# 4HCl(g) + O2 <---> 2H2O(g) + 2Cl2(g) HCl and O2 ratio of mole is 4:1. Cl2 concentration is 0.5 mol dm^-3 in equilibrium state, and the volume precentage of Cl2 is 20 %. DeltaH = -2231KJ...

4HCl(g) + O2 <---> 2H2O(g) + 2Cl2(g)

HCl and O2 ratio of mole is 4:1. Cl2 concentration is 0.5 mol dm^-3 in equilibrium state, and the volume precentage of Cl2 is 20 %. DeltaH = -2231KJ

What are the equilibrium concentrations of HCl(g), O2(g) , H2O(g) , Cl2(g) ?

The balanced chemical equation for the reaction is:

4HCl(g) + O2(g) `stackrel(larr)(->)` 2H2O(g) + 2Cl2(g)

Let, initially there were ‘a’ moles of oxygen, out of which x moles reacted according to the given reaction. Again, moles of HCl taken initially was 4*a = 4a.

As per the balanced equation, 1 mole of oxygen reacts with 4 moles of HCl.

So, x moles of oxygen should react with 4x moles of HCl.

Proceeding with similar logic, the concentration of various species at equilibrium will be given by:

4HCl(g) + O2(g) <---> 2H2O(g) + 2Cl2(g)

Initially 4a a 0 0

At eqm. (4a-4x) (a-x) 2x 2x

By the first condition of the problem 2x =0.5 mole*dm^-3 = 0.5 M

`rArr` x = 0.5/2 = 0.25 M

The total volume of all the gaseous species, at eqm.

= 4a-4x+a-x+2x+2x = (5a-x)

Volume percentage of Cl2 is `(2x*100)/(5a-x)`

By the second condition of the problem,

`(2x*100)/(5a-x) =20`

`rArr (2x)/(5a-x)=1/5`

`rArr 10x = 5a-x`

`rArr a =(11x)/5`

Putting the value of x, `a = (11*0.25)/5` = 0.55

Therefore, at eqm., [HCl] = 4a-4x = 4*(0.55-0.25) =**1.2 M**

[O2] =a-x = 0.55-0.25 = **0.3 M**

[H2O] =[Cl2] = 2x = 2*0.25 = **0.5 M**

=> answer.