Question

Question: for a reaction, ${k_p} > {K_c}$ . the forward reaction is favoured by
A) Low pressure
B) High pressure
C) High temperature
D) Low temperature

Answer 1

For some reversible reaction , by the law of mass action.
Rate of forward reaction ${r_f} = {K_f}{\left[ A \right]^a}{\left[ B \right]^b}$
Rate of backward reaction ${r_b} = {K_b}{\left[ N \right]^n}{\left[ M \right]^m}$
Where Kf and Kb be the rate constant of forward and backward reactions respectively.
At equilibrium ${r_f} = {r_b}$
${K_f}{\left[ A \right]^a}{\left[ B \right]^b} = {K_b}{\left[ N \right]^n}{\left[ M \right]^m}$
Or, $\dfrac{{{K_f}}}{{{K_b}}} = {K_{eq}} = \dfrac{{{{\left[ N \right]}^n}{{\left[ M \right]}^m}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$
Therefore Keq is an equilibrium constant which is the ratio between rate constants of forward reaction to backward reaction.

Complete Step by step answer: For the process considering concentration terms only, equilibrium constant
${K_c} = \dfrac{{{{\left[ N \right]}^n}{{\left[ M \right]}^m}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$
${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}$
Where $\Delta n$ = difference in no of moles of gaseous species only
${K_p} = {K_c}{\left( {RT} \right)^1}$, if $\Delta n > 0$
Then ${k_p} > {K_c}$
Number of moles of products is greater than reactants. Therefore, according to Le chatelier’s principle, low pressure favours forward reaction.
Since, the number of moles of products is larger, low pressure favours the reaction in a direction in which there is an increase in volume which is seen in the product's side. Therefore, low pressure favours forward reaction.

So, Option ‘A’ is correct.

Note: If pressure is increased, then process will move in that direction where number of moles of gaseous species are fewer and vice versa
Effect of addition of inert gas
-At constant volume – no effect
-At constant pressure – same as that of effect of volume increased