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29 August, 21:42

The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M?

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  1. 29 August, 22:02
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    C NO2 = 0.5644 M

    Explanation:

    The second order decomposition of NO2 (A):

    - rA = K (CA) ² = - δCA/δt

    ⇒ Kδt = - δCA/CA²

    ⇒ K∫δt = - ∫δCA/CA²

    ⇒ K*t = 1/CA - 1/CAo

    ∴ K = 0.255/M. s

    ∴ t = 4.00 s

    ∴ CAo = 1.33 M

    ⇒ 1/CA = K*t + 1/CAo

    ⇒ 1/CA = ((0.255/M. s) (4.00 s)) + (1/1.33 M)

    ⇒ 1/CA = 1.02/M + 0.752/M

    ⇒ 1/CA = 1.772/M

    ⇒ CA = 0.5644 M
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