lecture 7 application dta & dsc01

Lecture 7 Application DTA & DSC 01 1

APPLICATIONS OF DTA & DSC

(b) Measurement of chemical reaction:

dehydration decomposition polymer curing glass formation oxidation reaction

Application of the DTA/DSC methods are very wide and can be divided into two categories:

(a) Measurement of physical properties changes:

melting crystal phase changes changes in the conditions of

liquid and liquid crystal changes in the physical

properties of polymer phase diagram heat capacity glass transition thermal conductivity diffusivity emissivity


a) Determination of Changes in Physical Properties

Melting and Hfusion of Materials

Determination of melting point may be easier by using a simple apparatus, but determination of H of a fusion process is more complicated.

Although DTA may give the value of melting point (Tm) accurately, the H need to be determined by using a DSC method.

Figure 7.1 DSC peak for the melting of pure indium(Source: Haines: F_3.11)


H = K ∫T dt = K . (peak area)


a) Polymorphic : materials that have more than two shapes of crystal.

b) Enantiotropic : shapes system of a material that are stable in certain range of temperature and have specific transition temperature.

c) Monotropic : only one shape of the crystal is stable throughout the temperature range, while the other shapes are meta-stable. The shapes that are less stable tend to

transform into the more stable shapes.

CRYSTAL PHASE TRANSITION

The crystal shapes of a material may influence some properties of the material, such as solubility, density and the electrical properties.



Potassium Nitrate, KNO3 (Haines: F 3.12)

Heating - phase II to phase I at 128 oC and H = 5.0 kJ/mole

- melting of phase I at 334 oC with H = 10 kJ/mole

Cooling - the exothermic peak at 120 oC with H = -2.5 kJ/mole

- formation of another phase (phase III)

- at lower temperature, it slowly returns to phase II


Figure 7.2 DSC curve for potassium nitrate (4 mg, 5 K/min, nitrogen) (Source: Haines: F

3.12)


It is widely used as components of fertilizers and explosives

Ammonium Nitrate, NH4NO3 (Haines: Figure 3.13)

Transition from phase IV (rhombic II) to phase III (rhombic I) produces an endothermic peak at 40 oC

Transition from phase III to phase II (tetragonal) with a small endothermic peak at 80 oC

Transition from phase II to phase I (cubic) at 120 oC

Melting at 170 oC

Explosion of NH4NO3 at a temperature around 200 oC as shown by a large exothermic reaction

APPLICATION OF DTA & DSC

Figure 7.3 DTA curve for ammonium nitrate(Source: Haines: F 3.13)


POLYMORPHISM IN FOODS

Polymorphism is important in food because it influences the properties of the food materials in terms of:

Solubility Physiological activities Stability Bioavailability

The DSC is widely used in the study of polymorphism changes in the mixtures of fats in confectionery and the characterization of fats and oils.

Example:

Glycerides (glycerol esters) : CH2OH.CHOH.CH2OH

which show three phases:

shape – the most stable with the highest melting point

’ shape – a meta-stable shape shape – the most unstable shape with

the lowest melting point

Heating of ’ shape will produce a partial melting of ’ shape, followed by conversion to shape and melting of shape as shown by the dotted line (Figure 7.4)

If the liquid is cooled to a temperature several degrees above the melting point of shape and then allowed to remain at the same temperature for sometime until crystallization takes place, the ’ shape will be formed.


Conversion and crystalization of polymorph that is stable at 60 oC

Melting of shape at 54 oC

Solid line: first heating

Dotted line: second heating from 57 oC

Melting of shape at 73 oC

Figure 7.4 DSC curve of trystearin (gliceryl trystearate) : 4.3 mg, 5 K/min, nitrogen (Source: Haines: F 3.14)

Heating of ’ shape will produce a partial melting of ’ shape, followed by conversion to shape and melting of shape as shown by the dotted line

If the liquid is cooled to a temperature several degrees above the melting point of shape and then allowed to remain at the same temperature for sometime until crystallization takes place, the ’ shape will be formed.


