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

Explanation:

Moles of  $NH_3$ = 4.0 mole

Volume of solution = 2.0 L

Initial concentration of $NH_3$ = $\frac{moles}{Volume}=\frac{4.0}{2.0}=2.0M$

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 $NH_3$ = $\frac{moles}{Volume}=\frac{2.0}{2.0}=1.0M$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[x]\times [3x]^3}{[(2-2x)]^2}$

(2-2x) = 1.0

x= 0.5 M

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

$K_c=\frac{[0.5]\times [3\times 0.5]^3}{[(2-2\times 0.5)]^2}$

$K_c=1.6785$

Thus the value of K for this reaction is 1.6875