A2.00 mol sample of nitrogen dioxide was placed in an 80.0l vessel. at 200 degrees c, the nigrogen dioxide was 6.0% decomposed according to the equation: 2no2 in equilibrium 2no + o2 calculate the value of kc for this reaction at 200 degrees c.

The value of Kc at 200 degrees C is 3.06×10⁻⁶

Explanation:

2NO2 → 2NO + O2

Initially the concentration of NO2 is 0.025 M

C=n/v thus C = 2/80 = 0.025 M

from that concentration, only 6.0% decomposed to the two products

6% of 0.025 M = 0.025 x 6/100 = 1.5 x 10⁻³ M

Equilibrium concentration

NO2 = 0.025 - 1.5 x 10⁻³ = 0.0235 M (removing the amount decomposed)

NO = 1.5 x 10⁻³ M (ratio of NO and NO2 is 1, thus the amount decomposed equal to the amount produced)

O2 = 1/2 x 1.5 x 10⁻³ = 7.5 x 10⁻⁴ M (1/2 is the ratio of O2 and NO2 through stoichiometry)

Kc = [NO]²[O2]/[NO2]² = (1.5 x 10⁻³)²(7.5 x 10⁻⁴)/(0.0235)² = 3.06×10⁻⁶

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

c. orbital shape i hope this is correct