Initially 2.0 moles of n2(g) and 4.0 moles of h2(g) were added to a 1.0-liter container and the following reaction then occurred: 3h2(g) + n2(g) 2nh3(g) the equilibrium concentration of nh3(g) = 0.55 moles/liter at 700.°c. what is the value for k at 700.°c for the formation of ammonia?

Answer : The value of for the reaction is,

Solution :  Given,

Moles of = 2.0 mol

Moles of = 4.0 mol

Volume of solution = 1.0 L

Concentration of at equilibrium = 0.55 mol/L = 0.55 M

First we have to calculate the concentration of .

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

Initially conc.       2.0       4.0          0

At eqm.             (2.0-x)  (4.0-3x)     2x

The expression of will be,

      .......(1)

The concentration of at equilibrium = 0.55 M = 2x

So,

2x = 0.55

x = 0.27 M

Now put the value of 'x' in the above equation 1, we get:

Therefore, the value of for the reaction is,

2b2t

explanation:

2b2t