Stability Essays

Stability Essays Stability Essay Stability Essay Stable. Strong oxidizer contact with combustible material may cause fire. Incompatible with combustible materials, and strong reducing agents.ToxicologyHarmful if swallowed. May cause reproductive disorders.AimThe aim of this experiment is to determine the crystallisation temperature of the solution potassium nitrate at different concentrations and use information to find out the standard enthalpy of potassium nitrate.Equipment and reagents* Boiling tubes* Dark card* Bunsen burner* Thermometer* Weighing scale* Burette (50cm3)* Clamp* Stand* Potassium nitrate* Deionised water/ distilled waterSafety* Wear goggles for eye protection at all times.* Laboratory coats must be worn at all times.* Wear gloves to avoid chemical contact to skin, potassium nitrate.* Long hair was tied back when Bunsen burners were used.Procedure1. 10g of potassium nitrate were weighed out and placed into a boiling tube, and then the exact mass was noted in a result table.2. Exactly 8.0cm3 of deionised water was added to the boiling tube containing the potassium nitrate. This was done by using a burette.3. The tube was then heated gently until the crystals were dissolved and then the heat source was removed.4. The tubes were allowed to cool for safety reasons. The temperature at which the crystals were first appeared was noted. A dark card was used for this purpose. The results were recorded.5. A further 25cm3 of distilled water was added and steps 3-4 were repeated.6. A further 25cm3 of distilled water was added and steps 3-4 were repeated again.7. A further 25cm3 of distilled water was added and steps 3-4 were repeated again.In my own opinion I think the above procedure will be a suitable method as it will not take a lot of time to perform and also it is quite simple to carry out. However, it could be a bit imprecise as we have to watch and determine when the crystals of potassium nitrate forms and reforms, then try to read the temperature of the thermometer. This difficulty could result in a slight inaccurate measurements as valuable seconds are wasted between seeing the crystals reforming and reading the measurements on the thermometer, which means that the temperature would have increased further from when crystals first formed as the Bunsen burner would still be heating the boiling tube.Alternative methods1. compare my results with other class students or with txt books results:AdvantagesdisadvantagesThis would be an appropriate way to check if my results are rational. As if I had extremely different results to numerous other students, then It would be unlikely that all of them would be wrong.It is impossible to be 100% accurate in measuring temperature and reagents, therefore the results will differ slightly between students.2. Perform the experiment several times for each volume of water and take an average.AdvantagesdisadvantagesThis method would be a lot more precise.It would take too long to carry out3. Guess the crystallisation temperature.Advantagesdisad vantagesFor this method there wouldnt be any calculations other than for the enthalpy, therefore this method would be straightforward.This would be very inaccurate.I think my method didnt need to be altered in any way to adapt to the experiment because the method I used went easily.ResultsVolume of watercnR.In cnCrystallisation temperatureReciprocal of absolute temperature 1/T(cm3)(mol dm3)(R = 8.31 Jmol-1 K-1)( 0C) T(K)(K-1)8.012.3620.9062 3352.98 x 10-310.89.8919.0456 3293.04 x 10-312.08.2417.5347 3203.13 x 10-314.07.0616.2442 3153.17 x 10-3Table of values for the line of best fit.R.ln SReciprocal of absolute temperature 1/T(R = 8.31 J mol-1 K-1)(K-1)Y 1 and X121.350.00290Y 2 and X215.000.003255The estimation of S (the intercept) was noted from these pairs of values.If Y 1 = mX 1 + c and Y 2 = mX2 + cThen (Y 1 c) / X1 =( Y 2 c ) / X 2X2Y1 X2c = X1y2 X1cc = (X2Y1 X1Y2) / (X2 X1)The gradient, m = (Y1 Y2) / (X1 X2)The expression gradient = H was used to find H, the enthalpy of solution of potassium nitrate.Intercept = S = 73.2 J mol-1 K-1Gradient = H = 17.9 K j mol-1By comparing the values I obtained from my results to those of the line of best fit I was able to estimate the SD of the values I had calculated for S and H the following table shows the results.R.in cn1/T values (a)1/T values calculated from best fit line (b)Difference for each 1/T value, as a fraction of the best fit value.