Question

The values of \(\frac{{{K_p}}}{{{K_C}}}\) for the following reactions at 300 K are, respectively (At 300 K, RT = 24.62 dm³ atm mol^-1)

N₂ (g) + O₂ (g) ⇌ 2NO (g)

N₂ O_4 (g) ⇌ 2NO₂ (g)

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

1. 1; 24.62 dm³ atm mol^-1; 606.0 dm⁶ atm² mol^-2
2. 1; 24.62 dm³ atm mol^-1; 1.65×10^-3 dm^-6 atm^-2 mol²
3. 24.62 dm³ atm mol^-1; 606.0 dm⁶ atm^-2 mol²; 1.65 × 10^-3 dm^-6 atm^-2 mol²
4. 1; 4.1 × 10^-2 dm^-3 atm mol; 606.0 dm⁶ atm² mol^-2

Answer 1

Correct Answer - Option 2 : 1; 24.62 dm³ atm mol^-1; 1.65×10^-3 dm^-6 atm^-2 mol²

Concept:

We know that, the relationship between k_p and k_c of a chemical equilibrium state (reaction) is

\({k_p} = {K_c}{\left( {RT} \right)^{\Delta {n_g}}}\)

\(\Rightarrow \frac{{{K_p}}}{{{K_C}}}\; = {\left( {RT} \right)^{\Delta {n_g}}}\)

Where,

Δn_g = Σn_Products – Σn_Reactants  

Calculation:

(i) N₂ (g) + O₂ (g) ⇌ 2NO (g):

⟹ (RT)² – (1 + 1) = (RT)⁰ = 1

(ii) N₂O_4 (g) ⇌ 2NO₂ (g):

⟹ (RT)^2-1 = RT = 24.62 dm³ atm mol^-1

(iii) N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g):

⟹ (RT)^2-(3+1) = (RT)^-2

\( \Rightarrow {\left( {RT} \right)^{ - 2}} = \frac{1}{{{{\left( {24.62\;d{m^3}\;atm\;mo{l^{ - 1}}} \right)}^2}}}\)

∴ (RT)^-2 = 1.649 × 10^-3 dm^-6 atm^-2 mol²