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The atmospheric oxidation of NO 2NO (g) + O_(2) (g) hArr 2 NO_(2) (g) was studied with initial pressure of 1 atm of NO and 1 atm of O_2 . At equilibrium , partial pressure of oxygen is 0.52 atm . Calculate K_(p) of the reaction. |
Answer» Solution : As , `1 - x = 0.52` `x = 0.48` = At EQUILIBRIUM , `P_(NO) = 1 - 2x = 1 - 2(0.48) = 0.04` `P_(NO_(2)) = 2x= 2 (0.48) = 0.96` `K_(eq) = (P_(NO_(2))^(2))/(P_(NO)^(2) * P_(O_(2))) = (0.96 XX 0.96)/(0.04 xx 0.04 xx 0.52) = 11.07 xx 10^(2) (ATM)^(-1)` |
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