Consider the following reaction in a sealed vessel at equilibrium with concentrations: [N_2] = 3.0 × 10^-3 M , [O_2] = 4.2 × 10^-3 M , and [NO] = 2.8 × 10^-3 M . 2NO(g) leftharpoons N_2(g) + O_2(g) If 0.1 mol L^-1 of NO is taken in a closed vessel, what will be the degree of dissociation ( α ) of NO

1. First, find K_c from initial equilibrium data: K_c = [N_2][O_2]/[NO]^2 = (3.0 × 10^-3) × (4.2 × 10^-3)(2.8 × 10^-3)^2 = 12.6 × 10^-67.84 × 10^-6 ≈ 1.6 2. Let initial conc. of NO be C = 0.1 . 2NO leftharpoons N_2 + O_2 Equil. conc: C(1-α) , Cα/2 , Cα/2 . K_c = (α/2) (Cα/2)[C(1-α)]^2 = α^2/4(1-α)^2 1.6 = α^2/4(1-α)^2 √(1.6 × 4) = α/1-α √(6.4) = α/1-α 2.53 = α/1-α 2.53 - 2.53α = α 2.53 = 3.53α α ≈ 0.717

Consider the following reaction in a sealed vessel at equilibrium with concentrations: [N_2] = 3.0 × 10^-3 M , [O_2] = 4.2 × 10^-3 M , and [NO] = 2.8 - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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Consider the following reaction in a sealed vessel at equilibrium with concentrations: [N_{2}] = 3.0 \times 1 0^{- 3} M, [O_{2}] = 4.2 \times 1 0^{- 3} M, and [NO] = 2.8 \times 1 0^{- 3} M . 2 NO (g) ⇌ N_{2} (g) + O_{2} (g) If 0.1 mol L^{- 1} of NO is taken in a closed vessel, what will be the degree of dissociation (α) of NO (g) at equilibrium?

A $0.0889$
B $0.8889$
C $0.717$
D $0.0089$

Solution & Step-by-step Explanation

1. First, find K_{c} from initial equilibrium data: K_{c} = \frac{[ N _{2} ] [ O _{2} ]}{[ NO ] ^{2}} = \frac{( 3.0 \times 1 0 ^{- 3} ) \times ( 4.2 \times 1 0 ^{- 3} )}{( 2.8 \times 1 0 ^{- 3} ) ^{2}} = \frac{12.6 \times 1 0 ^{- 6}}{7.84 \times 1 0 ^{- 6}} \approx 1.6 2. Let initial conc. of NO be C = 0.1 . 2 NO ⇌ N_{2} + O_{2} Equil. conc: C (1 - α), \frac{C α}{2}, \frac{C α}{2} . K_{c} = \frac{( \frac{C α}{2} ) ( \frac{C α}{2} )}{[ C ( 1 - α ) ] ^{2}} = \frac{α ^{2}}{4 ( 1 - α ) ^{2}} 1.6 = \frac{α ^{2}}{4 ( 1 - α ) ^{2}} \Rightarrow 1.6 \times 4 = \frac{α}{1 - α} \Rightarrow 6.4 = \frac{α}{1 - α} 2.53 = \frac{α}{1 - α} \Rightarrow 2.53 - 2.53 α = α \Rightarrow 2.53 = 3.53 α \Rightarrow α \approx 0.717

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Consider the following reaction in a sealed vessel at equilibrium with concentrations: [N_{2}] = 3.0 \times 1 0^{- 3} M, [O_{2}] = 4.2 \times 1 0^{- 3} M, and [NO] = 2.8 \times 1 0^{- 3} M . 2 NO (g) ⇌ N_{2} (g) + O_{2} (g) If 0.1 mol L^{- 1} of NO is taken in a closed vessel, what will be the degree of dissociation (α) of NO (g) at equilibrium?

0.0889

0.8889

0.717

0.0089

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