(a-b)/bDifference2(R = 8.21 Jmol-1 K-1)K-1K-120.900.002980.0029250.01883.5344 x10-419.040.003040.003033.300 x10-31.089 x10-517.530.003130.0031135.461 x10-32.9822521 x10-516.240.003190.00319-6.2696 x10-33.93 x10-5CalculationsMolar mass (KNO) =K= 39N= 14O= 3 X16= 39 + 14 + 48= 101.11gNumber of moles :Amount = mass = 10.00 = 0.0989 mol dm-3Molar mass 101.11Concentration:Concentration = amountVolume(a) 8 cm3 : C = 0.0989 = 12.36 mol dm-38/1000(b) 10 cm3 : C = 0.0989 = 9.89 mol dm-310/1000(c) 12 cm3 : C = 0.0989 = 8.24 mol dm-312/1000(d) 14 cm3 : C = 0.0989 = 7.06 mol dm-314 /1000R.In cn:(a) 12.36 mol dm-3 : ln (12.36) = 2.514= R X ln (concentration)=8.31 x 2.51=20.90(b) 9.89 mol dm-3 : ln (9.89) = 2.2915= R X ln (concentration)=8.31 x 2.29=19.04(c) 8.24 mol dm-3 : ln (8.24) = 2.1090= R X ln (concentration)=8.31 x 2.11=17.53(d) 7.06 mol dm-3 : ln (7.06) = 1.954= R X ln (concentration)=8.31 x 1.95=16.24Crystallisation temperature in KelvinK = (0C ) + 273(a) 620C : 62+ 273 = 335K(b) 560C : 56 + 273 = 329K(c) 470C : 47+ 273 = 320K(d) 420C : 42+ 273 = 315KReciprocal 1/T(a) 20.90 : 1/335 = 2.98 X 10-3(b) 19.04 : 1/329 = 3.04 X 10-3(c) 17.53 : 1/320 = 313 X 10-3(d) 16.24 : 1/315 = 317 X 10-3GradientY = mx + cm = (y2 y1)(x2 x1)m = 15.00 21.350.003255 0.00290m = -6.35 .3.55 x 10-4m = -17,887.32394 Kj mol= -17.89 Kj molH = 17.89 Kj molS = (0.003255 x 21.35 0.00290 x 15.00)(0.003255 0.00290)S = 0.06949425 0.04353.55 x 10-4S = 73.223= 73.2Differences for each 1/T valueD = a-b / b(1) D = 0.00298 0.002925 = 0.01880.02925(2) D = 0.00304 0.00303 = 3.300 x 10- 30.00303(3) D = 0.00313 0.003113 = 5.461 x 10-30.003113(4) D = 0.00317 0.00319 = -6.2696 x 10-30.00319Differences2(1) (0.0188)2 = 3.5344 x 10-4(2) (3.300 x 10-3)2 = 1.089 x 10-5(3) (5.461 x 10-3)2 = 2.9822521 x 10-5(4) (-6.2696 x 10-3)2 = 3.93 x 10-5? (difference) 2 = 4.33 x 10-4Standard deviationSD = V? (- X X) 2 = V 4.33 X 10-4 = 0.01n 4S : % error = 73.2 + 0.01% error = 0.01 x 100 = 0.01%H : % error = 17.9 + 0.01% error = 0.01 x100 = 0.06%17.9DiscussionOn the whole I believe that my experiment went quite well because everything was conducted according to plan. I discussed my experiment and my results with my lecture and were said to be reasonably accurate. I also believe that any inaccuracy in my results was due to inexperience and human error, when measurements of weight, temperature and volume were made. My final results were obtained using results from a line of best fit placed on my graph this could have also caused a slight of inaccuracy in my results. These results could have been imprecise as extremely small numbers were used which were difficult to plot correctly on the graph and I had estimated where the points were to be placed. Also I had to estimate where to place my line of best fit. Additionally the inaccuracy could have been due to the thermometer I used, as there was a small gap between seeing the crystal and recording the temperature. For this reason I couldnt be as accurate as I would have been liked to be as because you had to be very quick in reading the temperature which was constantly changing. Furthermore, it was quite difficult to distinguish between small particles that had formed and air bubbles in the solution. Plus, it was hard to actually say what amount of particles formed, was the right amount to take the solution off the heat source and note the results.In my own opinion I think that my method was a suitable one because it gave me accurate results and didnt take up too much time to proceed.ImprovementsThe improvements that could have been made to make my experiment more accurate are to be more careful in measuring reagents and also to use a more accurate thermometer. In addition, a larger scaled graph could have been used as results would have been plotted more accurately.Confidently in future I would be more experienced in carrying out the experiment and would be more precise at spotting the crystals when they form.Conclusion and % errorThermometer 0.10620 : % error = 0.1 x 100 = 0.2%62560 : % error = 0.1 x 100 = 0.2%56470 : % error = 0.1 x 100 = 0.2%47420 : % error = 0.1 x 100 = 0.2%42Average % error = 0.2%Burette 0.1cm38cm3 : % error = 0.1 x 100 = 1.3%810cm3 : % error = 0.1 x 100 = 1%1012cm3 : % error = 0.1 x 100 = 0.8%1214cm3 : % error = 0.1 x 100 = 0.7%14Average % error = 1.3 + 1+ 0.8 +0.7 = 0.95%4Weighing scales + 0.01g% error = 0.01 x 100= 0.1% Show preview only