Question

In which of the following equilibrium, ​ $ {{K}_{p}}={{K}_{c}} $ ?
A) $ {{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}} $
B) $ 2N{{H}_{3(g)}}+5{{O}_{2(g)}}\rightleftharpoons 4N{{O}_{(g)}}+6{{H}_{2}}{{O}_{(g)}} $
C) $ {{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3(g)}} $
D) $ 2N{{O}_{(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{2(g)}} $

Answer 1

Hint: The equilibrium constant $ {{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 $ {{K}_{c}} $ can be given as: $ {{K}_{c}}=\dfrac{[product]}{[reac\tan t]} $
 $ {{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 $ {{K}_{p}} $ . It is a unitless number. $ {{K}_{p}}=\dfrac{{{p}_{products}}}{{{p}_{reac\tan t}}} $ .

Complete Step By Step Answer:
To solve this question we will know the relation between $ {{K}_{p}} $ and $ {{K}_{c}} $ . The relation can be given as: $ {{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}} $
Where $ \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 $ \Delta n $ . R is the gas constant and T is the temperature.
The value of $ {{K}_{p}} $ or $ {{K}_{c}} $ is constant if the temperature remains constant. Both will have different values and both the terms to be equal only if, $ \Delta n=0 $
We’ll find the reaction where $ \Delta n=0 $ and that is our correct answer:
A) $ {{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}} $ : Remember we will consider only gaseous products and reactants. $ \Delta {{n}_{reaction}}=2-(1+1)=2-2=0 $ . Therefore, $ {{K}_{p}}={{K}_{c}}{{(RT)}^{0}}\to {{K}_{p}}={{K}_{c}} $ and this is the correct answer
B) $ 2N{{H}_{3(g)}}+5{{O}_{2(g)}}\rightleftharpoons 4N{{O}_{(g)}}+6{{H}_{2}}{{O}_{(g)}} $ : $ \Delta {{n}_{reaction}}=(4+6)-(5+2)=10-7=3 $ . Therefore $ {{K}_{p}}={{K}_{c}}{{(RT)}^{3}}\to {{K}_{p}}={{K}_{c}}{{R}^{3}}{{T}^{3}} $ . This option is incorrect
C) $ {{N}_{2(g)}}+3{{H}_{2(g)}}\rightleftharpoons 2N{{H}_{3(g)}} $ : $ \Delta {{n}_{reaction}}=(2)-(3+1)=2-4=-2 $ . Therefore $ {{K}_{p}}={{K}_{c}}{{(RT)}^{-2}} $ . This option is incorrect
D) $ 2N{{O}_{(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{2(g)}} $ : $ \Delta {{n}_{reaction}}=(2)-(2+1)=2-3=-1 $ . Therefore $ {{K}_{p}}={{K}_{c}}{{(RT)}^{-1}} $ . This option is incorrect.

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