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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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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 $atm^{-1}$

Can you answer this question?
 
 

1 Answer

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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
$K_\rho = \frac{(\rho_{SO_3})^2}{(\rho_{O_2}) (\rho_{SO_3})^2} = \frac{(1)^2}{(0.5)(1)^2} = 2 atm^{-1}$

 

answered Nov 15, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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