Two forms of norfloxacin: form A and form B

They show different anti-microbe activities

Obtained from different sources

Transformation of Form B (to form A) is not seen when heating is repeated showing that the material is monotropic.

Polymorphisme in Pharmaceutical Materials

Norfloxacin : the derivative of4-quinolone carboxylic acid

Full characterization would require the use of DSC and other analytical instrumentation such as TG, XRD and spectroscopy

The form B experienced

transition (to Form A) at 195.6 0.2 oC

with H = 20 J/g (varies with the

heating rate)The Form A melts at

219.5 0.2 oC with H = 115 J/g

Figure 7.5 DSC curve for the two forms of polymorphic norfloxasin (10 K/min, nitrogen (Haines: F 3.15)


Transition of Liquid Crystals:

Liquid crystals:

Digital Display...Liquid Crystal Display

The molecules have:rod-like structurepolar end groupshighest order in the solid formvarious degrees of order in the liquid form:

various smectic phase, having plane like order nematic phase, having rod-like order cholesteric phase, these are the twisted nematic Isotropic phase, these are liquid of complete disorder

For the order changes in the liquid crystal transition, a greater entropy change is observed if disorder increased

Since S = H / T the enthalpy change, H is greater if a large change of order takes place


Figure 7.6 DSC curve for p-azoxianisol (15 mg, 16 K/min, static atmosphere) (Source: Haines: F 3.16)

Nematic crystal Nematic liquid

Nematic liquid True liquid

p-azoxianisol: CH3.C6H4.N=NO.C6H4.OCH3

The first transition at 119 oC has a large H because the material is transformed from a high order crystal phase into a disorder nematic liquid phase

The second (smaller) transition takes place at 135 oC when the material changes into a true liquid.


Figure 7.7 The DSC curve for N,N’-bis(4-octyloxibenzilydene)-p-phenylene diamine (OOBPD) (9 mg, 10 K/min, nitrogen) (Source: Haines: F 3.17)

A

B C

A. High order solid crystal phase at below 100 oCB. Peaks indicate the smectic phase changes

between 120 - 205 oC C. Stable nematic liquid crystal above 205 oC

When all crystal phases have been formed and the material is cooled down to 80 oC, two exothermic peaks are observed between 110 - 80 oC.

These are changes from the nematic liquid crystal phase to the smectic and from the smectic to the high order solid phase

Solid crystals


PHASE DIAGRAM

When more than one chemical components can co-exist

the thermodynamic will become complicated, and the melting behaviour will become more complex.

The cooling curve

Plot of T versus time for a mixture while it is being cooled from its melting temperature

It is the traditional method in the investigation of changes in a mixture of phases.

DTA/DSC It gives a lot of information about the sample The process is much faster than the traditional method The sample size is smaller than the traditional method provided that

heating is carried out at a very slow rate (less than 10 K/min)


EUTHECTIC POINT

When a pure solid A, completely insoluble in the pure solid B, the presence of B reduces the freezing point of A, producing a diagram as shown in Figure 7.8.

The lowest temperature where the liquid phase can exist is known as euthectic point, TE.

Below the euthectic point various mixture of crystals A and B can be found

Figure 7.8Euthectic phase diagram (Source: Haines: F 3.18)

When the temperature gets higher than PQR (Figure 7.8), the mixture begins to melt and an endothermic DSC peak is obtained, the size of which depends on the amount of the crystal that is melting at that temperature

If the euthectic composition is closer to the composition of Q, the TE peak will become larger

The mixture will slowly and continuously melt until it reaches the liquid line of UQW (Figure 7.8) where the last fraction will melt. This is shown as the last peak in the thermogram.


If A and B have a very similar structure, for example, cube shape crystal, they will form solid solution, where the crystal structures mix and there is an apparent intermediate crystal lattice between the structure A and the structure B.

This will produce a phase diagram as shown in Figure 7.9 or Figure 7.10.

