Lab 5 Report harith edit 8.32pm

THE UNIVERSITY OF BRITISH COLUMBIA

DEPARTMENT OF MATERIALS ENGINEERING

MTRL 359

LABORATORY 5: LEACHING

MUHAMMAD HARITH MOHD FAUZI

18204115

Introduction

Leaching is one of the core processes in hydrometallurgy. The purpose is to dissolve minerals of interest from an ore or concentrate in a suitable “lixiviant”. Once a solution of the metal(s) of interest is obtained it may be purified and treated to recover pure metal.

Results and Data

PbS screen size: 200 -230micron mesh size

Particle size: 63-75 micrometers

Experimental Conditions

ExperimentWeight of

FeCl3.6H2O (g)Leach solution

volume (ml)[Fe+3] (mg/L)

Temperature PbS (oC)

Weight PbS used (g)

1 21.9 1000 0.187 21 1.012 21.9 1000 0.187 30 0.983 21.9 1000 0.187 40 1.054 24.5 1000 0.130 21 1.02

Table 1

AA Calibration Data

[Pb] ppm Vol. flask mLVol. 100 ppm

Pb standard mLRequired HCl

g/LVol. 200 g/L

HCl mLBlank 100.0 0 10 5

5 100.0 5 10 510 100.0 10 10 520 100.0 20 10 525 100.0 25 10 5

Table 2

Standard (mg/L) Absorbance0 0.00045 0.084910 0.168720 0.320325 0.3941

Table 3

Graph 1

PbS Analysis Data

Weight PbS used 0.1003g

Digested PbS solution volume 100ml

Volumetric flask 250ml

Dilution factor 50

Pb AA result 17.266 mg/L

[Pb] 0.8633 mg/L

Weight% of Pb 86.1

Theoretical weight% Pb 86.6

Table 4

0 5 10 15 20 25 300

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45Absorbance vs [Pb]

[Pb]

Abs

orba

nce

Leaching Samples Analytical Data

Sample ID[exp. #-time min)]

Diluction factor

AA result [Pb] (mg/L)

Undiluted [Pb] (mg/L)

Corrected [Pb] in samples (g/L)

1-0

25

0.033 0.825 0.0001-5 3.637 90.925 0.0901-15 5.486 137.15 0.1361-25 8.399 209.975 0.2091-35

50

6.54 327 0.3261-45 8.477 423.85 0.4231-60 10.717 535.85 0.5351-80 13.278 663.9 0.663

Table 5


Diluction factor




2-025

0.399 9.975 0.0002-5 8.559 213.975 0.2042-11 13.257 331.425 0.3212-20

50

11.97 598.5 0.5892-30 13.836 691.8 0.6822-45 16.208 810.4 0.8002-60 17.351 867.55 0.8582-80 17.535 876.75 0.867

Table 6


Diluction factor




3-025

0.805 20.125 0.0003-3 24.269 606.725 0.587

3-6.5 19.84 496 0.4763-10

50

15.413 770.65 0.7513-15 17.266 863.3 0.8433-25 17.721 886.05 0.8663-35 17.817 890.85 0.8713-45 17.758 887.9 0.868

Table 7

Sample ID Diluction AA result [Pb] Undiluted [Pb] Corrected [Pb]

[exp. #-time min)] factor (mg/L) (mg/L) in samples (g/L)4-0

25

0 0 0.0004-5.5 4.535 113.375 0.1134-15 8.142 203.55 0.2044-25 11.596 289.9 0.2904-35

50

7.506 375.3 0.3754-45 9.2 460 0.4604-60 11.193 559.65 0.5604-80 13.063 653.15 0.653

Table 8

Mass of Pb Leached and α Values

Experiment 1

Time (min)

Sample vol. (mL)Undiluted [Pb]

(g/L)Corrected [Pb] (g/L)

Cumulative mass Pb in samples (g)

0

7

3.30E-05 0.000 0.0000005 3.64E-03 0.090 0.00063115 5.49E-03 0.136 0.00158525 8.40E-03 0.209 0.00241835 6.54E-03 0.326 0.00374745 8.48E-03 0.423 0.00524460 1.07E-02 0.535 0.00670680 1.33E-02 0.663 0.008387

