CHE471

CHEMICAL PROCESS PRINCIPLES 1

INTEGRATED PROJECT

GROUP MEMBERS

1) MOHD AZIZI BIN HUSAINI 20134597342) MARLISSA BINTI AZEMI 20138156183) MUHAMMAD HAZIQ BINTI MOHD IZHAR 2013426214 4) NASHRAH AYNA BINTI SHAH FIESAL 20134115145) NOR ATIKAH BINTI JEMBARI 2013261536

SUPERVISOR

DR NURUL FADHILAH BINTI KAMALUL ARIPIN

SUBMITTED TO

DR NORLIZA IBRAHIM

DATE SUBMITTED

06/06/2014

INTRODUCTION

This is a process of producing acetonitrile from a reaction of acetic acid with ammonia in vapor form. The vapors of acetic acid and ammonia is contacted with the presence of alumina catalyst consists of phosphoric acid that is impregnated on the alumina pellets. This is to produce a free ketone product and extremely high yields of acetonitrile. The process takes place in a reactor where the catalyst is placed inside the reactor tube. From the process, a production rate of 612.6 metric ton of acetonitrile is produced in a year.

I. Reactive Process.

In this process, the reactive process takes place inside a reactor tank where the reactants react with the presence of catalyst. Reactor function as a controlling and containing chemical reaction The reactants used are acetic acid and ammonia with the presence of alumina catalyst inside the reactor chamber. Mole ratio of ammonia to acetic acid is 1 : 1 ratio. The equation of reaction is :

CH3COOH + NH3 CH3CN + 2H2O

2.6 metric tons of acetic acid and 0.83 metric tons of ammonia were fed per day into the reactor. From the reaction, 95 % yield of acetonitrile and 95 % conversion acetic acid were produced. ( Hugh, J.H , Statman, M. , 1964 )

From the equation :

CH3COOH + NH3 CH3CN + 2H2O

Stoichiometry ratio :

nCH ₃ COOHnNH ₃

= 11

= 1

Real ratio :

CH ₃ COOHNH ₃

= 1.8 kmol2.0 kmol

= 0.9

Therefore, CH3COOH is a limiting reactant and NH3 is an excess

Feed :

Acetic acid, CH3COOH = 5700lb

dayx

1 kg2.205 lb

x1day24 hr

x1kmol60 kg

= 1.8 kmol / hr

Ammonia, NH3 = 1820lb

dayx

1 kg2.205 lb

x1day24 hr

x1kmol17 kg

= 2.0 kmol / hr

95 % acetic acid conversion: ( 0.05 unreacted )

n1 = ( 0.05 ) ( 1.8 kmol CH3COOH ) = 0.09 kmol

CH 3CN yield = 95 %

95 = n ₃x

x100 %

x = 1.8 kmol CH3COOH 1 mol CH3CN

1 mol CH3COOH

= 1.8 kmol

Therefore n3 = 1.71 kmol

By using Atomic Species balances

1) O balance : input = output

1.8 kmol CH3COOH 2 mol O = 0.09 kmol CH3COOH 2 mol O h 1 mol CH3COOH h 1 mol CH3COOH

+

n4 kmol H2O 1 mol O hr 1 mol H2O

n4 = 3.42 kmol/hr

2) N balance : input = output

2.0 kmol NH3 2 mol N = 1.71 kmol CH3CN 2 mol N h 1 mol NH3 h 1 mol CH3CN

+

n2 kmol NH3 1 mol N hr 1 mol NH3

n2 = 0.29 kmol/hr

Therefore, Total n = n1 + n2 + n3 + n4

= 0.09 + 0.29 + 1.71 + 3.42= 5.51 kmol

II. Non - Reactive Process.

In this process, one of the non – reactive processes occurs in the quench scrubber. Quench Scrubber act as a mixer where the water is adder to mix up with the mixture from the reactor in the process. Water is added into the quench scrubber together with gas product from the reactor to be condensed. 4.2 kmol of water is fed into the quench scrubber every hour. The product from the reactor enters the quench scrubber with a mole flow rate 5.51 kmol/hr.

Non – reactive process : Input = Output

5.51 kmol/hr + 4.2 kmol/hr = n1

n1 = 9.71 kmol / hr

CH 3COOH balances.

Input = Output 0.09 kmol CH3COOH/hr = ( X kmol CH3COOH /kmol)( 9.71 kmol/hr )

X = 0.009 kmol CH3COOH/kmol

NH 3 balances.

Input = Output 0.29 kmol NH3/hr = ( Y kmol NH3/kmol)( 9.71 kmol/hr )

Y = 0.03 kmol NH3/kmol

CH 3CN balances.

Input = Output 1.71 kmol CH3CN/hr = ( Z kmol CH3CN /kmol)( 9.71 kmol/hr )

Z = 0.176 kmol CH3CN/kmol

H 2O balances.

H2O = 1 – 0.009 – 0.03 – 0.176 = 0.785 kmol H2O/kmol

REFERENCES

1. Hugh, J.H, Statman, M. 1964. Acetonitrile Process, US Patent 3,161,6692. Lyon, G. 1974. Apparatus For Catalytic Reactions, US Patent 3,787,1883. Felder, R.M, Rousseau, R.W (2004). Elimentary Principles Of Chemical Processes. 3rd

ed. United States: WILEY. 85 - 153.