When a reactant A reacts in a batch reactor at 227 degree C and 10.1 kPa, the re

Question

When a reactant A reacts in a batch reactor at 227 degree C and 10.1 kPa, the reactant reaches an equilibrium conversion of 50% following the stoichiometry A 2B (reversible reaction). When the same reaction takes place at 727 degree C and 101.32 kPa, the equilibrium conversion is also 50 % and it follows the same stoichiometry. Using this information Calculate the heat of reaction assuming it is constant throughout the reaction (R = 1.987 cal/mol degree K) Calculate the equilibrium conversion this reaction takes place at 75 degree C and 10.1 kPa.

Explanation / Answer

Conversion 50% means that from 1 mol A at the beginning :

0.5 mol A is still present at equilibrium and 2x0.5=1 mol B was produced at equilibrium.

Thus the ratio B:A is 2:1 at equilibrium (as mol, partial pressure or concentration ratio).

At T1 = 227C and 10.1kPa:   

The partial pressures are

PA = 10.1/3 kPa = 3.37 kPa = 0.03326 atm

PB = (2/3)x10.1 kPa = 6.73 kPa = 0.06642 atm

The conversion to atm is critical to obtain the rigth and adimensional value for Kp.

At 227C and 10.1kPa:

Kp1 = (pB)2/pA = 0.066422/0.03326= 0.1326 (adimensional)

At T2 = 727C and 101.32 kPa:          

The partial pressures are

PA = 101.32/3 kPa = 33.77 kPa = 0.333 atm

PB = (2/3)x10.1 kPa = 6.73 kPa = 0.667 atm

The conversion to atm is critical to obtain the rigth and adimensional value for Kp.

At T2= 727C and 101.32kPa:

Kp2 = (pB)2/pA = 0.6672/0.333= 1.336      (adimensional)

To find dH use the van’t Hoff equation:

ln (Kp1/Kp2) = - (dH/R) ( 1/T2 – 1/T1)

ln (0.1326/1.336) = - (dH/R) ( 1/1000K – 1/500K)

- 2.31 = - (dH/1.987cal/mol.K) ( 0.001 – 0.002)

dH = - 4590 cal/mol = - 4.59 kcal/mol   

b.

Use the same equation. Use for comparison the first data set (same pressure)

ln (Kp1/Kp3) = - (dH/R) ( 1/T3 – 1/T1)

ln (Kp1/Kp3) = - (4590/1.987) ( 1/348K – 1/500K) =

                       = - 2310 (0.00287- 0.00200)

                        = - 2.01

Kp1/Kp3 = e-2.01 = 0.134

Kp3 = Kp1 / 0.134 = 0.99 round it to 1.0

Kp1 = (pB)2/pA = 1

Thus (pB)2 = pA     and also pB + pA = 0.1 atm

pB + (pB)2 = 0.1

pB = 0.0916

pA = 0.91

the ratio 0.091/0.91 = 1:10 may be used partial pressure or molar ratio.

The conversion is

(0.091/2) / (0.91 +0.091/2) = 0.5/10.5 = 0.0476 or 4.76%

Please verify all calculations.

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