General Chemistry 4th Edition . University Science Books presented by Sapling Le

Question

General Chemistry 4th Edition . University Science Books presented by Sapling Leaming Pc GR De Pc At 1700 C, the equilibrium constant, Kc, for the following reaction is 4.10 x10-4 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? Number 3.064 There is additional feedback available! View this feedback by clicking on the bottom divider bar. Click on the divider bar again to hide the additional feedback Clase Previous Give Up & View Solution Try Again Next Exit

Explanation / Answer

The reaction is, N2(g) + O2(g) <-------> 2NO(g)

Kc = [NO]2/[N2][O2] = 4.10 x 10-4

Initial: [N2] = 0.551 mol /0.777 L = 0.709 M

[O2] = 0.551 mol /0.777 L = 0.709 M

Let x = change in [N2]
So, at equilibrium we will have,

[N2] = 0.709-x ; O2 = 0.709-x

[NO] = x moles N2 react x (2 moles NO/1mole N2) = 2x

Now,

(2x)2/(0.709-x)(0.709-x) = 4.10 x 10-4

Taking the square root of both sides we will get  

=> 2x/(0.709-x) = 0.0202

=> 2x = 0.0202*(0.709–x)

=> 2x = 0.0143 – 0.0202x

=> 2.0202x = 0.0143

=> x = 0.00708

Percentage O2 that reacts =( 0.00708/0.709) x 100 % = 0.998 %

Or using moles we can get,

Moles O2 that react = 0.00708 mol/L x (0.777 L) = 0.00550 moles

Percentage O2 that reacts:(0.00550/0.551) x 100 % = 0.998 %

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