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

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.