Question

For the reaction: 2NO2(g)  2N2(g) + O2(g) a) If a plot of ln[N2O] as a function of time is linear, what is the rate law for the reaction? b) How many half-lives will it take for the [N2O] to reach 6.25% of its original concentration? At/A0 = (0.5)n where n = number of half-lives. This equation relates the amount of reactant remaining after t(At) amount initially present (A0).

Answer 1

a) The rate law for the reaction is rate = k[N2O]². b) Use the equation At/A0 = (0.5)^n to calculate the number of half-lives. This approach provides a quantitative understanding of the reaction kinetics.

Step-by-step explanation:

a) Determining the rate law for the reaction involves examining the correlation between the concentration of N2O and time.

If a plot of natural logarithm ln[N2O] against time yields a linear relationship, it indicates a second-order reaction.

In such a case, the rate law can be expressed as rate = k[N2O]², where k represents the rate constant.

b) Calculating the number of half-lives required for [N2O] to reach 6.25% of its initial concentration involves utilizing the equation At/A0 = (0.5)^n, where At is the final concentration and A0 is the initial concentration.

To solve for n, the equation can be rearranged as n = log(At/A0) / log(0.5).

By substituting the relevant values and performing the calculation, the number of half-lives necessary for the concentration of [N2O] to decrease to the specified level can be determined.

This approach provides a quantitative understanding of the reaction kinetics and the time required for the concentration to reach a given fraction of its initial value.

Answer 2

For the reaction 2NO2(g) → 2N2(g) + O2(g), the rate law is second-order with respect to NO2. The rate of formation of NO2 can be calculated using the given concentrations and rate constant.

Step-by-step explanation:

(a) To determine the rate law for the reaction, we can compare the changes in NO2 concentrations with the corresponding reaction rates. By comparing different experiments, we find that doubling the concentration of NO2 quadruples the reaction rate. This indicates that the reaction rate is proportional to [NO2]², leading to a second-order reaction.

(b) The balanced chemical equation shows that 4 mol of NO2 are produced for every 1 mol of O2. Using this information, we can calculate the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M. The rate constant for the reaction is given as 5.8 × 10-6 L² mol-² s-¹.