Question

5. A study of the system, 4 NH3(g) + 7 O2(g) 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [NH3] = [O2] = 3.60 M as the only components initially. At equilibrium, [N2O4] is 0.60 M. Calculate the value of the equilibrium constant, Kc, for the reaction.

Answer 1

Answer: The value of the equilibrium constant, Kc, for the reaction is 0.013

Step-by-step explanation:

Initial concentration of
`NH_3` = 3.60 M

Initial concentration of
`O_2` = 3.60 M

The given balanced equilibrium reaction is,

`4NH_3(g)+7O_2(g)\rightleftharpoons 2N_2O_4(g)+6H_2O(g)`

Initial conc. 3.60 M 3.60 M 0 M 0 M

At eqm. conc. (3.60-4x) M (3.60-7x) M (2x) M (6x) M

The expression for equilibrium constant for this reaction will be,

`K_c=([N_2O_4]^2* [H_2O]^6)/([NH_3]^4*[Cl_2])`

`[N_2O_4]=0.60M`

2x = 0.60 M

x= 0.30 M

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

`K_c=((2* 0.30)^2* (6* 0.30)^6)/((3.60-3* 0.30)^4* (3.60-7* 0.30)^7)`

`K_c=0.013`