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`N_(2)O_(4)` is dissociated to `33%` and `50%` at total pressure `P_(1)` and `P_(2)atm` respectively. The ratio of `P_(1)//P_(2)` is:A. `7//4`B. `7//3`C. `8//3`D. `8//5` |
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Answer» `underset(1-alpha)underset(2)(N_(2)O_(4))hArrunderset(2alpha)underset(0)(2NO_(2))` `K_(p)=((n_(NO_(2)))^(2))/((n_(N_(2)O_(4))))xx[(P)/(sumn)]^(1)` For `33%` dissociation: `K_(p)=((2xx33)^(2))/(67)xx[(P_(1))/(133)]^(1)` For `50%` dissociation: `K_(p)=((2xx50)^(2))/(50)xx[(P_(2))/(150)]^(1)` `:. (P_(1))/(P_(2))=(133xx67xx(2xx50)^(2))/((2xx33)^(2)xx150xx50)=2.72` |
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