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28 March, 06:41

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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  1. 28 March, 06:42
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    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
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