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Question: In which of the following equilibrium, ​ \( {{K}_{p}}={{K}_{c}} \) ? A) \( {{N}_{2(g)}}+{{O}_{2(g...

In which of the following equilibrium, ​ Kp=Kc{{K}_{p}}={{K}_{c}} ?
A) N2(g)+O2(g)⇌2NO(g){{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}
B) 2NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)2N{{H}_{3(g)}}+5{{O}_{2(g)}}\rightleftharpoons 4N{{O}_{(g)}}+6{{H}_{2}}{{O}_{(g)}}
C) N2(g)+3H2(g)⇌2NH3(g){{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3(g)}}
D) 2NO(g)+O2(g)⇌2NO2(g)2N{{O}_{(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{2(g)}}

Explanation

Solution

The equilibrium constant Kc{{K}_{c}} is the ratio of the equilibrium concentrations of the products to that of the concentrations of the reactants raised to the power of their respective stoichiometries. The Kc{{K}_{c}} can be given as: Kc=[product][reactan⁡t]{{K}_{c}}=\dfrac{[product]}{[reac\tan t]}
Kp{{K}_{p}} is the equilibrium constant given as the ratio of the partial pressures of the products to that of the reactants. The reaction equation is required to find the value of Kp{{K}_{p}} . It is a unitless number. Kp=pproductspreactan⁡t{{K}_{p}}=\dfrac{{{p}_{products}}}{{{p}_{reac\tan t}}} .

Complete Step By Step Answer:
To solve this question we will know the relation between Kp{{K}_{p}} and Kc{{K}_{c}} . The relation can be given as: Kp=Kc(RT)Δn{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}
Where Δn=nproducts−nreactan⁡ts\Delta n={{n}_{products}}-{{n}_{reac\tan ts}} i.e. the difference in the no. of moles of gaseous products and no. of moles of gaseous reactants. Pure solids and liquids are not included while determining Δn\Delta n . R is the gas constant and T is the temperature.
The value of Kp{{K}_{p}} or Kc{{K}_{c}} is constant if the temperature remains constant. Both will have different values and both the terms to be equal only if, Δn=0\Delta n=0
We’ll find the reaction where Δn=0\Delta n=0 and that is our correct answer:
A) N2(g)+O2(g)⇌2NO(g){{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}} : Remember we will consider only gaseous products and reactants. Δnreaction=2−(1+1)=2−2=0\Delta {{n}_{reaction}}=2-(1+1)=2-2=0 . Therefore, Kp=Kc(RT)0→Kp=Kc{{K}_{p}}={{K}_{c}}{{(RT)}^{0}}\to {{K}_{p}}={{K}_{c}} and this is the correct answer
B) 2NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)2N{{H}_{3(g)}}+5{{O}_{2(g)}}\rightleftharpoons 4N{{O}_{(g)}}+6{{H}_{2}}{{O}_{(g)}} : Δnreaction=(4+6)−(5+2)=10−7=3\Delta {{n}_{reaction}}=(4+6)-(5+2)=10-7=3 . Therefore Kp=Kc(RT)3→Kp=KcR3T3{{K}_{p}}={{K}_{c}}{{(RT)}^{3}}\to {{K}_{p}}={{K}_{c}}{{R}^{3}}{{T}^{3}} . This option is incorrect
C) N2(g)+3H2(g)⇌2NH3(g){{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3(g)}} : Δnreaction=(2)−(3+1)=2−4=−2\Delta {{n}_{reaction}}=(2)-(3+1)=2-4=-2 . Therefore Kp=Kc(RT)−2{{K}_{p}}={{K}_{c}}{{(RT)}^{-2}} . This option is incorrect
D) 2NO(g)+O2(g)⇌2NO2(g)2N{{O}_{(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{2(g)}} : Δnreaction=(2)−(2+1)=2−3=−1\Delta {{n}_{reaction}}=(2)-(2+1)=2-3=-1 . Therefore Kp=Kc(RT)−1{{K}_{p}}={{K}_{c}}{{(RT)}^{-1}} . This option is incorrect.

Note:
The magnitude of Kp{{K}_{p}} and Kc{{K}_{c}} determines the position of equilibrium. The larger the value of Kp{{K}_{p}} or Kc{{K}_{c}} , more the equilibrium will shift towards the right and the reaction will proceed to completion. Small value of Kp{{K}_{p}} or Kc{{K}_{c}} indicates the reaction will shift towards the left, i.e. backward reaction will occur.