Determine the crystallisation temperature of the solution potassium nitrate at different concentrations and use information to find out the standard enthalpy of potassium nitrate.

Alternative methods

1. compare my results with other class students or with txt books results:

2. Perform the experiment several times for each volume of water and take an average.

3. Guess the crystallisation temperature.

I think my method didn’t need to be altered in any way to adapt to the experiment because the method I used went easily.

Results

Table of values for the line of best fit.

The estimation of    S (the intercept) was noted from these pairs of values.

If Y 1 = mX 1 + c and Y 2 = mX2 + c

Then (Y 1 – c) / X1 =( Y 2 – c ) / X 2

X2Y1 – X2c = X1y2 – X1c

c = (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-1

Gradient =     H = 17.9 K j mol-1

By 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.

                 

Calculations

Molar mass (KNO) =

                K=        39

                N=        14

                O=        3 X16

                   =        39 + 14 + 48

                   =        101.11g

Number of moles :

Amount =    mass           =        10.00   = 0.0989 mol dm-3

             Molar mass               101.11

Concentration:

Concentration = amount

                        Volume

(a) 8 cm3 :                          C =   0.0989  _        = 12.36 mol dm-3

                                         8/1000

(b) 10 cm3 :                         C =   0.0989  _        = 9.89 mol dm-3

                                        10/1000

(c) 12 cm3 :                         C =   0.0989  _        = 8.24 mol dm-3  

                                        12/1000

(d) 14 cm3 :                         C =   0.0989  _        = 7.06 mol dm-3

                                        14/1000

R.In cn:

1. 12.36 mol dm-3     : ln (12.36) = 2.514

= R X ln (concentration)

=8.31 x 2.51

=20.90

2. 9.89 mol dm-3     : ln (9.89) = 2.2915

= R X ln (concentration)

=8.31 x 2.29

=19.04

3. 8.24 mol dm-3     : ln (8.24) = 2.1090

= R X ln (concentration)

=8.31 x 2.11

=17.53

4. 7.06 mol dm-3     : ln (7.06) = 1.954

= R X ln (concentration)

=8.31 x 1.95

=16.24

Crystallisation temperature in Kelvin

K = (0C ) + 273

1. 620C : 62+ 273 = 335K

2. 560C : 56 + 273 = 329K

3. 470C : 47+ 273 = 320K

4. 420C : 42+ 273 = 315K

Reciprocal 1/T

1. 20.90 : 1/335 = 2.98 X 10-3

2. 19.04 : 1/329 = 3.04 X 10-3

3. 17.53 : 1/320 = 313 X 10-3

4. 16.24 : 1/315 = 317 X 10-3

Gradient 

Y = mx + c

m =   (y2 – y1)

         (x2 - x1)

m =      15.00     –     21.35  

        0.003255 – 0.00290

m =         -6.35      .

        3.55 x 10-4

m = -17,887.32394 Kj mol

    = -17.89 Kj mol

 H =  17.89 Kj mol

 S =   (0.003255 x 21.35 – 0.00290 x 15.00)

                (0.003255 - 0.00290)

S  =   0.06949425 – 0.0435

                3.55 x 10-4

S  =  73.223

    =  73.2

Differences for each 1/T value

D = a-b / b

(1)  D =   0.00298 – 0.002925  = 0.0188

                0.02925

        (2)  D =   0.00304 – 0.00303    = 3.300 x 10-3

                                 0.00303

(3)  D =   0.00313 – 0.003113  = 5.461 x 10-3

                0.003113

(4)  D =   0.00317 – 0.00319    = -6.2696 x 10-3

                0.00319

Differences2

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 10-4

Standard deviation

SD =   √∑ (- X – X) 2           =    √ 4.33 X 10-4    =   0.01

                      n                                   4

   S : % error = 73.2 + 0.01

         % error = 0.01  x 100 = 0.01%

   H :  % error = 17.9 + 0.01

         

          % error =  0.01  x100 = 0.06%

                           17.9

Discussion

On 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 couldn’t 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 didn’t take up too much time to proceed.

Improvements

The 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 % error

Thermometer 0.10

620 : % error =  0.1  x 100 = 0.2%

                          62

560 : % error =   0.1  x 100 = 0.2%

                           56

470 : % error =   0.1  x 100 = 0.2%

                          47

420 : % error =   0.1  x 100 = 0.2%

                           42

Average % error = 0.2%

Burette 0.1cm3 

8cm3 : % error =  0.1  x 100 = 1.3%

                            8

10cm3 : % error =   0.1  x 100 = 1%

                               10

12cm3 : % error =   0.1  x 100 = 0.8%

                               12

14cm3 : % error =   0.1  x 100 = 0.7%

                               14

Average % error =   1.3 + 1+ 0.8 +0.7  = 0.95%

                                            4

Weighing scales + 0.01g

% error = 0.01 x 100= 0.1%