The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sample of NH3(g) is introduced into an evacuated vessel at 650 K, the equilibrium concentration of H2(g) is found to be 0.729 M. Calculate the concentration of NH3 in the equilibrium mixture. M

Concentration of in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of = 0.729 M

The given balanced equilibrium reaction is,

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

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

concentration of in the equilibrium mixture =

Thus concentration of in the equilibrium mixture is 0.31 M

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.