Question

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

Answer 1

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.