Print Calculator Perigdic Table Question 5 of 10 Available From Not Se Due Date:

Question

Print Calculator Perigdic Table Question 5 of 10 Available From Not Se Due Date: Points Possibl 10 Grade Category: Extra Map 12/8/2 General Chemistry 4th Edition University Science Books presented by Sapling Leaming At 1700 °C, the equilibrium constant, Kc, for the following reaction is 4.10 x10-4 2NO Policies: Homev What percentage of O2 will react to form NO if 0.551 mol of N2 and 0.551 mol of O2 are added to a 0.777-L container and allowed to come to equilbrium at 1700 °C? can check your a You nswe You can view solutions wh give up on any question. You can keep trying to ans Number until you get it right or giv You lose 5% of the points answer in your question attempt at that answer eTextbook Help With This Topic Web Help & Videos Previous Give Up & View Solution e Check Answer Next Exit Technical Support ar and B Hint

Explanation / Answer

initially:

[N2] = 0.551/0.777 = 0.7091

[O2] = 0.551/0.777 = 0.70913

[NO] = 0

in equilibrium

[N2] = 0.7091 - x

[O2] = 0.7091 - x

[NO] = 0 + 2x

substitute in K

K = [NO]^2 /([N2][O2)

4.10^-4 = (2x)^2 / (0.7091 - x)(0.7091 - x

sqrt(4.10^-4 )= 2x/(0.7091 - x)

0.059488 = 2x/(0.7091 - x)

(0.7091 - x) = 2/0.059488 x

33.62x = (0.7091 - x)

32.62x = 0.7091

x = 0.7091 / 32.62 = 0.02173

[O2] reacted =0.02173

then

% O2 reacted = 0.02173/0.7091*100

% O2 reacted = 3.064 %

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