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For the formation of NH(3) from N(2) and...

For the formation of `NH_(3)` from `N_(2)` and `H_(2)` at 500 K, the concentration of `N_(2), H_(2)` and `NH_(3)`at equilibrium are `1. 5xx10^(-2)` M and `1.2xx10^(-2)M,` respectively. The equilibrium constant for the reverse reaction is

A

`3.56xx10^(2)`

B

`2.81xx10^(-3)`

C

`3.56xx10^(-2)`

D

`2.81xx10^(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given: Conection of
`[N_(2)]= 1. 5xx10^(-2)M`
`[H_(2)]= 3. 0xx10^(-2)M`
`[NH_(3)]= 1. 2xx10^(-2)M`
The reaction for the formation of `NH_(3_)` is
`N_(2) + 3H_(2) hArr 2NH_(3)`
Thus, the reverse reaction will be
`2NH_(3) rArr N_(2)+3H_(2)and`
equilibrium constant `(K_(c))` will be
`K_(c)=([N_(2)].[H_(2)]^(3))/([NH_(3)]^(2))=([1.5xx10^(-2)][3.0xx10^(-2)]^(3))/([1.2xx10^(-2)]^(2))`
`((1.5xx10^(-2))(27.0xx10^(-6)))/(1.44xx10^(-4))=(40.5)/(1.44)xx10^(-4)`
`=28.125xx10^(-4)`
or `K_(c)= 2.81xx10^(-3)`
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