Expt 10, Thermochemistry. Name: Lab Partner: Date: Introduction. Part A. Determination of Heat Capacity of the Calorimeter

Transcription

1 Expt 10, Thermochemistry Name: Lab Partner: Date: Introduction Commented [1]: Use a nice title page better Commented [2]: What are you going to do and why are you doing it. Add a couple of sentences outlining what you expect the results to be. Part A Determination of Heat Capacity of the Calorimeter Initial temperature of calorimeter and 50 ml cold H 2O (T cold): Initial temperature of 50 ml warm H 2O (T hot) : Table 1: Time vs temperature data: Heat Capacity of the Calorimeter [Insert a suitable table here.] Plot Temp ( o C) y vs. Time (s) x. Extrapolate the graph back to the y axis to calculate T final, the final temperature of the calorimeter and the contents. Inert the (excel or equivalent) plot showing the extrapolation, don t forget title, axes labels, units and trendline. Commented [3]: Delete my comments, as always, include graph as the appendix. Don t forget how to calculate Tfinal, extrapolating the few points after the max temp back to the y axis - Heat lost by hot water = heat gained by cold water + heat gained by calorimeter [comment: note that you assume that the initial temperatures of the calorimeter and of the cold water are the same.] - (ml hot water x 1.00 g/ml x Thot x 4.18 /g o C) = (ml cold water x 1.00 g/ml x Tcold x 4.18 /g o C) + ( Tcold x Ccal) This equation can be rearranged to solve for C cal the heat capacity of the calorimeter If you obtain a negative value for C cal assume that the value of C cal is zero.. [..comment: but, if you do need to assume the heat capacity of the calorimeter is zero, mention this in the discussion and comment on why you think this assumption was necessary.] Show all your calculations of C cal Part B Commented [4]: Check you didn t do it wrong first before you assume this. Commented [5]: You can leave in the first equation in bold but the rest of the calculation should be your data. Don t forget to include the correct units which your equation should spit out. Determination of Heat of Neutralization 1

2 NaOH(aq) Molarity: HNO 3(aq) Molarity: _ Initial temperature of 50 ml NaOH(aq) and Calorimter (T b): Initial temperature of 50 ml HNO 3(aq) in grad cylinder (T a) : Table 2: Time vs temperature data: Heat of Neutralization Commented [6]: Make it more descriptive. [insert a suitable table here] Plot Temp ( o C) y vs. Time (s) x. Extrapolate the graph back to the time at which the liquids are mixed to calculate T final the final temperature of the calorimeter and the contents. Insert the (excel or equivalent) plot showing the extrapolation, units, titles, labels and trendline equation. Commented [7]: Delete my comments, as always, include graph as the appendix. Don t forget how to calculate Tfinal, extrapolating the few points after the max temp back to the y axis T acid = T final - T a T base = T final T b -q neutralization = q solution q solution = heat gained by acid + heat gained by base + heat gained by calorimeter q soln = (ml acid x 1.00 g/ml x Tacid x 4.18 /g o C) + (ml base x 1.00 g/ml x Tbase x 4.18 /g o C)+ ( Tbase x Ccal) Commented [8]: Remember the correct sign for the heat of neutralization is it exo or endothermic? Convert it to k and then divide it by the mol of base reacted to give an answer in units of k/mol Part C Determination of Heat of Reaction Mass of CuSO 4.5H 2O used : use the analytical balance to weigh approximately 5 g Mass of Zinc used : use the analytical balance to weigh approximately 6.5 g Measure as accurately as possible 95 ml of Distilled Water. Transfer the CuSO 4.5H 2O to the 95 ml of distilled H 2O in the calorimeter and dissolve completely. Commented [CC9]: delete Commented [CC10]: delete Commented [CC11]: delete Commented [CC12]: delete Initial temperature of 95 ml CuSO 4.5H 2O solution and Calorimter (T initial): Commented [13]: Leave this in and fill in the blanks 2

3 Add the Zinc to the calorimeter and begin counting time. Continue to measure the temperature at least 2 minutes after the maximum temperature has been reached. Table 3: Heat of Reaction Commented [14]: delete Commented [15]: jazz up the title heat of reaction for what? [insert a suitable table here.] *If your reaction has not reached a maximum temperature after 8 minutes. Stop and repeat.* Plot Temp ( o C) y vs. Time (s) x. Extrapolate the graph back to the time at which mixing uccurred to calculate T final the final temperature of the calorimeter and the contents. Insert the (excel or equivalent) plot showing the extrapolation, trendline, units, titles and axes labels Commented [16]: Delete from your report Commented [17]: It is important that only use the immediate 6 or 7 points after Tmax to extrapolate back to the y axis Commented [CC18]: delete T = T final T initial T = _ = - Commented [19]: I m leaving these in, make sure you understand all the steps. Fill in the blanks if you like or do your own calculation if you prefer and delete all mine. Zn(s) + CuSO 4(aq) Cu(s) + ZnSO 4(aq) + heat Note the stoichiometry - qrxn = heat gained by ZnSO4(aq) (I) + heat gained by Cu(s) (II)+ heat gained by excess Zn(s) (III) + heat gained by calorimeter(iv) - qrxn = (I) + (II) + (III) + (IV) (IV) Heat gained by calorimeter = C cal x T = o C x _/ o C = (II) Heat gained by Cu(s) (IV) = mol of CuSO 4.5H 2O = mass used g x 1 mol = mol* g mol copper produced = mol The Heat Capacity of both Copper and Zinc is 25 /mol o C Heat gained by copper = mol Cu x 25 /mol o C x T = (III) Heat Gained by excess Zinc (II) = mol zinc added = mass used g x 1 mol = mol 3

4 65.37 g mol zinc reacted = mol copper produced = mol moles excess zinc = mol zinc added mol zinc reacted = mol Heat absorbed by excess zinc = mol x 25 /mol o C x T = (I) Heat gained by ZnSO 4 solution mol of ZnSO 4 produced = mol of CuSO 4.5H 2O used = mol (III) = see * above Mass of ZnSO 4 produced = mol x g = _ g ZnSO 4 1 mol mol water from hydrate (5H 2O) = mol ZnSO 4 x 5 = mol mass of water from hydrate = _mol x g 1 mol = _ g H 2O from hydrate Mass of water added to calorimeter = (95) g g Total mass of ZnSO 4 solution = g + _ g from hydrate + g ZnSO 4 Commented [20]: Check this and get it right and then put it in the first space in the equation below. Your total mass should be around 100 g = g Heat gained by ZnSO 4 solution = _ g x /g o C x T = (I) = Therefore q rxn = - [ (I) + (II) + (III) + (IV) ] = = k Commented [21]: Add up all the parts I thru IV the qrxn will be this value with a negative sign Hrxn = qrxn = mol CuSO 4.5H 2O k/mol Commented [22]: Don t forget this will be negative Make sure sign is correct for the exothermic reaction Discussion Calculate the actual H o for the copper/zinc reaction using the data below, compare it to your result. Calculate your percent error. Comment and include 3 sources of error. You have calculated 4

5 the heat absorbed by the copper and the zinc separately; can you think of a simpler way to get the same final value? The standard enthalpies of formation of Zn 2+ (aq) and Cu 2+ (aq) from zinc and copper metals are -152 k/mol and 64.4 k/mol respectively. Use this data to calculate the H o the standard enthalpy of the reaction: Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) Conclusion Summarize the results of all three parts. 5