309,829 views
20 votes
20 votes
Study the reaction and read the statement.

Xe+3F2→XeF6
The rate constant for this reaction is 0.025, and the reaction is second order in Xe and first order in F2.

What is the rate of the reaction if [Xe] is 0.25 M, and [F2] is 0.4 M?

4.0 × 10–^4
2.5 × 10–^3
1.5 × 10–^3
6.3 × 10–^4

User Oluwakemi
by
2.9k points

1 Answer

7 votes
7 votes

Answer:

6.3 × 10–^4

Step-by-step explanation:

Hello there!

In this case, since the rate laws are written in terms of the rate constant and the concentration of the species contributing to the rate of reaction, as this one is second order in Xe and first order in F2, the rate law would be:


r=k[Xe]^2[F_2]

Thus, by plugging in the rate constant and concentrations, we obtain:


r=(0.025(1)/(M^2*s) )(0.25M)^2(0.4M)\\\\r=6.3x10^(-4)M/s

Best regards!

User Miw
by
3.2k points