Answer:
At [tex]P_{total} = 0.240\ atm[/tex]; 95.0% of the molecules will dissociate at this temperature
Explanation:
The chemical reaction of this dissociation is:
[tex]O_2 \leftrightarrow 2O_g[/tex]
The ICE table is as follows:
[tex]O_2 \ \ \ \ \ \ \ \leftrightarrow \ \ \ \ \ \ \ \ 2O_g[/tex]
Initial 100 0
Change -83 +166
Equilibrium 17 166
The mole fractions of each constituent is now calculated as:
[tex]Mole \ fraction \ of \ O(X_o) = \frac{166}{183}[/tex] = 0.9071
[tex]Mole \ fraction \ of \ O_2(X_o_2_}}) = \frac{17}{183}[/tex] = 0.0929
Given that the total pressure [tex]P_{total}[/tex] = 1.000 atm ; the partial pressure of each gas is calculated by using Raoult's Law.
[tex]Partial \ Pressure \ of \ O (P_o) = X_oP_{total}\\\\ = 0.9071 \ atm[/tex]
[tex]Partial \ Pressure \ of \ O_2 (P_o_2) = X_oP_{total}\\\\ = 0.0929 \ atm[/tex]
Now; we proceed to determine the equilibrium constant [tex]K_c[/tex]; which is illustrated as:
[tex]K_c = \frac{Po^2}{Po_2} \\ \\ K_c = \frac{(0.9072)^2}{0.0929} \\ \\ =8.86 \ atm[/tex]
Let assume that the partial pressure of [tex]O_2[/tex] be x ;&
the change in pressure of [tex]O_2[/tex] be y ; then
we can write that the following as the changes in concentration of species :
[tex]O_2 \ \ \ \ \ \ \ \leftrightarrow \ \ \ \ \ \ \ \ 2O_g[/tex]
Initial x 0
Change -y +2 y
Equilibrium x - y 2 y
From above; we can rewrite our equilibrium constant as:
[tex]K_c = \frac{(2y)^2}{x-y} \\ \\ 8.86 = \frac{(2y)^2}{x-y} ----- equation (1)[/tex]
From the question; we are told that provided that 95% of the molecules dissociate at this temperature. Therefore, we have:
[tex]\frac{y}{x}*100 = 95[/tex]% -------- equation (2)
Solving and equating equation 1 and 2 ;
x = 0.123 atm
y = 0.117 atm
Thus, the pressure required can be calculated as :
[tex]P_{total} = (x-y) +2y \\ \\ P_{total} = (0.123- 0.117)+ 2(0.123) \\ \\ \\ P_{total} = 0.240 \ atm[/tex]
