Chemistry: 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?

.72 (mol/L)/sExplanation:A P E X...

Initially 2.0 moles of n2(g) and 4.0 moles of h2(g)

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09.01.2022

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

Solution : Given,

Moles of N_2 = 2.0 mol

Moles of H_2 = 4.0 mol

Volume of solution = 1.0 L

Concentration of NH_3 at equilibrium = 0.55 mol/L = 0.55 M

First we have to calculate the concentration of $N_2\text{ and }H_2$ .

$\text{Concentration of }N_2=\frac{Moles}{Volume}=\frac{2.0mol}{1.0L}=2.0M$

$\text{Concentration of }H_2=\frac{Moles}{Volume}=\frac{4.0mol}{1.0L}=4.0M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

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

Initially conc. 2.0 4.0 0

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

The expression of K_c will be,

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

$K_c=\frac{(2x)^2}{(2.0-x)\times (4.0-3x)^3}$ .......(1)

The concentration of NH_3 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:

$K_c=\frac{(2\times 0.27)^2}{(2.0-0.27)\times (4.0-3\times 0.27)^3}$

$K_c=5.4\times 10^{-3}$

Therefore, the value of K_c for the reaction is, $5.4\times 10^{-3}$

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