Question

Question #f80a2

Answer 1

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.