Question

(d) 2 mol gaseous ethene (C2H4) and 1 mol gaseous water are combined in a reactor and react to form ethanol (C2HsOH) according to C2Hs (g) + H20 (g) C2HsOH (g) At a temperature of 500 K the equilibrium constant Kp is 1.256 x 10-2 bar1, Assume that all reactants and products behave like perfect gases (i) Calculate the equilibrium composition (in mol%) of the gas mixture at 500 K and a combined pressure of 1 bar. 8 marks (ii) what pressure is required to obtain 10 mol% of ethanol in the product mixture? [3 marks]

Answer 1

Part i

Let $\large \epsilon$ be the amount of water moles reacted

Condition	Ethene	Water	Ethanol	Total
Initial	2	1	0	3
At Equilibrium	2-ε	1-ε	ε	3-ε
Mole frac at eq.	3	3	$\large \frac{\epsilon }{3-\epsilon }$	1

At equilibrium the partial pressure of each component can be related as:

PPethanol ethene * PPwater

Where PP=Partial pressure of each of the component

P Yethanol P * Yethene* P* Ywater

Where P is the total pressure of the system

3 3

3 1 0.01256 * 1 3 3

Solving the above equation we get the value of $\large \epsilon$ =0.00829 moles

The mole % at equilibrium is

Condition	Ethene	Water	Ethanol
Mole %	66.57	33.15	0.28

Part ii

We require ethanol at equilibrium to be 10 mol%.

3-E-0.1

$\large \therefore$  $\large \epsilon$ = 0.2727

Condition	Ethene	Water	Ethanol	Total
Initial	2	1	0	3
At Equilibrium	2-ε	1-ε	ε	3-ε
Mole frac at eq.	3	3	$\large \frac{\epsilon }{3-\epsilon }$	1

Equilibrium Composition of other components are:

Condition	Ethene	Water	Ethanol	Total
Mole frac at eq.	0.6333	0.2667	0.1000	1.0000

P Yethanol P * Yethene* P* Ywater

0.1 0.01256*P- 0.6333 * 0.2667

$\large \therefore$P=47.13 bar