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For the formation of NH_3 from N_2 and H_2 , N_2(g)+3H_2(g)----->2NH_3(g),     K_p...

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roshan-rox | (Level 1) Valedictorian

Posted May 22, 2013 at 6:57 AM via web

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For the formation of NH_3 from N_2 and H_2 ,

N_2(g)+3H_2(g)----->2NH_3(g),     K_p = 4.34 *10^-3 at 300^0 C.

                       <-----

So what is the value of K_p for the reverse reaction ?

 

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 22, 2013 at 7:28 AM (Answer #1)

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We have given chemical equation

`N_2 uarr+3H_2uarr->2NH_3uarr`

We have given

Kp=4.34 x10 ^(-3)

Thus Kp for original equation is 4.34 x 10^(-3) .

We need to calculate  Kpr for the following equation

2NH_3 -> N_2 +3H_2

which is the reverse of the original reaction, Thus

Kpr=1/Kp

=1/(4.34 x 10^(-3))

=2.30 x10^2

 

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