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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

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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?

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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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