Question #f80a2

Given reversible gaseous rection

#color(red)(AB(g)rightleftharpoonsA(g)+B(g))#

ICD Table
Where
#I->"Initial state"#

#C->"Change to reach at desired state"#

#D->"Desired state i.e 50% dissociation"#

#color(blue)(" "AB(g)" "rightleftharpoons" "A(g)" "+" "B(g))#

#color(red)(I)" "" "1" mol"" "" "" "0" mol"" "" "" "0" mol"#

#color(red)(C)" "-alpha" mol"" "" "" "alpha" mol"" "" "alpha" mol"#

#color(red)(D)" "1-alpha" mol"" "" "" "alpha" mol"" "" "alpha" mol"#

Where #alpha# is the degree of dissociation of the reaction.The given value of #alpha=50%=0.5#

Now let #P# be the total pressure of the reaction when the desired 50% dissociation of #AB(g)# occurs.

In the desired state of 50% dissociation
the total number of moles in the reaction mixture is #N=1-alpha+alpha+alpha=1+alpha#

Mole fraction of #AB(g)=chi_"AB(g)"=(1-alpha)/(1+alpha)#

Mole fraction of #A(g)=chi_"A(g)"=alpha/(1+alpha)#

Mole fraction of #B(g)=chi_"B(g)"=alpha/(1+alpha)#

We know

#color(red)(p_i=chi_ixxP)#

Where

#p_i="partial pressure of ith component"#

#chi_i="mole fraction of ith component"#

#P="total pressure of the reaction mixture"#

So

Partial pressure of #AB(g)=p_"AB(g)"=((1-alpha)P)/(1+alpha)#

Partial pressure of #A(g)=p_"A(g)"=(alphaP)/(1+alpha)#

Partial pressure of #B(g)=p_"B(g)"=(alphaP)/(1+alpha)#

Now the Reaction Quotient in respect of pressure at the desired state of 50% dissociation.

#Q_p=(p_"A(g)"xxp_B(g))/p_"AB(g)"#

Inserting the values of partial pressure we get

#Q_p=((alphaP)/(1+alpha))^2/(((1-alpha)P)/(1+alpha))=(alpha^2P)/(1-alpha^2)#

Now inserting #alpha=0.5# we get

#Q_p=((0.5)^2P)/(1-(0.5)^2)=(P/4)/(1-1/4)=P/3#

So #P=3xxQ_p#

Hence the total pressure of the reaction mixture will be three times the reaction quotient at the desired state of 50% dissociation.