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?

Question

Chemistry

Griffin Norton

26 August, 09:31

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?

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Lilia · Answer 1

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