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

1 answer:

3 0

Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
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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Answer:

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3 years ago

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We have the final answer as

<h3>12.82 moles</h3>

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