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For a hypothetical chemical equilibrium A=>2B C it was found that initially only A and B are present with mole fraction B equal to 1/3. At equilibrium Initial molefraction of A become 1/2 then Kc is?
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For a hypothetical chemical equilibrium A=>2B C it was found that init...
Calculation of Kc for Chemical Equilibrium A=>2B+C


Given Information


  • Chemical equilibrium: A => 2B + C

  • Initial mole fraction of B: 1/3

  • Final mole fraction of A: 1/2



Calculating Equilibrium Concentrations

Let us assume that at equilibrium, the moles of A, B and C are a, b, and c respectively. Then we can write the expression for Kc as:


Kc = ([B]^2[C])/[A]


At the start of the reaction, we have:


  • [A] = a/V (where V is the total volume of the system)

  • [B] = b/V

  • [C] = c/V



At equilibrium, we have:


  • [A] = (a - x)/V, where x is the amount of A that has reacted

  • [B] = (b + 2x)/V, where x is the amount of A that has reacted

  • [C] = (c + x)/V, where x is the amount of A that has reacted



Calculating the Value of Kc

Substituting the equilibrium concentrations into the expression for Kc, we get:


Kc = ((b+2x)^2(c+x))/((a-x)V)


Since the initial mole fraction of B is 1/3, we can write:


b/a = 1/3


Solving for b and substituting into the expression for Kc, we get:


Kc = ((1/3 + 2x/a)^2(c+x))/((1/2 - x/a)V)


Since the initial mole fraction of A becomes 1/2 at equilibrium, we can write:


x/a = (1/2 - 1/2)/(1 + 2 + 1) = 1/12


Substituting this value into the expression for Kc, we get:


Kc = ((1/3 + 2(1/12))^2(c+1/12))/((1/2 - 1/12)V) = 49c/V


Therefore, the value of Kc for the chemical equilibrium A => 2B + C is 49c/V.
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For a hypothetical chemical equilibrium A=>2B C it was found that init...
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For a hypothetical chemical equilibrium A=>2B C it was found that initially only A and B are present with mole fraction B equal to 1/3. At equilibrium Initial molefraction of A become 1/2 then Kc is?
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