Question

At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction below. 2 NH3(g) equilibrium reaction arrow N2(g) 3 H2(g) At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction

Answer 1

Answer: The value of K for this reaction is 1.6875

Step-by-step explanation:

Moles of
`NH_3` = 4.0 mole

Volume of solution = 2.0 L

Initial concentration of
`NH_3` =
`(moles)/(Volume)=(4.0)/(2.0)=2.0M`

The given balanced equilibrium reaction is,

`2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)`

Initial conc. 2 M 0 M 0 M

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

Equilibrium concentration of
`NH_3` =
`(moles)/(Volume)=(2.0)/(2.0)=1.0M`

The expression for equilibrium constant for this reaction will be,

`K_c=([x]* [3x]^3)/([(2-2x)]^2)`

(2-2x) = 1.0

x= 0.5 M

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

`K_c=([0.5]* [3* 0.5]^3)/([(2-2* 0.5)]^2)`

`K_c=1.6785`

Thus the value of K for this reaction is 1.6875