INTERNAL ASSESMENT

INTERNAL ASSESMENT ENTHALPY CHANGEBy Aldo M Hamka Class 11.ASubject ChemistryTeacher : Ms. Nandini Priyanti

13/14/2011

Internal Assesment (IA) Topic: Enthalpy change Research Questions: what is the enthalpy change of a dissolution Ammonium Nitrate (NH4NO3) Aims : to determine the enthalphy change of dissolution of Ammonium Nitrate Equipment :1. 2 Polystyrene cup (stacked)2. 5 g of Ammonium nitrate (NH4NO3)3. 125 ml of water 4. 1 thermometer5. 1 Stiring rod 6. 1 25 ml graduated cylinder Method :1. Pour the water slowly into the 25 ml graduated cylinder until it hit 25 ml and after that pour it on the stacked polystyrene cup 2. Measure the temperature of the water 3. Put in the 5 g of ammonium nitrate ( can be more or less) on the stacked polystyrene cup and stir it with the stiring rod 4. Observe and see what happen with the temperature that showed in the thermometer.5. Do this experiment 5 times to collect the necessary data

Data collection and processing1. Data collection

2. ProcessingA.) The enthalphy changes in kj/molFormula : Where : M = water solution + NH4NO3 : C = 4.18 kj : T= water temperature temperature changeschanges in kj/mol : Q/1000 moles of reactant

B.) UncertaintyNo.1 (0.025/25)*100% = 0.1%(M) (0.05/10)*100%= 0.5% (T)Q: 0.1%+0.5% = 0.6%

No .2(0.025/25)*100% = 0.1%(M) (0.05/10.9)*100%= 0.45% (T)Q: 0.1%+0.45% = 0.55%

No .3(0.025/25)*100% = 0.1%(M) (0.05/9.8)*100%= 0.51% (T) Q: 0.1%+0.51% = 0.61%

No .4 (0.025/25)*100% = 0.1%(M) (0.05/10.9)*100%= 0.45% (T)Q: 0.1%+0.45% = 0.55%

No .5(0.025/25)*100% = 0.1%(M) (0.05/10.5)*100%= 0.47% (T)Q: 0.1%+0.47% = 0.57%

C.) Presenting data

i.) Research question answer:1.) Enthalphy changes in mole = 19.589 kj/mol-1 0.6%2.) Enthalphy changes in mole = 21.775 kj/mol-1 0.55%3.) Enthalphy changes in mole = 19.610 kj/mol-1 0.61%4.) Enthalphy changes in mole = 21.456 kj/mol-1 0.55%5.) Enthalphy changes in mole = 20.838 kj/mol-1 0.57%ii.) Pecentage error ( accepted value = +25.69 kj/mol-1)1.) (I(25.69 kj/mol-1 -19.589 kj/mol-1)I/25.69 kj/mol-1)*100%= 23.74%2.) (I(25.69 kj/mol-1 -21.775 kj/mol-1)I/25.69 kj/mol-1)*100%= 15.23%3.) (I(25.69 kj/mol-1 -19.610 kj/mol-1)I/25.69 kj/mol-1)*100%= 23.67%4.) (I(25.69 kj/mol-1 -19.456 kj/mol-1)I/25.69 kj/mol-1)*100%= 24.27%5.) (I(25.69 kj/mol-1 -20.838 kj/mol-1)I/25.69 kj/mol-1)*100%= 18.88%iii.) Average data 1. Enthalpy changes in mole :(19.589 kj/mol-1 +21.775 kj/mol-1 +19.610 kj/mol-1 +21.456 kj/mol-1 +20.838 kj/mol-1 )/5= 20 .6536 kj/mol-12. Uncertainty : (make percentage uncertainty into absolute uncertainty first )(0.117+0.119+0.119+0.118+0.118)/5= 0.118

Conclusion and Evaluation1. Conclusion :There is a average precision within the data this was caused by the realisation that there are many random error within my experiment not only that the room temperature are also taken effect to the ammonium nitrate so it did not get the maximum effect on the dissolution and as the result the decreasing of the temperature when i calculated it are below the acepted value. 2. Evaluations:There is a average precision within the data this was caused by the realisation that there are many random error within my experiment not only that the room temperature are also taken effect to the ammonium nitrate so it did not get the maximum effect on the dissolution and as the result the decreasing of the temperature when i calculated it are below the acepted value. Although the enthalphy change are close with each other ,in this experiment i found many mistakes that may cause the percentage error become around 20 % . First i didnt measures the ammnonium nitrate in the exact 5 grams , second i didnt close the polystyrene cup when i stir it and may cause the solution that having a decrease in temperature being heated up by the room temperature ,but why i didnt close the cup this is based on the reasons that if i close the cup it will be hard to see what happening with the ammonium nitrate am i already stir it or is it already melted and i did not notice it , the third is the calculations and how i read the uncertainty of the measurement .3. Improving investigation As i learn from this experiment there is many error in how i do the experiment and maybe i can be more focus on do the experiment . i have to learn more about the enthalpy change again so i did not get lost when i calculating the solution and i have to be more precise of determine the ammonium nitrate or any other solid based stuff .

Sheet1NO.water solution ( ml 0.025)NH4NO3 ( g 0.01)water temperature(c 0.05)water +NH4NO3( temperature drop)temperature changes (c)enthalpy changes enthalpy changes in KJ /MOL1255.1526.816.8101260.2719.5897347652255.0326.91610.91368.2268621.77520807093255.0226.8179.81229.7392819.61017706034255.122716.110.91372.3274421.45655395065255.0727.51710.51319.772320.8383455936

Sheet1NO.water solution ( ml 0.025)NH4NO3 ( g 0.01)water temperature(c 0.05)1255.1526.82255.0326.93255.0226.84255.12275255.0727.5