The equilibrium constant kc for the reaction pcl3(g) + cl2(g) ⇌ pcl5(g) is 49 at 230°c. if 0.70 mol of pcl3 is added to 0.70 mol of cl2 in a 1.00-l reaction vessel at 230°c, what is the concentration of pcl3 when equilibrium has been established? a) 0.049 mb) 0.11 mc) 0.59 md) 0.30 me) 0.83 m

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration and .

The given equilibrium reaction is,

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of will be,

Now put all the given values in the above expression, we get:

By solving the term x, we get

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of at equilibrium = x = 0.59 M

Therefore, the concentration of at equilibrium is 0.11 M

idk what the question is about

a electronegativity decreases beacuse the size of nucleus is bigger.