Question

In: Chemistry

At 1700 degrees C, the equilibrium constant, Kc, for the following reaction is 4.10 × 10^-4....

At 1700 degrees C, the equilibrium constant, Kc, for the following reaction is 4.10 × 10^-4.

N2 (g) +O2 (g) <---> 2NO (g)

What percentage of O2 will react to form NO if 0.713 mol of N2 and 0.713 mol of O2 are added to a 0.521-L container and allowed to come to equilbrium at 1700 degrees C?

Solutions

Expert Solution

Molarity of O2 = Molarity of N2 = moles/volume in L = 0.713/0.521 = 1.368M

                    N2 (g) +O2 (g) <---> 2NO (g)

Initial          1.368    1.368             0

Final          (1.368-x) (1.368-x)      2x

Taking underoot both sides we get

2.024 * 10^(-2) = 2x/(1.368-x)

assuming 1.368-x is approximately equal to 1.38

2x = 2.024 * 1.38 * 10^(-2)

x = 1.39656 * 10^(-2)

Percentage of O2 reacted = 1.39656 * 10^(-2)/1.368 * 100 = 1.0208%

queen_honey_blossom answered 2 years ago
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