Table 9

Experiment 2

Time (min)


(g/L)Corrected [Pb] (g/L)


0

7

3.99E-04 0.000 0.0000005 8.56E-03 0.204 0.00142811 1.33E-02 0.321 0.00367820 1.20E-02 0.589 0.00637030 1.38E-02 0.682 0.00889245 1.62E-02 0.800 0.01037660 1.74E-02 0.858 0.01160680 1.75E-02 0.867 0.012070

Table 10

Experiment 3

Time (min)


(g/L)Corrected [Pb]

(g/L)


0

7

8.05E-04 0.000 0.0000003 2.43E-02 0.587 0.004106

6.5 1.98E-02 0.476 0.00743710 1.54E-02 0.751 0.00858515 1.73E-02 0.843 0.01115625 1.77E-02 0.866 0.01196435 1.78E-02 0.871 0.01215745 1.78E-02 0.868 0.012170

Table 11

Experiment 4

Time (min)


(g/L)Corrected [Pb]

(g/L)


0

7

0.00E+00 0.000 0.0000005.5 4.54E-03 0.113 0.00079415 8.14E-03 0.204 0.00221825 1.16E-02 0.290 0.00345435 7.51E-03 0.375 0.00465645 9.20E-03 0.460 0.00584760 1.12E-02 0.560 0.00713880 1.31E-02 0.653 0.008490

Table 12

Experiment 1

Time(min)

Sample vol. (ml)Cumulative mass Pb in samples (g)

Leach vol. remaining (ml)

Mass Pb in the leach solution (g)

0

7

0.000000 993 0.00005 0.000631 986 0.089515 0.001585 979 0.135025 0.002418 972 0.205735 0.003747 965 0.318545 0.005244 958 0.410560 0.006706 951 0.515580 0.008387 944 0.6343

Table 13

Experiment 2

Time (min)

Sample vol. (mL)Cumulative mass Pb in samples (g)



0

7

0.000000 993 0.00005 0.001428 986 0.202611 0.003678 979 0.318420 0.006370 972 0.578430 0.008892 965 0.666945 0.010376 958 0.777260 0.011606 951 0.827280 0.012070 944 0.8303

Table 14

Experiment 3

Time (min)




0

7

0.000000 993 0.00003 0.004106 986 0.5825

6.5 0.007437 979 0.473310 0.008585 972 0.738115 0.011156 965 0.824825 0.011964 958 0.841535 0.012157 951 0.840245 0.012170 944 0.8313

Table 15

Experiment 4

Time (min)




0

7

0.000000 993 0.00005.5 0.000794 986 0.112615 0.002218 979 0.201525 0.003454 972 0.285235 0.004656 965 0.366845 0.005847 958 0.446560 0.007138 951 0.539480 0.008490 944 0.6251

Table16

Experiment 1

Time(min)

Total mass Pb leached (g)

α 1-(1-α)1/3 1-2/3 α – (1-α)2/3

0 0.0000 0.0000 0 0.00005 0.0895 0.1092 0.037826113 0.005415 0.2245 0.1649 0.058295516 0.014425 0.4302 0.2512 0.091912365 0.057335 0.7487 0.3889 0.151392776 0.082845 1.1592 0.5012 0.206942471 0.128160 1.6748 0.6294 0.281727687 0.156580 2.3091 0.7745 0.39133154 0.1585

Mass PbS used (g) 1.01Mass Pb added (g) 0.81902

Table 17

Experiment 2

Time(min)


α 1-(1-α)1/3 1-2/3 α – (1-α)2/3

0 0.0000 0.0000 0 0.00005 0.2026 0.2104 0.07571907 0.072611 0.5209 0.3307 0.125256438 0.043320 1.0994 0.6007 0.263650047 0.142730 1.7662 0.6926 0.325102985 0.213545 2.5434 0.8072 0.422281921 0.233760 3.3706 0.8591 0.479625371 0.232080 4.2009 0.8624 0.483679262 0.2210


Table 18

Experiment 3

Time(min)


