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7 December, 09:26

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

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  1. 7 December, 09:32
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    Missing question: What is the rate constant for the reaction?

    [RS2] (mol L-1) Rate (mol / (L·s))

    0.150 0.0394

    0.250 0.109

    0.350 0.214

    0.500 0.438

    Chemical reaction: 3RS₂ → 3R + 6S.

    Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:

    rate = k·[RS₂]².

    k = 0,438 : (0,500) ².

    k = 1,75 L/mol·s.
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