 475 kJ mol-1
Part B
The ionic equation for the reaction taken place
CO32-(aq.) + 2 H+(aq.) → CO2(g) + H2O(l)
Heat evolved by one mole of calcium carbonate.
Heat change = (100/1000)(4.2)(1.5)
= 0.63kJ
Number of mole of calcium carbonate = 3/(40.1+12+16*3)
= 0.02997 mol
Heat evolved by one mole of calcium carbonate
=0.63/0.02997
= 21kJ mol-1
Enthalpy change of formation of calcium carbonate
= (enthalpy change of formation of calcium dioxide + enthalpy change of part A) - (enthalpy change of part B + enthalpy change of formation of water)
= (-475) + (-393) + (-286) – (21)
= -1133 kJ mol-1
Discussion: The mass of calcium metal is the limiting factor, the concentration of the hydrochloric acid just affects the rate of the reaction and should be excess. So, it is not important to the exact concentration. And also, we should not use the measuring cylinder, the scale of the measuring cylinder is far from accurate, we should use the pipette. There are some major sources of inaccuracy in the experiment, like heat loss to surrounding by evaporation, conduction and radiation. The thermometer absorbed some energy. There are two assumptions which are the specific heat capacity and the densities of the solution are the same as those of the water. Some of the samples stick in the weighing bottle. The reading of the thermometer is not accurate enough. In order to make the experiment more accurate, we can have the following improvement, like uses the vacuum flask calorimeter instead of the polystyrene foam cup or uses the cotton wool to cover the outer surface of the cup for reduce heat loss to surroundings, use a more accurate reading thermometer. Find out the specific heat capacity of the solution and the thermometer. Use the method of “weighing by difference” to find out the amount of the samples used. The law which is used is conservation of energy and Hess’s Law, Hess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place, The law of conservation of energy states that energy can neither be created nor destroyed but can be changed from one form to another. The importance of Hess’s law lies in the fact that enthalpy changes can be calculated for reactions that cannot be carried out in the laboratory, or reactions that are too slow or involve the formation of side products. So, I think it is useful.
End of report
