what is the rate law for NO+O3=NO2+O2 in second order reaction if [NO] Doubled
To determine the rate law for a reaction, we need to examine the effect of changing the concentration of reactants on the reaction rate. In this case, we are given that the reaction is second order with respect to NO.
For a second-order reaction, the rate law equation follows the form:
Rate = k * [A]^n * [B]^m
Where:
- Rate is the reaction rate
- k is the rate constant
- [A] and [B] are the concentrations of reactants A and B
- n and m are the reaction orders for A and B, respectively
Based on the given information that the reaction is second order with respect to NO, the reaction rate can be expressed as:
Rate = k * [NO]^2 * [O3]
If [NO] is doubled, we can plug in the new concentration into the rate law equation and observe the effect.
Let's assume the initial rate of the reaction is R1 when [NO] is at its original concentration. When [NO] is doubled, the new rate (R2) can be determined using the rate law equation:
R2 = k * (2[NO])^2 * [O3]
= 4 * k * [NO]^2 * [O3]
Comparing R2 to R1, we can observe that the new rate (R2) becomes 4 times the original rate (R1). Thus, doubling the concentration of NO leads to a four-fold increase in the reaction rate.
Therefore, the rate law for the reaction NO + O3 → NO2 + O2 in terms of the concentration of NO is second order.