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# 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 is the Reaction Mixture at Equilibrium? If Not, What is the Direction of the Net Reaction? - CBSE (Science) Class 11 - Chemistry

ConceptApplications of Equilibrium Constants Predicting the Direction of the Reaction

#### Question

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 × 102

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

#### Solution 1

The reaction is N_2(g) + 3 H_2(g) ‏⇌ 2NH_3(g)

Concentration quotient (Q_c) = [NH_3]^2/([N_2][H_2]^3)  = (8.13/20 "mol L"^(-1))^2/((1.57/ 20 "mol L"^(-1)) xx ("1.92 / 20" "mol L"^(-1))^3)

= 2.38 xx 10^3

The equilibrium constant (K_c) for the reaction = 1.7 xx 10^(-2)

As Q_c != K_c this means that the reaction is not in a state of equilibrium

#### Solution 2

The given reaction is:

N_(2(g)) + 3H_(2(g)) ↔ 2NH_(3(g))

The given concentration of various species is

[N_2] = 1.57/20 "mol L"^(-1)  [H_2] = 1.92/20 mol L^(-1)

[NH_3] = 8.13/20 mol L^(-1)

Now, reaction quotient Qc is:

Q_c = ([NH_3]^2)/[N_2][H_2]^3

= ((8.13)/20)^2/((1.57/20)(1.92/20)^3)

= 2.4 xx 10^3

Since Q_c != K_C the reaction mixture is not at equilibrium.

Again, Q_c > K_c Hence, the reaction will proceed in the reverse direction.

Is there an error in this question or solution?

#### APPEARS IN

Solution 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 is the Reaction Mixture at Equilibrium? If Not, What is the Direction of the Net Reaction? Concept: Applications of Equilibrium Constants - Predicting the Direction of the Reaction.
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