Question

1. Read the write up through the Data Analysis section. A student carried out the experiment described there. The student used 0.4894 g of Mg in the first reaction and found that the temperature of the solution went from 22.1 to 42.7'C. In the second reaction, 0.803 g of Mgo was used and the solution went from 22.7 to 28.7*C. Calculate the following. Show all calculations. AH, The enthalpy of the reaction of magnesium with acid. AT = Mass of final solution = moles of Mg = AH =

Answer 1

Enthalpy of formation of MgO

(A) The enthalpy of the reaction of magnesium with acid,

Mg + 2 HCl ---> MgCl₂ + H₂

▵T = 42.7 - 22.1 = 20.6 ^oC

Mass of final solution = 100 + 0.4894 g = 100.4894 g [HCl = 100 mL or 100 g as density = 1 g/mL]

q = m C_p ▵T = 100.4894 x 4.184 x 20.6 = 8.661 kJ

Moles of Mg = 0.4894 g/24 g/mol = 0.0204 mole

▵H_f = -8.661 kJ/0.0204 mol = -424.73 kJ/mol

(B) The enthalpy of the reaction of magnesium oxide with acid

MgO + 2HCl --> MgCl₂ + H2O

▵T = 28.7 - 22.7 = 6.0 ^oC

Mass of final solution = 100 + 0.803 g = 100.803 g [HCl = 100 mL or 100 g as density = 1 g/mL]

q = m C_p ▵T = 100.803 x 4.184 x 6.0 = 2.530 kJ

Moles Mg = 0.803 g/40 g/mole = 0.0201 mole

▵H_f = -2.53 kJ/0.0201 mole = -125.87 kJ/mole

(C) Enthalpy of formation of MgO

Mg + 2HCl --> MgCl₂ + H₂            ▵H_f = -424.73 kJ/mol

MgCl₂ + H₂O ---> MgO + 2 HCl     ▵H_f = +125.87 kJ/mol

H₂ + 1/2O₂ ------> H₂O                 ▵H_f = -285.8 kJ/mol

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Mg + H₂O ----> MgO + H₂             ▵H_f = -584.7 kJ/mol

% error = (593.3 - 584.7) x 100/593.3 = 14.5 %