Experiment 25 Essay

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Experiment 9: Calorimetry & Hess’s Law Date Performed: 11/01/2012 Name: Bria Burris
Date Submitted: 11/8/2012 Partner Name: Bianca Idas Chemistry 1045L-Sec#46
Zhang, Xiaoguang

Purpose
The purpose of this experiment was to test and expand on the knowledge of Calorimetry and how energy and heat moves from a system to the surroundings and vice versa. The purpose of this experiment is also to closely study the concept of enthalpy and specifically the enthalpy of a reaction.
Experimental Procedure
Dillon, Stephanie. “Calorimetry & Hess’s Law.” Laboratory Manual. Pearson Publishing, 2012, pp. 163-177.
Data and Results
Part A Data Part A | Trial 1 | Trial 2 | Exact volume of cold water | 50mL | 50 mL | Temperature of cold water | 22◦C | 22◦c | Exact Volume of Hot water | 50mL | 50mL | Temperature of Hot water | 46.5◦C | 57◦C |
Part A Calculations Trial 1 Trial 2 Mass of cold water | 50g | 50g | Tf from graph by extrapolation | 29.8℃ | 31.2℃ | ∆T of Cold water | 7.8℃ | 9.2℃ | ∆T ofHot Water | -16.7℃ | -25.8℃ | qHW | -3490.3 J | -5392 J | qCW | 1630.2 J | 1922.8 J | qcal | 1860.1 J | 3469.2 J | Ccup | 238.5 J | 377.1 J |
What ∆T should be used? Cold Water
Part B Data Part B | Trial 1 | Description of sample | Clear HCl solution and silver colored Mg strips. After experiment Mg strips no longer present. | Exact Volume of HCl | 100mL | Initial Temperature of HCl | 22◦C | Exact Mass of Mg | 0.112g |

Part B Calculations Tffrom graph by extrapolation | 25.2℃ | Mass of HCl solution | 100g | ∆Tcw for HCl | 3.2℃ | qHCl | 1337.6 J | qcal | 985 J | qrxn | 2322.6 J | ∆Hrxn for Mg | 503.8 kJ/mol |

Balanced Equation Part B:
2H⁺ (aq) + Mg(s) ->Mg²⁺ (aq) + H₂ (g)

Part C Data Part C | Trial 1 | Description of sample | Sample is a clear color. No color | Exact Volume of 100 mL HCl | 100mL | Initial Temperature of HCl | 22◦C | Exact mass of MgO | .5g |

Part C Calculations Tf from graph by extrapolation | 23.5℃ | ∆Tcw for HCl | 1.5℃ | qHCl for the cup | 627 J | qcal for the cup | 461.7 J | qrxn | 1088.7 J | ∆Hrxn for MgO | 87.8 kJ/mol |

Balanced Equation Part C
2H⁺ (aq) + MgO(s) ➝Mg²⁺(aq) + H₂O(g)

Graphs

Calculations
Part A Calculations
*Using Trial 1 data for calculations
Mass of Cold Water:
Mass of Cold Water = Volume of Cold water ×Density of Water
50mL×1g1ml=50g
∆THW for Cold water:
∆THW for Cold water= Tf-TI
29.8℃-22℃=7.8℃
∆THW for Hot water:
∆THW for Cold water= Tf-TI
29.8℃-46.5℃=-16.7℃
qHW for Hot Water: qHW for Hot Water=mc∆THW
50g4.18Jg℃-16.7℃=-3490.3 J qCW for Cold Water: qCW for Cold Water=mc∆Tcw
50g4.18Jg℃7.8℃=1630.2 J qcal for the cup: qcal= qHW-qCW
-3490.3-1630.2=1860.1 J
Ccup for the cup:
Ccup for the cup= qcal∆T
1860.1J7.8=238.5 J
Part B Calculations
Mass of HCl solution: Volume of Cold water ×Density of Water
100mL×1 g1mL=100g
∆TCW for HCl:
Tf-TI=∆T for HCl
25.2℃-22℃=3.2℃
qHCl for HCl: mc∆T=qHCl 100g4.18Jg℃3.2℃=1337.6 qcal for HCl:
Ccup× ∆THCl=qcal
307.8J×3.2=985J
qrxn: qHCl+qCal=qrxn 1337.6+985=2322.6J
∆Hrxn for Mg: qrxnnMg=∆Hrxn 2322.6J0.00461 mol=503817J=503.8Kj/mol nmol=0.112g×1 mol of Mg24.31g of Mg=0.00461 mol
Part C Calculations
∆TCW for HCl:
Tf-TI=∆T for HCl
23.5℃-22℃=1.5℃
qHCl for HCl: mc∆T=qHCl 100g4.18Jg℃1.5℃=627J qcal for the cup:
Ccup×∆THCl=qcal
307.8×1.5=461.7J qrxn: qHCl+qCal=qrxn
627J+461.7J=1088.7J
∆Hrxn for MgO: qrxnnMg=∆Hrxn 1088.7J0.0124 mol=87798J=87.8kJ/mol nmol=0.5g×1 mol of MgO40.31g of MgO=0.0124 mol

Conclusion To conclude, Part A, Part B, and Part C all went successfully they all cooled in the end with the temperature going from high to low. In Part A cold water was…