A reaction has a standard free-energy change of ?4.21 kJ/mol at 25 ?C. What are

Question

A reaction has a standard free-energy change of ?4.21 kJ/mol at 25 ?C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? My problem is that I do not understand how the quadratic equation answer is x2-0.88x+0.12. It says "substituting the value of K and doing some algebra (including dividing by K) yeilds the following quadratic equation" Please explain how to get the formula.

I need to know because I am trying to solve the same problem with a different K value.

Explanation / Answer

after setting up the ICE table

we got

[A]eq = 0.3-x

[B}eq = 0.4 -x

[C]eq = x

now

K= [C]/ [A] [B]

so

K = x / ( 0.3-x) ( 0.4-x)

x=K ( 0.3-x) (0.4-x)

x/K = ( 0.3-x) ( 0.4 -x)

x/K = 0.12 - 0.3x - 0.7x + x2

x/K = 0.12 - 0.7x + x2

x2 -0.7x + 0.12 = x/K

x2 -0.7 x - x/K + 0.12 = 0

x2 - ( 0.7 + (1/K) ) x + 0.12 =0

given

K = 5.556

so

x2 - ( 0.7 + ( 1/5.556) ) x + 0.12 = 0

x2 -- ( 0.7 + 0.18) x + 0.12 = 0

x2 -0.88x + 0.12 = 0

so the equation becomes

x2 - 0.88x + 0.12 = 0

now

solving the quadratic equation

we get

x= 0.17

now

[A]eq = 0.3-x = 0.3-0.17 = 0.13 M

[B]eq = 0.4-x = 0.4 -0.17 = 0.23 M

[C]eq = x = 0.17 M

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