# Question f80a2

##### 1 Answer
Jan 25, 2017

Given reversible gaseous rection

$\textcolor{red}{A B \left(g\right) r i g h t \le f t h a r p \infty n s A \left(g\right) + B \left(g\right)}$

ICD Table
Where
$I \to \text{Initial state}$

$C \to \text{Change to reach at desired state}$

$D \to \text{Desired state i.e 50% dissociation}$

$\textcolor{b l u e}{\text{ "AB(g)" "rightleftharpoons" "A(g)" "+" } B \left(g\right)}$

$\textcolor{red}{I} \text{ "" "1" mol"" "" "" "0" mol"" "" "" "0" mol}$

$\textcolor{red}{C} \text{ "-alpha" mol"" "" "" "alpha" mol"" "" "alpha" mol}$

$\textcolor{red}{D} \text{ "1-alpha" mol"" "" "" "alpha" mol"" "" "alpha" mol}$

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 $A B \left(g\right)$ 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 $A B \left(g\right) = {\chi}_{\text{AB(g)}} = \frac{1 - \alpha}{1 + \alpha}$

Mole fraction of $A \left(g\right) = {\chi}_{\text{A(g)}} = \frac{\alpha}{1 + \alpha}$

Mole fraction of $B \left(g\right) = {\chi}_{\text{B(g)}} = \frac{\alpha}{1 + \alpha}$

We know

$\textcolor{red}{{p}_{i} = {\chi}_{i} \times P}$

Where

${p}_{i} = \text{partial pressure of ith component}$

${\chi}_{i} = \text{mole fraction of ith component}$

$P = \text{total pressure of the reaction mixture}$

So

Partial pressure of $A B \left(g\right) = {p}_{\text{AB(g)}} = \frac{\left(1 - \alpha\right) P}{1 + \alpha}$

Partial pressure of $A \left(g\right) = {p}_{\text{A(g)}} = \frac{\alpha P}{1 + \alpha}$

Partial pressure of $B \left(g\right) = {p}_{\text{B(g)}} = \frac{\alpha P}{1 + \alpha}$

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

${Q}_{p} = {\left(\frac{\alpha P}{1 + \alpha}\right)}^{2} / \left(\frac{\left(1 - \alpha\right) P}{1 + \alpha}\right) = \frac{{\alpha}^{2} P}{1 - {\alpha}^{2}}$

Now inserting $\alpha = 0.5$ we get

${Q}_{p} = \frac{{\left(0.5\right)}^{2} P}{1 - {\left(0.5\right)}^{2}} = \frac{\frac{P}{4}}{1 - \frac{1}{4}} = \frac{P}{3}$

So $P = 3 \times {Q}_{p}$

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