Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will have reacted when the rate falls to one-half of its initial value?

Explanation:

given reaction

                        2O₃        ⇄      3O₂

 initial                1                      1

    final           1-x                   1+ 1.5x

rate of reaction = k  [ O₃ ]² / [ O₂ ]

initial rate = k . 1² / 1 = k

final rate = k (1-x )² / (1+ 1.5x) = k/2

(1-x )² / (1+ 1.5x) = 1/2

2x² - 5.5 x + 1 = 0

x = 0.196

So fraction of ozone reacted = .196

since electrons are negative the answer is neagtaive

b) nuclear ive learned that in a pedtrodic table