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25 December, 20:03

Consider the following reaction 2 N2O (g) = > 2 N2 (g) + O2 (g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, calculate the concentration of N2O remaining after 2.0 min if k = 3.4 x 10-3 s-1.

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  1. 25 December, 21:29
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    After 2.0 minutes the concentration of N2O is 0.3325 M

    Explanation:

    Step 1: Data given

    rate = k[N2O]

    initial concentration of N2O of 0.50 M

    k = 3.4 * 10^-3/s

    Step 2: The balanced equation

    2N2O (g) → 2 N2 (g) + O2 (g)

    Step 3: Calculate the concentration of N2O after 2.0 minutes

    We use the rate law to derive a time dependent equation.

    -d[N2O]/dt = k[N2O]

    ln[N2O] = - kt + ln[N2O]i

    ⇒ with k = 3.4 * 10^-3 / s

    ⇒ with t = 2.0 minutes = 120s

    ⇒ with [N2O]i = initial conc of N2O = 0.50 M

    ln[N2O] = - (3.4*10^-3/s) * (120s) + ln (0.5)

    ln[N2O] = - 1.101

    e^ (ln[N2O]) = e^ (-1.1011)

    [N2O} = 0.3325 M

    After 2.0 minutes the concentration of N2O is 0.3325 M
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