A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction

N2(g)+3H2(g) ⇌ 2NH3(g) is $$1.7\times 10^{2}$$

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Asked by Pragya Singh | 1 year ago |  80

##### Solution :-

The given reaction is:

N2(g)+3H2(g) ⇌ 2NH3(g)

The given concentration of various species is

$$[N_2]= \dfrac{1.57}{20}mol\, L^{-1} [H_2]$$

$$= \dfrac{1.92}{20}mol\, L^{-1}[NH_3]$$

$$= \dfrac{8.31}{20}mol\, L^{-1}$$

Now, reaction quotient Qc is:

$$Q=\dfrac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\$$

$$=\dfrac{(\dfrac{(8.13)}{20})^{2}}{(\dfrac{1.57}{20})(\dfrac{1.92}{20})^{3}}\\ \\$$

$$=2.4\times 10^{3}$$

Since, Qc ≠ Kc, the reaction mixture is not at equilibrium.

Again, Qc > Kc. Hence, the reaction will proceed in the reverse direction.

Answered by Abhisek | 1 year ago

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