Consider the following equilibrium in a closed container:
N2O4 (g) ⇋ 2NO2 (g)
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant Kp and degree of dissociation α?
Consider the following equilibrium in a closed container:
N2O4 (g) ⇋ 2NO2 (g)
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant Kp and degree of dissociation α?
neither 
nor 
changes
both 
and 
changes
changes, but 
does not change
does not change, but 
changes
The correct option is D does not change, but changes
The Equilibrium is established at a particular temperature T, for which the equilibrium constant is Kp
Let us assume there were initially n moles of N2O4 (g) and there were no products.
N2O4(g)NO2(g)t=0n0t=eqn(1−α)2α
At Equilibrium, total number of moles = n(1+α)
Partial pressure of NO2 (g) = mole fraction of 2NO2(g) × Total Pressure P
Partial pressure of N2O4 (g) = mole fraction of N2O4 (g) × Total Pressure P
Using ideal gas equation, PV=n(1+α)RT. Can we use boyle's law here??
Boyle's law cannot be used here because there are three variables P, V and α. As a special case, boyle's law can be used under one exceptional condition. When Δn = 0 { Δn = sum of stoichiometric coefficients of gaseous products - sum of stoichiometric coefficients of gaseous reactants}
In this case, Δn=1>0; For a homogeneous equilibrium which has the Δn = 0, the number of moles in the system is constant at all times, even when equilibrium is not reached. In such cases, Kp=Kc
For a given temperature, the value of Kp does not change. In the above ideal gas equation, P, V and α are all variables while n R and T are all constants. Can we say that the value nRT is constant? Definitely!
The product on the right hand side is constant since Kp is a constant ( T =constant). So if V is halved would be doubled (increased). As α (which is less than 1) increases, (1−α) decreased. So its reciprocal increases.
Consequently, increases as α increases. So definitely α changes. If V decreases, α also decreases. Similarly if V increases, α also follows a similar pattern. This could be generally extended to systems which follow Δn> 0. What can you comment about systems which have Δn>0?
does not change, but changes
The Equilibrium is established at a particular temperature T, for which the equilibrium constant is Kp
Let us assume there were initially n moles of N2O4 (g) and there were no products.
N2O4(g)NO2(g)t=0n0t=eqn(1−α)2α
At Equilibrium, total number of moles = n(1+α)
Partial pressure of NO2 (g) = mole fraction of 2NO2(g) × Total Pressure P
Partial pressure of N2O4 (g) = mole fraction of N2O4 (g) × Total Pressure P
Using ideal gas equation, PV=n(1+α)RT. Can we use boyle's law here??
Boyle's law cannot be used here because there are three variables P, V and α. As a special case, boyle's law can be used under one exceptional condition. When Δn = 0 { Δn = sum of stoichiometric coefficients of gaseous products - sum of stoichiometric coefficients of gaseous reactants}
In this case, Δn=1>0; For a homogeneous equilibrium which has the Δn = 0, the number of moles in the system is constant at all times, even when equilibrium is not reached. In such cases, Kp=Kc
For a given temperature, the value of Kp does not change. In the above ideal gas equation, P, V and α are all variables while n R and T are all constants. Can we say that the value nRT is constant? Definitely!
The product on the right hand side is constant since Kp is a constant ( T =constant). So if V is halved would be doubled (increased). As α (which is less than 1) increases, (1−α) decreased. So its reciprocal increases.
Consequently, increases as α increases. So definitely α changes. If V decreases, α also decreases. Similarly if V increases, α also follows a similar pattern. This could be generally extended to systems which follow Δn> 0. What can you comment about systems which have Δn>0?
