Question

One mole of N2 and 3 moles of H2 are mixed in a closed vessel of 1dmcube. At equilibrium of vessel contains the total of 2.4 moles. Calculate equilibrium constant Kc for reaction N2 + 3H2 ---> 2NH3

Answer 1

The value of equilibrium constant (K) is, 59.3

Explanation :

First we have to calculate the concentration of $N_2,H_2$  and total concentration at equilibrium.

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{1mol}{1L}=1M$

and,

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{3mol}{1L}=3M$

and,

$\text{Concentration at equilibrium}=\frac{\text{Moles at equilibrium}}{\text{Volume of solution}}=\frac{2.4mol}{1L}=2.4M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                     $N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

Initial conc.    1            3                0

At eqm.        (1-x)     (3-3x)            2x

Total concentration at equilibrium  = 2.4 M

(1-x) + (3-3x) + 2x = 2.4

4 - 2x = 2.4

x = 0.8 M

The expression for equilibrium constant is:

$K=\frac{[NH_3]^2}{[N_2][H_2]^3}$

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

$K=\frac{(2x)^2}{(1-x)\times (3-3x)^3}$

Putting value of x = 0.8

$K=\frac{(2\times 0.8)^2}{(1-0.8)\times (3-3\times 0.8)^3}$

$K=59.3$

Thus, the value of equilibrium constant (K) is, 59.3

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