InterviewSolution
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InterviewSolution
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For the reaction , ` N_(2) (g) + 3 H _(2) (g) hArr 2 NH_(3) (g),` the partial pressures 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 atmospheres. What is `K_(p)` for the above reaction ? |
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Answer» 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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