Question

mixture of SO3, SO2 and O2 gases is maintained at equilibrium in 10 litre flask at a temp in which Kc for the reaction 2SO2 + O2 gives 2SO3. Kc= 100 .at equilibrium-
1. if no. of moles of SO3 and SO2 in flask are same then how many moles of O2 are present.
2.no. of moles of SO3 is twice of SO2, then how many moles of O2 are present.

Answer 1

Answer:- (1) 0.1 mole and (2) 0.4 moles.

Solution:- The given balanced equation is:

$2SO_2+O_2\leftrightarrow 2SO_3$

The equilibrium expression is written as:

$Kc=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$

(1) From given information, moles of sulfur trioxide and sulfur dioxide are same. Let's say at equilibrium, we have n moles of sulfur trioxide and m moles of oxygen. Moles of sulfur dioxide will also be n since sulfur trioxide and sulfur dioxide has same moles.

Kc is given as 100. Divide the moles of the liters to get the concentrations and plug in the values in the equilibrium expression.

$[SO_3]=\frac{n}{10}$

$[SO_2]=\frac{n}{10}$

$[O_2]=\frac{m}{10}$

$100=\frac{[\frac{n}{10}]^2}{[\frac{n}{10}]^2[\frac{m}{10}]}$

$100=\frac{10}{m}$

$m=\frac{10}{100}$

m = 0.1

So, there are 0.1 moles of $O_2$ at equilibrium.

(2) Let's say there are n moles of sulfur dioxide and m moles of oxygen. Then moles of sulfur trioxide will be 2n as the moles of it are twice of sulfur dioxide.

$[SO_3]=\frac{2n}{10}$

$[SO_2]=\frac{n}{10}$

$[O_2]=\frac{m}{10}$

$100=\frac{[\frac{2n}{10}]^2}{[\frac{n}{10}]^2[\frac{m}{10}]}$

$100=\frac{40}{m}$

$m=\frac{40}{100}$

m = 0.4

So, there are 0.4 moles of $O_2$ at equilibrium.

Answer 2

Explanation:

I hope this will help you