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?
rate = k(NO)(O3)
rate = 1.5E7(NO)(O3)
To find the rate of oxidation of NO, we can use the rate law equation for a first-order reaction:
rate = k * [NO] * [O3]
Given that the rate constant (k) is 1.5 * 10^7 M^(-1)·s^(-1), and the concentrations of NO and O3 are both 5.0 * 10^(-7) M, we can substitute these values into the rate equation:
rate = (1.5 * 10^7 M^(-1)·s^(-1)) * (5.0 * 10^(-7) M) * (5.0 * 10^(-7) M)
rate = 3.75 * 10^(-6) M·s^(-1)
Therefore, the rate of oxidation of NO is 3.75 * 10^(-6) M·s^(-1).
To calculate the rate of oxidation of NO in M·s-1, we can use the rate equation for a first-order reaction:
rate = k[NO][O3]
Given that the rate constant (k) is 1.5 × 10^7 M-1·s-1 and the concentrations of NO and O3 are both 5.0 × 10^-7 M, we can substitute these values into the rate equation:
rate = (1.5 × 10^7 M-1·s-1)(5.0 × 10^-7 M)(5.0 × 10^-7 M)
Simplifying this expression, we get:
rate = 3.75 × 10^7 M·s-1
Therefore, the rate of oxidation of NO is 3.75 × 10^7 M·s-1.