Nitrogen dioxide decomposes according to the reaction: 2 NO 2 (g) --> <-- 2 NO (

Question

Nitrogen dioxide decomposes according to the reaction: 2 NO2 (g) --> <-- 2 NO (g) + O2(g) where Kp = 4.48 x 10-13 at a certain temperature.if 0.75 atm of NO2 is added to a container and allowedto come to equilibrium , what are the equilibrium partial pressuresof NO (g) and O2 (g)? please show me step by step how to solve this problem,including the solution. my answer keeps being wrong and i'm notsure what else to do. Thanks in advance :) Nitrogen dioxide decomposes according to the reaction: 2 NO2 (g) --> <-- 2 NO (g) + O2(g) where Kp = 4.48 x 10-13 at a certain temperature.if 0.75 atm of NO2 is added to a container and allowedto come to equilibrium , what are the equilibrium partial pressuresof NO (g) and O2 (g)? please show me step by step how to solve this problem,including the solution. my answer keeps being wrong and i'm notsure what else to do. Thanks in advance :)

Explanation / Answer

                  2NO2 (g) --> <-- 2 NO (g) + O2 (g) Initial               0.75atm               0               0 Change               -2x                    +2x            +x Equilibrium   (0.75-2x)                  2x               x Kp = 4.48 * 10-13 = 2x. x / (0.75 - 2x) Solving, we get x = 4.098 * 10-7 So partial pressure of NO at equilibrium = 2x = 2 (4.098 *10-7 )                                                                      =8.19 * 10-7 atm Partial pressure of O2 = x = 4.098 * 10-7 atm

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