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Yesterday, 21:15

The average rate of disappearance of ozone in the reaction 2o3 (g) → 3o2 (g) is found to be 7.25*10-3 atm over a certain interval of time. what is the rate of appearance of o2 during this interval?

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  1. Yesterday, 22:10
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    Answer;

    1.0875 x 10-2 atm

    Explanation;

    2O3 (g) → 3O2 (g)

    rate = - (1/2) ∆[O3]/∆t = + (1/3) ∆[O2) / ∆t

    The average rate of disappearance of ozone ... is found to

    be 7.25 * 10-3 atm over a certain interval of time.

    This means (ignoring time)

    ∆[O3]/∆t = - 7.25 * 10^-3 atm

    (it is disappearing, thus the negative sign)

    rate = - (1/2) ∆[O3]/∆t

    rate = - (1/2) * (-7.25 * 10^-3 atm)

    = 3.625 * 10^-3 atm

    Now use the other part of the expression:

    rate = + (1/3) ∆[O2) ∆t

    3.625 * 10-3 atm = + (1/3) ∆[O2) / t

    ∆[O2) / ∆t = (3) * (3.625 * 10^-3 atm)

    = 1.0875 x 10-2 atm over the same time interval
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