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23 November, 05:39

The reaction described by the equation O 3 (g) + NO (g) ⟶ O 2 (g) + NO 2 (g) has, at 310 K, the rate law rate of reaction = k [ O 3 ] [ NO ] k = 3.0 * 10 6 M - 1 ⋅ s - 1 Given that [ O 3 ] = 7.0 * 10 - 4 M and [ NO ] = 5.0 * 10 - 5 M at t = 0, calculate the rate of the reaction at t = 0.

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  1. 23 November, 09:30
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    Equation of the reaction:

    O3 (g) + NO (g) ⟶ O2 (g) + NO2 (g)

    From the Rate Law:

    R = k[O3][NO]

    The order of NO is 1 and that of O3 is also 1, therefore order of reaction = 1 + 1

    = 2

    Given:

    Concentration of O3, [O3] = 7.0*10^-4M

    Concentration of NO, [NO] = 5.0*10^-5M

    Rate constant, k = 3.0 * 10^6M^-1s^-1

    R = 3.0*10^6 * 7.0*10^-4 * 5.0*10^-5

    = 0.105 M/s
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