DETERMINING THE ENTHALPY OF FORMATION OF CaCO 3 Standard Enthalpy Change Standard Enthalpy Change for a reaction, symbolized as H 0 298, is defined as The enthalpy change when the molar quantities of reactants shown in a balanced chemical equation completely react to from products under standard conditions. Standard conditions are as follows: - a pressure of 1 atmosphere - a temperature of 298 K - if solution is used, 1M is specified. - Elements or compounds in their normal stable state Example H 2(g) + 1/2 O 2(g) H 2 O (l) H 0 = -285.9 kj mol -1 When one mole of hydrogen and half mole of oxygen react to form one mole water, 285.9 kj of heat are evolved under standard conditions. This is an exothermic reaction. All the substance involved must be in their normal physical states under standard conditions. The enthalpy of all elements in their most stable form and standard state is conventionally taken as zero. For example, H 0 298 [O 2(g) ] = 0 H 0 298[Na (s) ]=0 Therefore, the standard enthalpy of formation of a compound represents the enthalpy content of the compound. Hess s Law The law of conservation of energy states that energy can be neither be created or destroyed, though one form of energy can be converted to another form. This law therefore implies that the total energy content ( in the system ) is constant. Based on it, Hess s law arises. Hess s Law: The standard enthalpy change of reaction depends only on the difference between the standard enthalpy of the reactants and the standard enthalpy of the products, and not no the route by which the reaction occurs. In other words, for processes involving several stages, the enthalpy change for the reaction is equal to the algebraic sum of the enthalpy change for each intermediate stage. Page 1
The enthalpy change for the oxidation of carbon (graphite) to carbon dioxide is the same whether it is carried out Route A ( 1 stage ) or Route B ( 2 stages) Route A: C (s) + O 2(g) CO 2(g) H a = -393.5 kj mol -1 Route B: C (s) + 1/2 O 2(g) CO (g) H b1 = -110.5 kj mol -1 CO (g) + 1/2 O 2(g) CO 2(g) H b2 = -283.0 kj mol -1 H b1 + H b2 = (-110.5) + (-283.0) = -393.5 kj mol -1 The Use of Hess s Law in determining Enthalpy Changes Some reactions proceed very slowly or involve formation of side products. Hess s Law can be used to determine the enthalpy changes of such reactions which cannot be determined directly by calorimetry experiments. Page 2
The molar enthalpy of formation, H f, is defined as the enthalpy change when 1 mole of a compound is formed directly from its elements. Although some compounds (e.g., MgO) can be formed by direct synthesis from the elements, many cannot. Calcium carbonate is one of them. Below is the chemical equation representing the formation of 1 mole of CaCO 3 :- Ca (s) + C (s) + 1½ O 2 (g) CaCO 3 (s) The enthalpy change for this reaction can be found indirectly from reactions which can actually be carried out in the lab and then applying Hess s Law. Both calcium metal and calcium carbonate react with dilute hydrochloric acid to form a solution of calcium chloride. In addition, calcium metal forms H 2 gas and calcium carbonate forms CO 2 and H 2 O. The enthalpy changes for these two reactions can be measured relatively easily in the lab. For the calculation of H f for CaCO 3, two more pieces of data are needed, i.e., the enthalpies of formation of water and carbon dioxide. Page 3
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METHOD 1) H 1 for Ca + HCl Weigh out accurately about 1g of calcium metal into a weighing bottle and record the mass. Weigh a polystyrene calorimeter cup and record the mass. Use a measuring cylinder to add 100cm 3 of 1M HCl. Reweigh the cup with the acid and record the mass. Place the polystyrene cup inside a beaker for support, as shown in the diagram opposite. Place the thermometer in the acid and start the stop-clock. Record the temperature every minute for 3 minutes. On the 4 th minute, quickly add all the calcium metal and stir to mix thoroughly, keeping the lid on top of the cup. Record the temperature every minute from the 5 th minute for a further 10 minutes. Stir the solution before each reading. Empty and rinse the cup and thermometer and dry them with paper towel. 2) H 2 for CaCO 3 + HCl Weigh out accurately between 2 and 3g of calcium carbonate powder into a clean weighing bottle and record the mass. Using the same polystyrene cup as before, add a fresh lot of 100cm 3 of 1M HCl using the measuring cylinder, weigh again and record the mass. Repeat the steps in experiment 1, recording the temperature every minute and adding the solid on the 4 th minute. Empty and rinse the cup and thermometer and place them on the wet tray. Clear away all the apparatus and leave your bench clean and tidy. Page 5
EXPERIMENT 1 EXPERIMENT 2 Mass of calcium metal Mass of calcium carbonate Mass of empty cup Mass of cup plus acid Mass of empty cup Mass of cup plus acid Mass of acid 1 (M 1 ) Mass of acid 2 (M 2 ) TIME (minutes) 1 2 3 TEMPERATURE ( O C) EXPERIMENT 1 EXPERIMENT 2 4 ADD THE SOLID 5 6 7 8 9 10 11 12 13 14 15 FINDING THE TEMPERATURE CHANGES Plot two separate graphs of temperature against time, one for each experiment. Extrapolate the points find out the instantaneous temperature change at the 4 th minute in each reaction. Call these t 1 and t 2. Page 6
CALCULATING THE ENERGY CHANGES Take the Specific Heat Capacity (c) for the solutions to be 4.18 J.g -1. O C -1 Take the mass of the solutions to be the mass of acid used in each experiment Ignore the sign of the temperature change (rise or drop) just use the numerical value 1) For Ca metal Heat change = mass of solution 1 c t 1 =. J = kj Moles of Ca = mass of Ca Atomic Wt. (At.Wt. for Ca = 40.1) = Energy per mole = Heat change number of moles =. kj.mol -1 Now decide whether the reaction was exothermic (temperature rise) or endothermic (temperature drop). Use the correct sign convention to write your value for the enthalpy change for this reaction below:- H 1 =.. kj.mol -1 2) For CaCO 3 Heat change = mass of solution 2 c t 2 =. J = kj Moles of CaCO 3 = mass of CaCO 3 Molar Mass (Molar Mass = 100.1) = Energy per mole = Heat change number of moles =. kj.mol -1 Now decide whether the reaction was exothermic or endothermic. Use the correct sign convention to write your value for the enthalpy change for this reaction below:- H 2 =.. kj.mol -1 3) Use Hess Law and your values for H 1 and H 2 and the enthalpies of formation for water and carbon dioxide to calculate the enthalpy of formation for CaCO 3. H 2 (g) + ½O 2 (g) H 2 O (l) H f = -285.8 kj.mol -1 (call this H 3 ) C (s) + O 2 (g) CO 2 (g) H f = -393.5 kj.mol -1 (call this H 4 ) Page 7