Question

For the reaction N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) ΔH = -ve : 
1. K_p = K_c (RT)^-2
2. K_p = K_c
3. K_p = K_c RT
4. K_p = K_c (RT)^-1

Answer 1

Correct Answer - Option 1 : K_p = K_c (RT)^-2

Concept:

Haber's process:

• This process is used in large scale preparations of ammonia.

N2 + 3H2 → 2NH3

• During the process, nitrogen and hydrogen are used in the ratio 1:3.
• The process is exothermic ΔH = -ve  in nature means heat is produced in the process.
• According to the Le-chateliar principle, the temperature is kept high to speed up the process.
• The gaseous ammonia produced is converted to liquid ammonia to remove the products formed.
• This drives the reaction forward.

Equilibrium Constants:

• ​The constants K_p and K_c are both equilibrium constants.
• K_p is used when the concentration terms are given in partial pressures i.e, in gaseous reactions.
• K_c is used when the reaction terms are expressed in molarities.
• The relation between K_p and K_c is given by:

\({K_p} = {K_c} \times {\left( {RT} \right)^{\Delta n}}\) where R = Universal gas constant, T = Temperature, and \(\triangle n\) = change in moles of gases in the reaction.

Calculation:

• ​The relation between K_p and K_c is

\({K_p} = {K_c} \times {\left( {RT} \right)^{\Delta n}}\)

For the reaction

N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)

\(No.\;of\;moles\;of\;products = 2\)

\(No.\;of\;moles\;of\;reactants = 3 + 1 = 4\)

\(Change\;in\;number\;of\;moles\;of\;gases = {n_{products}} - {n_{reactants}} = \;\Delta n\)

\( = 2 - 4 = - 2\)

Hence,

\({K_p} = {K_c} \times {\left( {RT} \right)^{-2}}\)

Hence, or the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) ΔH = -ve, \({K_p} = {K_c} \times {\left( {RT} \right)^{-2}}\)