# For the equilibrium $2SO_2 + O_2 \rightleftharpoons 2SO_3$ we start with 2 moles of $SO_2$ and 1 mole of $O_2$ at 3 atm. When equilibrium is attained, pressure changes to 2.5 atm. Hence, $K_p$ is:

(a) 3 $atm^{-1}$

(b) 2.5 $atm^{-1}$

(c) 2 $atm^{-1}$

(d) 0.5 $atm^{-1}$

Toolbox:
• PV=nRT

Answer: 2 $atm^{-1}$

Initiallly, P= 3 atm
Total no. of moles = 2 + 1 = 3

PV = nRT
3 $\times$ V = 3RT
V = RT

$2SO_2 + O_2 \rightleftharpoons 2SO_3$
__2_____1__________at t=0
(2-2x)__(1-x)___(2x)__at t=t'

Total no. of moles at equilibrium = 3-x

PV = nRT
$\Rightarrow$ 2.5 $\times$ V = (3-x)RT
$\Rightarrow$ 2.5 $\times$ RT = (3-x)RT
$\Rightarrow$ x = 0.5

No. of moles at equilibrium
$SO_2$ = 2(1 - x) = 2(1 - 0.5) = 1
$O_2$ = 1 - x = 1 - 0.5 = 0.5
$SO_3$ = 2x = 2 $\times$ 0.5 = 1

$\therefore$ Total no. of moles at equilibrium = 2.5

We know $\rho_A$ = $x_A$P
$\rho_{SO_2} = \frac{1}{2.5} \times$ 2.5 = 1 atm
$\rho_{O_2} = \frac{0.5}{2.5} \times$ 2.5 = 0.5 atm
$\rho_{SO_3} = \frac {1}{2.5} \times$ 2.5 = 1 atm
$K_\rho = \frac{(\rho_{SO_3})^2}{(\rho_{O_2}) (\rho_{SO_3})^2} = \frac{(1)^2}{(0.5)(1)^2} = 2 atm^{-1}$

edited Nov 23, 2013