Part 9: Measuring the enthalpy of the reaction of Mgo(s)+2 HCI Mass of Mgo 262 m
ID: 1043115 • Letter: P
Question
Part 9: Measuring the enthalpy of the reaction of Mgo(s)+2 HCI Mass of Mgo 262 mL Volume of HCI O Table 7: Temperature Data for Mgo+ HCI Time (s) | Temperature (°C) | Time (s) | Temperature (°C) | Time (s) | Temperature (°C) 0 13I.C 120 23. 270 245 9C 30023.2 180 | 25.-7 ? | 330 |25.3? 150.s 60 90123- | Addition of MgO| 240 | 27.3"C | 390 28,4 °C Part 10: Calculating the enthalpy of the reaction of MgO(s)+ 2 HCI(aq) Table 8: Comparison of Experimental and Calculated AH(6) Mass of MgO used (g) o.63 Mols of MgO used AT (C) q (J) # mol MgO reacted Experimental ??%.pr(6) (kJ/mol) Theoretical ??.eo(6) (kJ/mol) Calculated in Prelab 23.1° % error between measured and calculated ??96)- -96Explanation / Answer
Ans. # Assuming density of HCl solution to be 1.00 g/ mL-
# Total mass of reaction mixture = 100.0 g + 0.63 g = 100.63 g
# Moles of Mg used = Mass / MW = 0.63 g / (24.305 g/ mol) = 0.02592 mol
# Heat released = q = - (m s dT) - equation 1
Where,
q = released during reaction
m = mass of reaction mixture
s = specific heat of sample = 4.184 J g-10C-1
dT = Final temperature – Initial temperature
dT = Highest temperature attained during reaction- Initial temperature
= 28.40C – 23.10C = 5.300C
Putting the values in equation 1-
q = - (100.63 g x 4.184 J g-10C-1 x 5.300C) = -2231.490376 J
# Experimental dH0exp = Total heat released / Moles of Mg
= (-2231.490376 J) / 0.02592 mol
= -86091.45 J/mol
= 86.091 kJ/mol
# Experimental dH0exp using dT = 23.10C
Experimental dH0exp = - (100.63 g x 4.184 J g-10C-1 x 5.300C) / 0.02592 mol
= -375.228 kJ/mol
# Theoretical dH0theo= = - 462.0 kJ/mol
# % error = [(dH0theo - dH0exp) / dH0theo] x 100
= [(-462.0 kJ/mol + 375.228 kJ mol-1) / (-462.0 kJ/mol) ] x 100
= 18.78 %
Note: The actual dT as in table seems to be 5.300C. If you get 23.10C, please re-assure it. Using 5.300C would give a must higher % error.