The oxidation of NO by O3 is first order in each of the reactants, and its rate constant is 1.5 ´ 107 M-1·s-1. If the concentrations of NO and O3 are each 5.0 ´ 10-7 M, what is the rate of oxidation of NO in M·s-1?

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To find the rate of oxidation of NO in M·s-1, we need to use the rate law equation and plug in the given concentrations.

The rate law equation for this reaction is:
Rate = k [NO]^1 [O3]^1

Where:
Rate = rate of the reaction in M·s-1
k = rate constant
[NO] = concentration of NO in M
[O3] = concentration of O3 in M

Given:
k = 1.5 × 10^7 M^-1·s^-1
[NO] = 5.0 × 10^-7 M
[O3] = 5.0 × 10^-7 M

Substituting the given values into the rate law equation, we get:
Rate = (1.5 × 10^7 M^-1·s^-1) × (5.0 × 10^-7 M)^1 × (5.0 × 10^-7 M)^1

Simplifying the equation, we get:
Rate = (1.5 × 5.0 × 5.0) × (10^7 × 10^-7 × 10^-7) M·s^-1

Rate = 3.75 × 10^2 × 10^-7 M·s^-1

Finally, we can convert scientific notation to regular notation:
Rate = 3.75 × 10^2 × 10^-7 M·s^-1
Rate = 3.75 × 10^2 × 10^-7 M·s^-1
Rate = 3.75 × 10^-5 M·s^-1

Therefore, the rate of oxidation of NO in M·s^-1 is 3.75 × 10^-5 M·s^-1.