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In: Chemistry

Ethanol is produced commercially by the hydration of ethylene: C2H4+H2O-->C2H5OH Some of the product is converted...

Ethanol is produced commercially by the hydration of ethylene:

C2H4+H2O-->C2H5OH

Some of the product is converted to diethyl ether in side reaction

2C2H5OH-->(C2H5)2O+H2O

The feed to the reactor contains ethylene,steam,and an inert gas. A sample of the reactor effluent gas is analyzed and found to contain 43.3 mole% ethylene, 2.5% ethanol, 0.14% ether, 9.3% inerts, and the balance water.

a. Take as a basis 100 mol of effluent gas, draw and label a flowchart, and do a degree-of-freedom analysis based on atomic species to prove that the system has zero degrees of freedom?

b. Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol, and the selectivity of ethnaol production relative to ether production?

Expert Solution

Step1: Formula:

C2H4+H2O------------->C2H5OH

some products make: 2C2H5OH------------>(C2H5)O(C2H5) +H2O

A. Degreeof freedom analysis basic on anatomic species:

B:molar composition of the reactor feed:

Step 1: Know data:

43.3 moles % etilene, 2.5% ethanol, 0.14% ether, 9.3% inerts

Etilene: 1.18Kg/mL, ethanol= 789kg/mL, etilic ether= 713kg/mL

Density= 1.18x10-3g/mL, ethanol= 0.789g/mL, etilic ether= 0.713g/mL

PM=Etilene: 28.05g/mol, ethanol= 46.06g/mol, etilic ether= 74.12 g/mol

M=mol solute/L solvent

(etilene) 1.18x10-3g etilene/mL1g/1000mL= 1.18g etilene*43.3%solute /100%solvente=0.51094g etilene

mol etilene= 0.51094g/28.05g/mol=0.018mol etilene M etilene= 0.018mol/1=0.018M

Ethanol=0.789g/mL1g/1000mL=789g ethanol*2.5%solute/100%solvente=19.725g ethanol

mol eti=19.725g etilene/46.06/mol=0.70mol ethanol M0.70 mol ethanol/1L= 0.428 M ethanol

ether= 0.713g/mL1g/1000mL=713g ether*0.14%solute/100%solvent=0.9982g

mol ether: 0.9982g/74.12g/mol= 0.0134mol ether M=0.0134mol/L=0.013M

Note: if the problem dont say volume. we can say that is 1 L

queen_honey_blossom answered 3 months ago

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