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Hess' Law

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  • December 14, 2013
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Hess’ Law

To find out the enthalpy change of Mg+ ½O2=MgO, we used a calorimeter, thermometer, 0.2 g of Mg, 0.2 g of MgO, and 2.0 M of HCl. We used a thermometer to measure the initial and final temperatures in Celcius. We recorded the initial temperature of the HCl. After we put the Mg or MgO into the calorimeter, we put in the HCl and covered it using a lid, mixed it around with the thermometer, and recorded the final temperature. We also used this method to find the enthalpies of the equations for the reactions. Data:

Trial #
Mass of Mg
Mass of HCl
Initial Temperature
Final Temperature
1
0.2 g
10.01 g
23.67 °C
88.19 °C
2
0.2 g
10.11 g
24.71 °C
84.51 °C

Trial #
Mass of MgO
Mass of HCl
Initial Temperature
Final Temperature
1
0.2 g
9.79 g
22.97°C
40.27°C
2
0.2 g
9.91 g
23.19°C
38.43°C

Calculations: ** The -Qreaction=Qsolution**

Mg:
Trial 1:
q=mCΔT
q=10.01g x 4.184 J/g°C x (88.19-23.67)°C= 2699.52 J =2.70 kJ 0.2g Mg x1 mole Mg/24.31 Mg=0.0082 mol Mg
-2.70 kJ/0.0082 mol = -329.27 kJ/mol = ΔH
Trial 2:
q=mCΔT
q=10.11g x 4.184 J/g°C x (84.51-24.71)°C=2529.55 J =2.53 kJ 0.2g Mg x1 mole Mg/24.31 Mg=0.0082 mol Mg
-2.53 kJ/0.0082 mol = -308.54 kJ/mol = ΔH
*Average of the ΔH for Mg= -318.9 kJ/mol
MgO:
Trial 1:
q=mCΔT
q= 9.79g x 4.184 J/g°C x (40.27-22.97)°C= 708.63 J = 0.71 kJ 0.2 MgO x 1 mole MgO/40.31 MgO=0.005 mol MgO
-0.71 kJ/0.005 mol = -142.0 kJ/mol = ΔH
Trial 2:
q=mCΔT
q= 9.91g x 4.184 J/g°C x (38.43-23.19) °C = 631.90 J = 0.63 kJ 0.2 MgO x 1 mole MgO/40.31 MgO=0.005 mol MgO
-0.63 kJ/ 0.005 mol = -126 kJ/mol = ΔH
*Average of the ΔH for MgO =-134 kJ/mol
-------------------------------------------------------------------------------------------------- Given: Mg+ ½O2MgO
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (l)ΔH= -318.905 kJ/mol MgO (s) + 2HCl (aq)  MgCl2 (aq) + H2O (l)ΔH= -134 kJ/mol H2 (l) + ½ O2 (g)  H2O (l) + heatΔH= -285.8 kJ/mol
---------------
Given: Mg+ ½O2MgO
Mg (s) + 2HCl (aq) ...