Ethylene oxide is produced by the catalytic oxidation of ethylene: c2h4 g 1 2 o2 g ! c2h4o g an undesired competing reaction is the combustion of ethylene to co2. 544 chapter 9 balances on reactive processes the feed to a reactor contains 2mol c2h4/mol o2. the conversion and yield in the reactor are respectively 25% and 0.70mol c2h4o produced/mol c2h4 consumed. a multiple-unit process separates the reactor outlet stream components: c2h4 and o2 are recycled to the reactor, c2h4o is sold, and co2 and h2o are discarded. the reactor inlet and outlet streams are each at 450°c, and the fresh feed and all species leaving the separation process are at 25°c. the combined fresh feedrecycle stream is preheated to 450°c. (a) taking a basis of 2 mol of ethylene entering the reactor, draw and label a flowchart of the complete process (show the separation process as a single unit) and calculate the molar amounts and compositions of all process streams. (b) calculate the heat requirement (kj) for the entire process and that

2b2t

explanation:

2b2t

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point