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Initially 2.0 moles of N2(g) and 4.0 moles of H2(g) were added to a 1.0-liter container and the following reaction then occurred: 3H2(g) + N2(g) 2NH3(g) The equilibrium concentration of NH3(g) = 0.55 moles/liter at 700.°C. What is the value for K at 700.°C for the formation of ammonia?

Answer :

Answer : The value of [tex]K_c[/tex] for the reaction is, [tex]5.4\times 10^{-3}[/tex]

Solution :  Given,

Moles of [tex]N_2[/tex] = 2.0 mol

Moles of [tex]H_2[/tex] = 4.0 mol

Volume of solution = 1.0 L

Concentration of [tex]NH_3[/tex] at equilibrium = 0.55 mol/L = 0.55 M

First we have to calculate the concentration of [tex]N_2\text{ and }H_2[/tex].

[tex]\text{Concentration of }N_2=\frac{Moles}{Volume}=\frac{2.0mol}{1.0L}=2.0M[/tex]

[tex]\text{Concentration of }H_2=\frac{Moles}{Volume}=\frac{4.0mol}{1.0L}=4.0M[/tex]

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

                           [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]

Initially conc.       2.0       4.0          0

At eqm.             (2.0-x)  (4.0-3x)     2x

The expression of [tex]K_c[/tex] will be,

[tex]K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

[tex]K_c=\frac{(2x)^2}{(2.0-x)\times (4.0-3x)^3}[/tex]       .......(1)

The concentration of [tex]NH_3[/tex] at equilibrium = 0.55 M = 2x

So,

2x = 0.55

x = 0.27 M

Now put the value of 'x' in the above equation 1, we get:

[tex]K_c=\frac{(2\times 0.27)^2}{(2.0-0.27)\times (4.0-3\times 0.27)^3}[/tex]

[tex]K_c=5.4\times 10^{-3}[/tex]

Therefore, the value of [tex]K_c[/tex] for the reaction is, [tex]5.4\times 10^{-3}[/tex]

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