Answered

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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

Sagot :

sebassandin

Answer:

6.3 × 10–^4

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:

[tex]r=k[Xe]^2[F_2][/tex]

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

[tex]r=(0.025\frac{1}{M^2*s} )(0.25M)^2(0.4M)\\\\r=6.3x10^{-4}M/s[/tex]

Best regards!

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