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The partial pressures of `N_(2)O_(4) "and" NO_(2) "at" 40^(@)C` for the following equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)` are `0.1 "atm and" 0.3` atm respectively. Find `K_(P)` for the reaction. |
Answer» Correct Answer - `0.9"atm"` `P_(N_2O)_(4)=0.1 "atm" P_(NO_(2))=0.3` atm. reaction is `N_(2)O_(4)(g)hArr2NO_(2)(g)` `K_(P)=((P_NO_(2)))/((P_(N_(2)_O_(4))))=((P_(NO_(2)))^(2))/((P_(N_(2)_O_2)))=(0.3)^(2)/((0.1))=0.9` atm |
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