Question

Chlorine is produced from hydrogen chloride and oxygen via the following reaction at 700K and 1bar:

4HCl_(g) +0_2(g) --> 2H₂O_(g)+2Cl_2(g)

The data for the standard for the standard Gibbs free energy and the standard heat of formation for each species in the rxn at 298.15K are given in table below.

	Standard Gibbs Free Energy (j/mol)	Standard Heat of Formation (J/mol)
HCl(g)	-95299	-92307
H2O(g)	-228572	-241818

a)Estimate the equilibrium production rate (kmol/s) for chlorine if the initial flow rate for hydrogen chloride and oxygen are 20kmol/s and 40kmol/s, respectively.Assume ideal gas condition is valid.(14m)

b)Based on part (a), propose the new equlibrium temperature if the production rate for chlorine is 6kmol/s, assuming ideal gas condition with the reaction pressure of 2bar.(5M)

c) An engineer has proposed to add 5kmol/s of chlorine into the reactor together with hydrogen chloride and oxygenduring the reaction to increase the chlorine production on part (a).Justify the feasibility of this proposal with calculation and explanation.(6m)

Answer 1

Given reaction

Species (kmole/hr)	HCl	O2	H2O	Cl2
t=0	20	40	0	0
t=tequilibrium	20-x	40-(x/4)	2x	2x

Total gas moles at t=0 = 60 (kmoles/s)

Total gas moles at t=t, = 60-(11x/4)

We can calculate

As we know that

Now, we can write

P=1 bar (given)

kmol/s

Therefore in product

Mole flowrate of HCl = 20-x = 20-10.48 = 9.52 (kmol / sec)

Moles of oxygen = 40-(x/4) = 40-(10.48/4) = 37.38 (kmol/s)

Mole of water = 2*10.48 = 20.96 (kmol/s)

Mole of chlorine = 2*10.48 = 20.96 (kmol/s)

b)

Now if product flowrate is 6 kmol/s

Therefore we can write

So, we can write

Now as we know that

c)

Now if we add 5 kmol/s of chlorine in the fees together with HCl and water. As we know that number of gaseous moles on product side is less than that of reactant side. Therefore according to Le-Chatelier principle, on adding more gaseous moles, product gas flowrate will increase as number of gaseous mole in product side is less.  

We can see by calculation also after adding 5 kmol/s of chlorine

Given reaction

Species (kmole/hr)	HCl	O2	H2O	Cl2
t=0	20	40	0	5
t=tequilibrium	20-x	40-(x/4)	2x	2x+5

Total gas moles at t=0 = 60+5 = 65 (kmoles/s)

Total gas moles at t=t, = 60+5-(11x/4)

We can calculate

As we know that

Now, we can write

P=1 bar (given)

kmol/s

Therefore in product

Mole of chlorine = 2*9.65+5 = 24.3 (kmol/s)

Hence we can see that flowrate of chlorine in product has increased from 20.96 kmol/s to 24.3 kmol/s