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.

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

### 1 Answer | Add Yours

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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