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`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction isA. `4`B. `27`C. `4//27`D. `1//27` |
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Answer» Correct Answer - A `{:(,NH_(4)COOH_(2)(s)hArr,2NH_(3)(g),+CO_(2)(g)),("Initial moles",a,0,0),("Equilibrium moles",a-x,2x,x):}` Partial pressure `= ("Moles fraction") x ("Total pressure")` `:. P_(NH_(3))= (2x)/(3x)P_(t)= (2)/(3)P_(t)` `P_(CO_(2))=(x)/(3x)P_(t)= (1)/(3)P_(t)` where `P_(t)` is the total pressure at equilibrium. (Note that we do not consider moles of `NH_(4)COONH_(2)` as it is a solid.) According to the law of chemical equilibrium, we have `K_(P)=P_(NH_(3))^(2)P_(CO_(2))` `= ((2)/(3)P_(t))^(2)((1)/(3)P_(t))` Since `P_(t)= 3` atm, we have `K_(P)= ((2)/(3).3)^(2)((1)/(3).3)` `=2^(2)` `=4` |
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