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 A reaction has a forward rate constant of 2.3 × 10^6s–1 and an equilibrium...

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lak-86 | Student, Undergraduate | Salutatorian

Posted June 22, 2013 at 4:33 PM via web

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 A reaction has a forward rate constant of 2.3 × 10^6s–1 and an equilibrium constant of 4.0 × 10^8. What is the rateconstant for the reverse reaction?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 22, 2013 at 4:49 PM (Answer #1)

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You need to use the following equilibrium ratio to evaluate the rate constant for the reverse reaction, such that:

`K = (K_f)/(K_r)`

`K_f` represents the rate constant of forward reaction

`K_r` represents the rate constant of reverse reaction

Since the problem provides the equilibrium constant and the rate costant of forward reaction, you need to replace the given values in equation above, such that:

`4.0 × 10^8 = (2.3 × 10^6)/(K_r) => K_r = (2.3 × 10^6)/(4.0 × 10^8)`

`K_r = 0.575 x 10^(6 - 8)`

`K_r = 0.575 x 10^(-2)`

Hence, evaluating the rate constant of reverse reaction, under the given conditions, yields `K_r = 0.575 x 10^(-2).`

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