Figure 7.9 Phase diagram of a series of solid solution (Source: Haines F 3.19)


Figure 7.10 Phase diagram for a partially miscible system of solid α and solid β (Source: Haines F 3.20)


Figure 7.11An overlap of the euthectic phase diagrams and the DTA curves (Source: Haines F 3.21)

QP R

When the temperature gets higher than PQR (Figure 7.8), the mixture begins to melt and an endothermic DSC peak is obtained, the size of which depends on the amount of the crystal that is melting at that temperature

If the euthectic composition is closer to the composition of Q, the TE peak will become larger

The mixture will slowly and continuously melt until it reaches the liquid line of UQW (Figure 7.8) where the last fraction will melt. This is shown as the last peak in the thermogram.

UW

Euthectic system produces two DTA peaks


Figure 7.12An overlap of solid solution phase diagram and the DTA curves (Source: Haines F 3.22)

Solid solution system produces single broad DTA peaks

A partially miscible system (Figure 7.10) produces both DTA thermograms (two peaks and single broad peak)

The phase diagram becomes more complicated when components are interacting to each other.

The DSC/DTA systems have been widely used for the study of phase diagrams for metals, liquid crystal and pharmaceutical materials.


COMPATIBILITY OF PHARMACEUTICAL MIXTURES

Compatibility refers to situation where two components of a pharmaceutical dose can be mixed without any interaction to each other as long as the shelf life duration of the dose has not expired.

Each pharmaceutical dose normally contains a drug active component together with other inactive materials as the diluent, binder or dispersant. Any chemical or physical interaction between these components will affect the pharmaceutical properties of the mixture.

If the drug reacts with the diluent component, the shape of the mixture thermogram would be different from both the original respective thermograms.

This means that any changes or appearance of new characteristics of the thermogram curve would indicate that there is some interaction between the mixtures (Figure 7.13).

E.g.: mixture of stearic acid with pheniciline shows similar changes obtained by DSC method and stability test for 8 weeks at 50 oC.


Figure 7.13DTA thermograms of materials compatibility (Source: Haines F 3.23)

Mixture of glycerides show euthectic behaviour, but this behaviour changes in the presence of ketoprophene.


Sometimes alteration in the treatment of the materials may change the mixture stability as shown by the mixture of the drug trimethoprim (TMP) with the liquid diluent consisting of the cross-linked polyvinylpirolidone (PVP-XL):

shows thermogram characteristics similar to those of the original

shows different thermogram characteristics when the mixture is ground.

Figure 7.14The DSC curve for (a) PVP-XL, (b) TMP, (c) physical mixture of 75 % PVP-XL / 25 %

TMP, (d) ground mixture 75 % PVP-XL / 25 %

TMP (Source: Haines F 3.24)


PHASE CHANGES IN THE THERMOPLASTIC POLYMERS

Polymers that melt before they decompose might be: transformed into crystals

completely or retained as amorphous

inhomogeneous solid or a glass

Finally, the polymer melts within a wide range of temperature to become liquid (C).

When heated, the changes of glass like polymer into a plastic like polymer can be seen through an increase in the heat capacity (A). This process is called Glass Transition.

The polymer molecules can now move more freely and rearranged again to become a tidier crystal structure. This process is exothermic (B).

(A)

(B)

Figure 7.15The DSC curve to show changes in poly(ethylene terefthalate), PET, (10 mg, 10 K/min, nitrogen) (Source: Haines F 3.25)

(C)


Some polymers show multiple endothermic melting processes, meaning that the relevant portion of the sample had crystallized at different temperatures during the previous heat treatment applied to the sample

Polymer samples normally do not perfectly crystallize simultaneously. The percentage of crystallization can be determined by comparison of the thermogram peak area of the sample with that of a polymer standard of known crystality.

Area of the sample melting peak x (% standard)% crystallization =

Area of the standard melting peak

The components of a polymer blend might be melting at different temperatures and these could be use to determine each of the components, qualitatively as well as quantitatively.


DETERMINATION OF HEAT CAPACITY

Heat Capacity (Cp,s J/K) is defined as the amount of heat required to raise the temperature of a body by 1 K.

Cp,s = (∂qs / ∂T)p

Heat Capacity is an important parameter in the industrial polymers, insulators and construction materials

In a DSC system, the y-axis is calibrated as the integration of the supplied heat, or the Δenergy or (dΔq/dt). Since heating (dT/dt) is at a constant rate, deviation from the y-axix baseline without the sample divided by the rate of heating will give the difference in the heat capacity of the sample.

