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The equilibrium constant for the reaction `N2O4(g)harr 2NO2(g)`  is `6.10* 10^(-3)` ...

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lak-86 | Student, Undergraduate | Salutatorian

Posted June 24, 2013 at 7:10 PM via web

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The equilibrium constant for the reaction

`N2O4(g)harr 2NO2(g)`  is `6.10* 10^(-3)`  at 25˚C.

Calculate the value of K for this reaction:

`NO2(g)harr 1/2N2O4(g)`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 24, 2013 at 7:21 PM (Answer #1)

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`N_2O_4 harr 2NO_2`

Equilibrium constant for the above is as follows.

`K = [NO_2]^2/[[N_2O_4]]`

`6.1xx10^(-3) = [NO_2]^2/[[N_2O_4]] ` -----(1)

 

Let us consider the new reaction.

`NO_2 harr 1/2N_2O_4`

 

`K = [N_2O_4]^(1/2)/[[NO_2]]`

`K^2 = [[N_2O_4]]/[NO_2]^2` -----(2)

 

` (1)xx(2)`

 

`6.1xx10^(-3)xxk^2 =[NO_2]^2/[[N_2O_4]]xx[[N_2O_4]]/[NO_2]^2`

`K^2 = 1/(6.1xx10^(-3))`

`K = sqrt (1/(6.1xx10^(-3)))`

`K = 12.8`

 

So the answer is K = 12.8

 

 

 

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