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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

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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 kp and kc 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,

Δng = ΣnProducts – ΣnReactants  

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

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