Cp,s = K . Δy / (dT/dt)

Where K = the calorimetric sensitivity, determined by calibration

Calibration is normally carried out using sapphire, i.e. pure crystal of Al2O3 of known Heat Capacity.

Larger deviation from the baseline is observed when heating rate (dT/dt) is increased.


Heat capacity of sapphire (Al2O3) at 445 K is 0.997 J/(K g). Using data from Figure 7.16, determine the heat capacity (Cp) of polyethylene (PE). Heating rate is 20 oC/min.

Figure 7.16 DSC curve for the measurement of heat capacity of a polyethylene (PE) melt (Source:

Haines: F 3.26)

Cp (sapphire) = 0.997 x 133.6 x 10-3

= 0.1332 J/KΔy (sapphire) = 75.0 mmCp (sapphire) = (K x 75)/20

= 0.1332Hence, K = 3.552 x 10-2 J/(min mm)For PE: Δy = 34 mmTherefore, Cp(PE) = (3.552 x 10-2 x 34) / 20

= 0.0604 J/KPE mass = 23.14 mg Cp(PE) = 23.14 x 0.0604

= 2.609 J/(K g)


GLASS TRANSITION TEMPERATUREWhen a materiaql is heated, a small increase in volume and energy until Tg is reached where the polymer chain becomes more mobile and the polymer behaves like plastics or rubber.

Figure 7.17 The energy transition scheme for materials undergoing glass formation (Source: Haines: F 3.27)

Direct cooling from the melting will cause some portion of the polymer to crystallize at temperature Tm.

At Tg, the heat capacity increased due to increase mobility of the polymer chain. Hence, ΔCp and the expansion of the sample appears to increase.

At temperatures below the glass transition, Tg, the polymer molecules do not have enough energy to rotate or rearrange. Materials under this condition is like glass and brittle.

Tg is phenomenon that depends on time. Polymer may slowly changes in shape (distort) at lower temperature and appears brittle under a sudden bent.

Although Tg can be determined under certain range of equilibrium temperature, the value depends on the heating and cooling rate used in the DSC system.

Further heating leads to the crystallization and melting processes


Figure 7.18 The DSC curves to show glass transition of several polymers (20 mg, 10 K/min, nitrogen). PVC = poly(vynil chloride), PS = polystirene, EP = cured epoxy resin, PC = polycarbonate (Source: Haines: F 3.28).


A sample that has been cooled at a low cooling rate (0.1 K/min) undergoes Tg at 80 oC. If the sample is, then heated up at a high heating rate (20 K/min), it will not under glass transition until the temperature reaches 85 oC. This means that the sample must have absorbed more energy to become rubber-like material. As a result, an endotherm is observed overlapping with the glass transition step as shown in Figure 7.19.

Figure 7.19 The DSC curve for polystyrene that has been cooled at the rate of 0.2 K/min and then heated up at a rate of 5 K/min (Source: Haines F 3.29)

While the rate of heating and cooling must be stated, Tg also depends on the :

• molecular weight• degree of cure• amount of plastisizer

Glass transition with an endothermic phenomenon


NEASUREMENT OF PURITY

Purity test is very important in the industry such as pesticides, pure organics and metals,.

Figure 7.20 The portion of the Euthectic Phase diagram that is close to the Pure A (Source: Haines: F 3.30)

Assumption: two effective components that will produce an euthectic phase (Figure 7.20). When A is not pure and contains a small amount of impurity, B (xB mol %), the van’t Hoff equation shows that the freezing point will reduce from T0,A for pure A to Tm, the melting temperature of the mixture.

ΔT = (T0,A – Tm) = (RT 20,A / ΔHfus,A) . xB

Although the mole fraction of B is small and ΔT is also very small the DSC curve of impure material is broader than the pure sample due to gradual melting of the impurity concerned.

Since melting starts from the euthectic point, the line PQR can be divided into solid : liquid ratio as follows:

The phase line can be considered almost linear and a similar triangle can be constructed as shown in Figure 7.20.

