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Example 1
Simulated reaction of 2nd order of the type A + B -> C
| Parameter
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log A/(s-1 l mol-1) = 5, EA = 50 kJ mol-1, Qr = - 37 kJ mol-1
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| Molar masses / g mol-1
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A = 100, B = 50, C = 150
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| Densities / g ml-1
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A = 1, B = 1.5, C = 2, Solvent = 1.25
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| Heating rates / K min-1
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0.5 , 1, 2, 5, 10
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| Concentration relationship A : B
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1:5, 1:4, 1:3, 1:2, 1:1, 5:32, 9:1
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Characteristics:
- extreme variation of the relationship of the starting concentrations of A and B
- presence of strongly different solvent quantities,
strongly different densities.
- perfect adjustment to the simulated curves
- model parameters are practically identical to the defaults:
lg A = 5.001 s-1(l/mol),
EA = 50 kJ mol-1,
Qr = -36.97 kJ mol-1
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Examination of the calculated concentration profile for curve (b = 1K mi-1); based on the initial values (first column) the final concentrations should correspond to those from the last column.
| Start amount of substance
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Final amount of substance
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Final mass
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Final Volume
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Final concentration
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| nA,0: 5 mol
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nA,f: 3 mol
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mA,f: 300 g
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VA: 300 ml
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cA,f: 3.3962 mol l-1
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| nB,0: 2 mol
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nB,f: 0 mol
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mB,f: 0 g
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VB: 0 ml
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CB,f: 0 mol l-1
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| nC,0: 0 mol
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nC,f: 2 mol
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mC,f: 300 g
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VC: 150 ml
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cC,f: 2.2642 mol l-1
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VSolvent: 433.33 ml
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Vtotal: 883.33 ml
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- The final concentrations for A and C (dashed lines) correspond exactly to the calculate values
- The volume change during reaction is taken into account correctly
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Other application examples
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