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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?

Text Solution

Verified by Experts

The reaction is : `N_(2) (g) + 3H_(2) (g) hArr 2 NH_(3) (g)` ,
We are given that at equilibrium , `p_(N_2) = 0.80 " atmosphere ", p_(H_2) = 0.40 " atmosphere "`
` p_(N_2) + p_(H_2) + p _(HN _(3)) = 2. 80 " atmosphere " :. p_(NH_(3)) = 2.80- (0.80 + 0.40) = 1.60 " atmosphere ".`
Applying the law of chemical equilibrium , we get (taking pressures with respect with respect to standard state pressure of 1 atm)
`K_(p) =(P_(NH_(3))^(2))/(P_(N_2) xx P_(H_(2))^(3)) = (1.60)^(2)/(0.80 xx(0.40)^(3))=50.0`
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