α 1-(1-α)1/3 1-2/3 α – (1-α)2/3

0 0.0000 0.0000 0 0.00003 0.5825 0.6592 0.301514377 0.0021

6.5 1.0558 0.5357 0.225638787 0.007010 1.7939 0.8353 0.451872826 0.014815 2.6187 0.9335 0.594791191 0.025925 3.4602 0.9524 0.637504283 0.041035 4.3005 0.9509 0.633799175 0.065345 5.1318 0.9409 0.610384899 0.0968


Table 19

Experiment 4

Time(min)


α 1-(1-α)1/3 1-2/3 α – (1-α)2/3

0 0.0000 0.0000 0 05.5 0.1126 0.1321 0.046117414 0.00174187315 0.3141 0.2364 0.085970339 0.00627836825 0.5993 0.3346 0.126978907 0.01648471135 0.9661 0.4303 0.171020643 0.03484493245 1.4127 0.5238 0.219111021 0.05349399160 1.9520 0.6327 0.283865835 0.08668891380 2.5771 0.7333 0.35628571 0.130044604


Table 20

Discussion

1. (i) Plot the following

Leach solution [Pb] vs Time

0 10 20 30 40 50 60 70 80 900

100

200

300

400

500

600

700

800

900

1000Leach solution [Pb] versus Time

Test 1Test 2Test 3

Time (min)

[Pb]

(mg/

L)

Graph 2 Test 1, 2, and 3

0 10 20 30 40 50 60 70 80 900

100

200

300

400

500

600

700

800Leach solution [Pb] versus Time

Test 1Test 4

Time (min)

[Pb]

(mg/

L)

Graph 3 Test 1 and 4

o 1-(1-α)1/3 versus Time

0 10 20 30 40 50 60 70 80 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

f(x) = 0.00438203366430559 x + 0.0157399913326297

f(x) = 0.019349855053937 x

f(x) = 0.00766513669047629 x

f(x) = 0.00482321865040963 x − 0.00734055927634691

1-(1-α)1/3 versus time

Test 1Logarithmic (Test 1)Linear (Test 1)Test 2Linear (Test 2)Test 3Linear (Test 3)Test 4Linear (Test 4)

Time (min)

1-(1

-α)1

/3

Graph 4 1-(2/3)α – (1-α)2/3 versus Time

0 10 20 30 40 50 60 70 80 900

0.020.040.060.08

0.10.120.140.160.18

f(x) = 0.00000000913934 x⁴ − 0.00000240729 x³ + 0.000169587 x² − 0.000883419 x + 0.00185122

1-(2/3)α - (1-α)2/3 versus Time

Series2Polynomial (Series2)

Time (min)

1-(2

/3)α

- (1

-α)2

/3

Graph 5 Test

0 10 20 30 40 50 60 70 80 900

0.05

0.1

0.15

0.2

0.25f(x) = 0.0000000455918 x⁴ − 0.00000696003 x³ + 0.000249888 x² + 0.00382476 x + 0.0121388

1-(2/3)α - (1-α)2/3 versus Time


Time (min)

1-(2

/3)α

- (1

-α)2

/3

Graph 6 Test 2

0 5 10 15 20 25 30 35 40 45 500

0.020.040.060.08

0.10.12

f(x) = 0.0000000336527 x⁴ − 0.00000253216 x³ + 0.0000712615 x² + 0.00102718 x − 0.000806893

1-(2/3)α - (1-α)2/3 versus Time


Time (min)1-(2

/3)α

- (1

-α)2

/3

Graph 7 Test 3

0 10 20 30 40 50 60 70 80 900

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.000009562062443 x² + 0.000482712067 x − 0.001307242458

1-(2/3)α - (1-α)2/3 versus Time


Time (min)

1-(2

/3)α

- (1

-α)2

/3

Graph 8 Test 4

The alpha function should not be constrained to the origin.

(ii) Leaching completion

From the Pb analysis data, we can predict that the leaching of PbS did go to completion in experiment 2 and 3. This is because the last 2-3 samples have the roughly the same [Pb] which indicates that all the PbS has reacted before the end of the test.

However, the condition of monosized particles is quite false and the simple leaching model thus fails. This affects the curve of the concentration versus time plot. This can lead to the wrong extraction of information from the graph. A good model should produce a smooth curve that close to the line of best fit for any types of equation of line.

