Question

The standard free energy change of the reaction is__________.

• A. $7.2kJmo{l}^{-1}$
• B. $8.98kJmo{l}^{-1}$
• C. $9.08kJmo{l}^{-1}$
• D. $10.5kJmo{l}^{-1}$

Answer 1

The correct option is A $7.2kJmo{l}^{-1}$

The ideal gas equation $PV=nRT=\frac{W}{M}RT$
$PM=\frac{W}{V}RT$

$PM=\rho RT$

The average molecular weight $M=\frac{\rho RT}{P}$

$M=\frac{1.2gd{m}^{-3}\times 0.08206L\cdot atm/mol\cdot K\times 700K}{100kPa\times \frac{1atm}{101.325kPa}}$

$M=69.84g/mol$

The molecular weights of $COC{l}_2$, $CO$ and $C{l}_2$ are 98.9 g/mol, 28 g/mol and 71 g/mol respectively.

Suppose initially, we start with 1 mole of $COC{l}_2$ and n moles of $COC{l}_2$ dissociate to reach equilibrium.

Then we have $(1-n)$ moles of $COC{l}_2$, n moles of $CO$ and n moles of $C{l}_2$ at equilibrium.

The average molecular weight of mixture $M=\frac{98.9(1-n)+28n+71n}{(1-n)+n+n}$

But $M=69.84$ g/mol

Hence,

$69.84=\frac{98.9(1-n)+28n+71n}{(1-n)+n+n}$

$69.84=\frac{98.9-98.9n+28n+71n}{1+n}$

$69.84(1+n)=98.9$

$69.84+69.84n=98.9$

$69.84n=29.06$

$n=0.42$

The equilibrium number of moles of
$COC{l}_2,CO,C{l}_2$ are $1-0.42=0.58$, 0.42, 0.42 respectilvely.

Total equilibrium pressure $P=100kPa\times \frac{1atm}{101.325kPa}=0.987atm$

Equilibrium partial pressures are,

$P_{COC{l}_2}=\frac{0.58mol}{1.42mol}\times 0.987atm=0.403atm$, $P_{CO}=P_{Cl2}=\frac{0.42mol}{1.42mol}\times 0.987atm=0.291atm$

The equilibrium constant $K_p=\frac{P_{CO}\times P_{Cl2}}{P_{COCl2}}=\frac{0.291\times 0.291}{0.403}=0.210$

$\Delta G=-RTln{K}_p$

= $-8.314\times {10}^{-3}\times 700\times ln\left(0.210\right)$

= +9.08 kJ/mole