Chemistry, asked by TJ26, 1 year ago

Answer the question in the picture( 19 A)

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Answered by kvnmurty
2
 K_c / K_p = 1/(RT) for this reaction.

General relation between K_p and K_c :

  n = Total number of moles of the products - total number of moles of reactants
           K_p / K_c = (RT)^n

====================  DERIVATION

Suppose that we had initially 2X moles of SO3. At equilibrium we have X-2+2+1 = X+1 moles. Let the volume of the mixture be V liters. T = 700K.  Assuming ideal gas behavior,

      P V = n R T        => Total pressure P = (X+1) R T/V

  2 SO3 (g) <==> 2 SO2 (g) + O2 (g),   T = 700 K,  K_p = 1.8 * 10^-5 bar

    X-2                        2              1   moles

K_c = ?  per liter..         K_p =  [SO2]^2 [O2] / [SO3]^2


Partial pressure of SO3 = (X-2)/(X+1) * P = (X-2) RT/V

Partial pressure of SO2 = 2/(X+1)*P = 2 RT/V

Partial pressure of O2 = RT/V

K_p = 4 R^2 T^2/V^2 * RT / V * V^2 /[(X-2)^2*R^2 T^2]

        = 4 RT /[ V (X-2)^2 ]     pressure units

 

Concentration equilibrium constant is defined in terms of molarity ?

Concentration of SO3 :  (X-2)/V   Molar

Concentration of SO2:  2/V   moles/litre

Concentration of O2:    1/V  moles/litre

K_c = (2/V)^2 * (1/V) * V^2/(X-2)^2

        = 4 / [ V * (X-2)^2 ]     volume^-1 units

 

K_c / K_p =  1 / (R T)

K_c = K_p /(R T) = 1.8 *10^-5 /(8.314 * 700) = 3.093 * 10^-9 /m^3

       = 3.093 * 10^-6 /Litre


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