The rate law is first-order in N2O5. At 45 °C the rate constant is 6.08 x 10–4 s–1. What is the rate of the reaction when [N2O5] = 0.100 M, and what happens to the rate when the concentration of N2O5 is doubled to 0.200 M?

Explanation:

Rate constant = 6.08 x 10–4 s–1

What is the rate of the reaction when [N2O5] = 0.100 M..?

A first-order reaction depends on the concentration of one reactant, and the rate law is:

r= −dA / dt = k[A]

Rate of reaction = k [ A ]

where [A] = 0.100 M

Rate of reaction = (6.08 x 10–4) x 0.100

Rate of reaction = 6.08 x 10–5 Ms–1

what happens to the rate when the concentration of N2O5 is doubled to 0.200 M?

Rate of reaction = k [ A ]

[A] = 0.200

Rate of reaction = (6.08 x 10–4) x 0.200

Rate of reaction = 1.216 x 10–4 Ms–1

The rate of the reaction increases to 1.216 x 10–4 Ms–1

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

i think that the answer is c