Ethylene and chlorine react to form ethylene dichloride, like this: CH2 CH2(g)+C

Question

Ethylene and chlorine react to form ethylene dichloride, like this: CH2 CH2(g)+Cl2(g) CH2 CH2 C2(g) Use this chemical equation to answer the questions in the table below. Suppose 235, mmol of CH2CH2 and | None. 235. mmol of Cl2 are added to an Some, but less than 235. mmol. empty flask. How much CH2CH2C 235. mmol. will be in the flask at equilibrium? More than 235. mmol. Suppose 200. mmol of CH,CH,CI2 None are added to an empty flask. How Some, but less than 200. mmol. much CH2 CH2C12 will be in the200. mmol. flask at equilibrium? O More than 200. mmol.

Explanation / Answer

Ans. Note the following points-

I. Following stoichiometry, 1 mol ethylene reacts with 1 mol chlorine.

II. At equilibrium, all the constituent chemical species of the reaction must be presents. That is, all the three compounds, CH2CH2, Cl2 and CH2CH2Cl2 must be present at the equilibrium – through their relative amount would be determined by equilibrium constant.

III. Le Chatelier’s principle states “if a dynamic equilibrium is disturbed by changing the conditions (Concentration, Volume, Pressure, temperature, etc.), the position of equilibrium shifts to counteract the change to reestablish an equilibrium”.

# Coming to the question:

#1. Correct option- B. Some, but less than 235 mmol.

The reaction proceed to the right to form CH2CH2Cl2 and to establish equilibrium – during which some of the reactants would be converted into product. Note that, as mentioned in #II above, all the three chemical species must be present at equilibrium. So, the amount of CH2CH2Cl2 would always be less than 235 mol because not all the reactants are converted into product – so, consumption of less than 235 mmol reactants form less than 235 mmol product (CH2CH2Cl2).

#2. Correct option- B. Some, but less than 200 mmol.

When CH2CH2Cl2 alone is present in the reaction vessel, some of it would decompose into CH2CH2 and Cl2 in order to establish equilibrium. However, as mentioned in #II above, some of CH2CH2Cl2 would always be there in the reaction mixture at equilibrium. However, since CH2CH2Cl2 decomposes to establish equilibrium, its equilibrium amount be less than 200 mmol.

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