Tuesday 2 June 2015

For the reaction A(g) ⇔ 2 B(g), K = 0.010. What is the equilibrium constant for B(g) ⇔ 1/2 A(g)?

The equilibrium constant K for a chemical reaction is equal to the product of the concentrations of reactants, raised to the power of their coefficients, divided by the product of the concentrations of products raised to the power of their coefficients. Here's what this looks like for the reactions in your question:


`A(g) -> 2 B(g)`     `K = [B]^2/([A])`


Reversting the equation gives:


`2 B(g) -> A(g)`     `K = ([A])/[B]^2`


`B(g) ->...

The equilibrium constant K for a chemical reaction is equal to the product of the concentrations of reactants, raised to the power of their coefficients, divided by the product of the concentrations of products raised to the power of their coefficients. Here's what this looks like for the reactions in your question:


`A(g) -> 2 B(g)`     `K = [B]^2/([A])`


Reversting the equation gives:


`2 B(g) -> A(g)`     `K = ([A])/[B]^2`


`B(g) -> 1/2 A(g)`   `K = [A]^(1/2)/([B])`


These equilibrium constant expressions show the following relationships for equilibrium constants in general:


1. If a reaction is written in reverse, its K is the reciprocal of the K for the forward reaction.


2. If the coefficients of a chemical equation are multiplied by a factor, the K is equal to the original K raised to the power of that factor. In this case the coefficients are multiplied by 1/2.


The K for the second reaction in the question is the reciprocal of the square root of the K for the first reaction, because the equation is both reversed and multiplied by 1/2:


 `K_2 = 1/(K_1)^(1/2) = 1/sqrt(0.010) = 10`


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