At a certain concentration of n2 and o3, the initial rate of reaction is 0.770 m / s. what would the initial rate of the reaction be if the concentration of n2 were halved? be sure your answer has the correct number of significant digits.

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

At a certain concentration of and , the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of were halved? Be sure your answer has the correct number of significant digits.

Answer : The initial rate of the reaction will be 0.192 M/s

Explanation :

Rate law expression for the reaction:

As we are given that:

Initial rate = 0.770 M/s

Expression for rate law for first observation:

....(1)

Expression for rate law for second observation:

....(2)

Dividing 1 by 2, we get:

Therefore, the initial rate of the reaction will be 0.192 M/s

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.