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Calculate `K_(p)` for the following reaction if partial pressures of `NH_(3), N_(2) and H_(2)` are 0.4,0.3,0.2, atm, respectively. `2NH_(3) hArr N_(2)+3H_(2)` |
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Answer» `K_(p)=(p^(3)H_(2).pH_(2))/(P^(2)NH_(2))` `K_(p)=((0.2)^(3)*(0.3)("atm")^(4))/((0.4)^(2)("atm")^(2))` `K_(p)=0.015 "atm"^(2)` |
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