If the concentration of N2 is halved, then the concentration term [N2]^3 would be reduced to (1/2)^3 = 1/8 of its original value.
Since the rate of the reaction is directly proportional to [N2]^3, if [N2] is reduced to 1/8, the rate will also be reduced to 1/8 of its original value.
Therefore, the initial rate of the reaction would be 0.780 M/s / 8 = 0.098 M/s.
Rounded to three significant digits, the initial rate of the reaction would be 0.098 M/s.
The rate of a certain reaction is given by the following rate law:
rate=k[N2]^3[O3]
Use this information to answer the question below.
At a certain concentration of N2 and O3, the initial rate of reaction is 0.780 M / s. What would the initial rate of the reaction be if the concentration of N2 were halved? Round your answer to 3 significant digits.
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