L OBJECTIVES 1. To measure heat produced in chemical reactions. 2. To apply Hess
ID: 505400 • Letter: L
Question
L OBJECTIVES 1. To measure heat produced in chemical reactions. 2. To apply Hess' Law to measured enthalpies. 2. To determine the enthalpy of formation ofMgo(s). To understand temperature changes as a result of chemical reactions. 3. IL INTRODUCTION Energies for chemical reactions are generally measured and recorded as enthalpy changes. An efficient way to tabulate enthalpy changes for chemical reactions is to record the standard enthalpy of formation for each compound, where AHindicates enthalpy change (products reactants), findicates formation and indicates standard conditions, which for thermochemical reactions means of pressure. The standard enthalpy of formation of a enthalpy change for the formation of one mole of compound (at standard conditions) from the elements in their standard thermochemical states. In this experiment we are interested in determining experimentally the standard enthalpy of formation of Mgo0s, AH [Mgo(s which is equal to the enthalpy change for the reaction: Mg(s) O2(g) Mgo(s) at 298.15 K and 1 atm. The elements that make up magnesium oxide are magnesium, which at 25 °C is Mg(s), and oxygen, which at 25 °C and 1 atm exists naturally aso20g og(g is a form of the pure element in nature, but O20g) is the standard state. It is to measure the enthalpy change for this reaction directly, so we will measure enthalpy changes for the reactions: AH.. Mg(s) 2 HCl (aq) (2) Mgoos 2 HCl(aq) These along with the standard enthalpy of formation for water: can be added together by Hess' Law to give AH for: AHExplanation / Answer
Ans. Amount of heat gained by water for recorded temperature increases is given by-
q = m x s x dT - equation 1
Where,
m = mass in gram,
s = specific heat of water = 4.184 J g-10C-1
dT = change in temperature
#1. For Reaction 1:
Heat gained by water, q
=100.0 g x (4.184 J g-10C-1) x 16.60C = 6945.44 J = 6.945 kJ
The amount of heat lost during reaction of Mg(s) with HCl must be equal to the amount of heat gained by water,
Or, q’ = - 6.945 kJ ; [-ve sign indicates heat loss during reaction]
#2. Mass of Mg(s) reacted with HCl = 0.452 g
Moles of Mg(s) = Mass/ Molar mass = 0.452 g/ (24.305 g/ mol) = 0.0186 mol
Heat lost per mol of Mg(S), dH = q’ / moles of Mg(s)
Or, dH = -6.945 kJ / 0.0186 mol
Hence, dH = -373.39 kJ/ mol
#3. For Reaction 2:
Heat gained by water, q
=100.0 g x (4.184 J g-10C-1) x 10.10C = 4225.84 J = 4.226 kJ
The amount of heat lost during reaction of MgO(s) with HCl must be equal to the amount of heat gained by water,
Or, q’ = - 4.226 kJ ; [-ve sign indicates heat loss during reaction]
#4. Mass of MgO(s) reacted with HCl = 1.32 g
Moles of MgO(s) = Mass/ Molar mass = 1.32 g/ (44.3044 g/ mol) = 0.0297 mol
Heat lost per mol of MgO(S), dH = q’ / moles of Mg(s)
Or, dH = -4.226 kJ / 0.0297 mol
Hence, dH = -142.29 kJ/ mol
#5. Reaction 4 can be written as sum of reaction 1, 3 and reverse of reaction 3 as follows-
Mg(s) + 2HCl(aq) --------> MgCl2(aq) + H2(g) , dH = -373.39 kJ/ mol - 1
H2(g) + ½ O2 (g) ---------> H2O(l) , dHf0 = -285.8 kJ/ mol - 3
H2O(l) + MgCl2(aq) ------> MgO(s) + 2HCl(aq) , dH = -532.24 kJ/ mol - (-2)
Mg(s) + ½ O2(g) ---------> MgO(s) - 4
Using Hess law, dH of reaction 4 is given by-
dH(4) = dH(1) + dH(3) – dH(2)
= (-373.39 kJ/ mol) + (-285.8 kJ/ mol) – (142.29 kJ/ mol)
= - 516.95 kJ/ mol
#6. Theoretical value for reaction 4 = - 601.6 kJ/ mol
% error = { [- 601.6 kJ/mol – (- 516.95 kJ/ mol)] / (- 601.6 kJ/mol) } x 100
= 14.07 %
The larger % error might be due to “ignoring the heat capacity of calorimeter” or other experimental errors like incomplete reaction, etc.
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