Samples that should be omitted from the alpha function plots.

Experiment 2

Time (min) Undiluted [Pb] (mg/L)

45 810.4

60 867.55

80 876.75Table 21

Experiment 3

Time (min) Undiluted [Pb] (mg/L)

25 886.05

35 890.85

45 887.9Table 22

These are some of the points that should be omitted since the undiluted concentration are roughly the same the same towards the end reaction. This indicates that all the PbS has fully reacted at the end of the experiment.

Based on the leaching results, the function 1-(1-α)1/3 best fit the leaching results. This is because the values from the function produce curves that are closer to linear line of best fit compares to another function. Thus, the function 1-(1-α)1/3 is more reliable in showing the experimental data obtained from the experiments.

2. (i) Calculate rate constants.

The rate constant is equal to the slope of the linear best fit of the chosen function.

0 10 20 30 40 50 60 70 80 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

f(x) = 0.00438203366430559 x + 0.0157399913326297

f(x) = 0.019349855053937 x

f(x) = 0.00766513669047629 x

f(x) = 0.00482321865040963 x − 0.00734055927634691

1-(1-α)1/3 versus time

Test 1Logarithmic (Test 1)Linear (Test 1)Test 2Linear (Test 2)Test 3Linear (Test 3)Test 4Linear (Test 4)

Time (min)

1-(1

-α)1

/3

Graph 9

Summary of rate constant for each experiment.

Experiment Rate constant (slope), k

1 0.0044

2 0.0077

3 0.0193

4 0.0048

Table 23

(ii) Plot the natural logarithm of the rate constants versus 1/T.

0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345

-6

-5

-4

-3

-2

-1

0

f(x) = − 7195.76663182717 x + 18.9519302640634

Plot of natural logarithm of K vs (1/T)

ln (k)Linear (ln (k))

1/T

ln k

Graph 10

Calculation of pre-exponential factor A

ln k=ln A−EaR

( 1T

)

From this formula, ln A is the y-intercept. Thus to find the value for A, equate it with the y- intercept from the line equation.

ln A=18.952

A=e18.952

A=1.701∗108min (−1)

Calculation of activation energy in kJ/mol.

In this case, the activation energy is equal to the slope of the line equation. EaR

=7195.8

Ea=7195.8∗8.314

¿59825.88Jmol

¿59.826kJmol

(iii) The activation energy is consistent with the chosen leaching model. The activation energy of chemical reaction is between the range of 42 -105 kJ/mole. Our activation energy falls within this range.

3. (i) Proving.

k s=Vmk ' f (CB)

ro

Since we are assuming that Vm and CB are constants, we eliminate these from the equation and get:

k s=k '

ro

Rate constant for a PbS particle of initial size ro1:

k s1=k '

ro1

k '=ks1∗ro1

Rate constant for a PbS particle of initial size ro2:

k s2=k '

ro2

Therefore, substitute for k’ and get

k s2=ks1∗ro1

ro2

(shown )

(ii) Value of ks at 32 degree Celcius.

(ii) Value of ks at 32 degree Celsius.

Using the equation [27] and our own data values,

k s=Ae−Ea/RT

k s=(1.701∗108)e( −59825.884

8.314∗(32+273))

k s=0.009649

(iii) Time taken

From equation 21:

t=[1−(1−α )1 /3 ]∗roV mk

' f (CBo )

=[1−(1−α )1 /3]

k s2

Substituting equation from part i),

k s2=ks1∗ro1

ro2

k s1=Ae−Ea /RT

k s1=(1.701∗108 )e−(59825.884)/(8.314∗(28+273))

k s1=0.0070522

To find ro1

ro1=0.00004

2=0.00002m

And ro2,

ro2=0.000063+0.000075

4=0.0000345m

Coming back to the other equation,

k s2=ks1∗ro1

ro2

k s2=0.0070522∗0.00002

0.0000345=0.00408

Based on equation:

t=[1−(1−α )1 /3]

k s2

For α = 0.95,

t=[1−(1−0.95 )1/3]

0.00408=154.8min

For α = 0.99,

t=[1−(1−0.99 )1/3]

0.00408=192.29min

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Lab 5 Report harith edit 8.32pm