Liquid fraction / solid fraction = PQ / QR or Melting fraction, F = PQ / QR

From the similar triangle concept, a ration on the y-axis is obtained:F = (T0,A – T) / (T0,A – Tm) = (T0,A – T) / ΔT

Replace into the van’t Hoff eqn.: (T0,A – T) = (1/F) . (RT 20,A / ΔHfus,A) . xB

Rearranged: T = T0,A - (1/F) . (RT 20,A / ΔHfus,A) . xB The plot of temperature

versus (1/F) will produce a straight line with the slope, b depends on xB, T0,A and ΔHfus,A.


The effect of thermal lag on the temperature, T can be corrected in two ways:

1. Using the software of the DSC system, or2. Using the measurement of the DSC peak for pure indium (Figure 7.21)

Figure 7.21 DSC peak for the melting of pure indium(Source: Haines: F_3.11)


From the DSC curve (Figure 7.22) that was obtained using the lowest heating rate, the ΔHfus,A and T0,A can be estimated and the portion of the sample that has melted at the temperature T oC is obtained from the ratio of the area a (up to the temperature T oC) divided by the area of the peak, A.

Figure 7.22DSC curve for an impure material showing the measurement of the related area and melting temperature correction (Source: Haines: F 3.31)

The melted fraction, F = a/A

The eqn. F = a/A should give a straight line. However, non-linear relationship is obtained due to several factors:

1. disregard of any pre-melting factor prior to the measurement of the peak area;

2. non-equilibrium conditions, due to the dynamic of heating;

3. the formation of solid solution.

Correction for these factors may be carried out in two ways:

a) Using an addition factor, da, or

b) Removal of non-equilibrium effects by using discrete steps of heating


Correction for factors leading towards non-linear relationship of the equation F = a/A:

a) Using an addition factor, da

F ´ = (a + da) / (A + da)andΔHfus,A = K . (A + da)

This can be related to the corrected van’t Hoff equation for the solid solution

ΔT ´ = (1 – k) · (RT 20,A / ΔHfus,A) . xB

where k = the distribution constant in the solid solution

If the heating steps of less than 1 K is used and the system is then allowed to achieve an equilibrium, peaks of equal areas will be formed until melting begins to take place.

The peak size will increase until the melting has been completed.

The total sum of the peak areas, after correction for the heat capacity, will give the value of a and A.

b) Removal of non-equilibrium effects by using discrete steps of heating


A sample of pesticide (MW = 365) is melted in a DSC system at the rate of 1 K/min and the results are as follows:

Example:

A = 10.5 cm2

K = 10.0 mJ/cm2

T (K) a (cm2) F

427 1.78 0.170

428 2.31 0.220

429 3.21 0.306

430 4.58 0.436

431 6.33 0.603

Calculate the values of T0,A, ΔHfus,A, xB


The plot of T versus 1/F produces curves that meet/cross on the axis at T = 434 K (Figure 7.23)

Figure 7.23 Plot of T (K) versus 1/F and the corrected 1/F. (a) without correction, (b) 5 % correction, (c) 10 % correction, (d) 19 % correction.(Source: Haines: F 3.32 )


The data can be treated as follows:

1. Continue to add small amount of a and A until a straight line is obtained

The consecutive additions of 0.5 cm2 to the area results in the best straight line when the correction is about 2.0 cm2 and this give a total sum of corrected areas to 12.5 cm2, and the corrected ΔH to 15.2 kJ/mole and the slope –2.2 which gives the value of xB = 2.1 mole % (97.9 % A purity)

2. The plot of (1/(T0,A – T)) versus Fthe slope = k

If the equation is rearranged, it is found that the plot of 1/(T0,A – T) versus F gives a straight line with the slope of (ΔH / RT2

0,A) / xB and the intersection is related to the correction of k. If T0,A = 434 K, then xB = 0.028 and the purity is 97.2 %.

3. Carry out regression analysis to the curve

This is the best method which will give the correction factors of 0.163 and the purity of 97.9 %.

lecture 7 application dta & dsc01

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