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For the reaction, N(2)(g)+3H(2)(g) hAr...

For the reaction,
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
the partial pressure of `N_(2)` and `H_(2)` are `0.80` and `0.40` atmosphere, respectively, at equilibrium. The total pressure of the system is `2.80` atm. What is `K_(p)` for the above reaction?

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The correct Answer is:
A, B
The reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
We are given that at equilibrium
`p_(N_(2))=0.80 "atm" p_(H_(2))=0.40 "atm"`
`p_(N_(2))+p_(H_(2))+p_(NH_(3))=2.80 "atm"`
`:. P_(NH_(3))=2.80-(0.80+0.40)=1.60 "atm"`
Applying the law of chemical equilibrium,
`K_(p)=(p_(NH_(3))^(2))/(P_(N_(2))xxp_(H_(2))^(3))=((1.60)^(2))/(0.80xx(0.40)^(3))=50.0 "atm"^(-2)`
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