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Report Sheet and Data Analysis - CHM 111 Determine the Enthalpy of Chemical Reaction - Hess's Law
General Chemistry I (CHEM 211)
George Mason University
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Chem assignmentPreview text
Revised 11/15/
NAME_ ASIFA PARVEEN STATION #_B_DATE__04/13/ CHM 111 DETERMINE THE ENTHALPY OF CHEMICAL REACTION - HESS’S LAW REPORT SHEET AND DATA ANALYSIS PART I, II, AND III Reaction 1 Reaction 2 Reaction 3 Maximum temperature (°C) 36°C 16°C 31°C Initial temperature (°C) 22°C 22°C 22°C Temperature change (∆ T ) 14°C -5°C 9°C Heat, q (kJ) 6 -2 4 ∆ H (kJ/mol) 62/mol -23/mol 40/mol
DATA ANALYSIS 1. Calculate the amount of heat energy, q , produced in each reaction. Use 1 g/mL for the density of all solutions. Use the specific heat of water, 4 J/(g•°C), for all solutions. Reaction 1 : 100ml1/ml 103g4/g.°C * 14°C = 6242/103= 6 Reaction 2 : 100ml1/ml 103g4/g.°C * -5°C = -2324/103= -2 Reaction 3: 100ml1/ml 103g4/g.°C * 9°C = 4047/103= 4 2. Calculate the enthalpy change, ∆ H , for each reaction in terms of kJ/mol of each reactant. React 1 : 6.24KJ/0 = 62/mol React 2: 2.32Kj/0 = -23/mol React 3: 4.05KJ/0 = 40/mol 3. Use your answers from 2 above and Hess’s law to determine the experimental molar enthalpy for Reaction 3. 62+ (-23) = 39/mol = ∆ H the above equation agrees with the experiment value in equation 3 = (40/mol) from Hass law: balanced reaction equation for 3rd Reaction. 4. Use Hess’s law and the accepted values of Δ H in the Pre-Lab Exercise to calculate the Δ H for Reaction 3. How does the accepted value compare to your experimental value? I got 40/mol for ΔH for reaction 3and by hess law 85+40= 126/mol 85.6KJ/mol-40/mol = 45.1/126 KJ/mol= 0*100 = 35.
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50/ 103ml = 0*2m = 0.
Revised 11/15/
- Does this experimental process support Hess’s law? Suggest ways of improving your results. (Use another paper if needed).
Yes, this experiment supports Hess’s Law. The enthalpy of reactions 1 and 2 add up to the calculated enthalpy for reaction 3, even though it's not the same as the accepted value. If the experiment took place in a closed system or in our lecture lab, then I think that no heat would be released into the surroundings. Instead of using a thin Styrofoam cup, one could use a thermos flask or calorimeter with multiple layers that has a vacuum between two of the layers. Vacuums have a 0-thermal conductivity, so no heat would be lost resulting in a more accurate set of data.
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Report Sheet and Data Analysis - CHM 111 Determine the Enthalpy of Chemical Reaction - Hess's Law
Course: General Chemistry I (CHEM 211)
University: George Mason University
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