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Answer 1 · 2017-01-25T17:46:38

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.