Rate Law from Initial Rates (A) I-(aq) + OCI-(aq) rightarrow 10-(aq) + Cl-(aq) T

Question

Rate Law from Initial Rates (A) I-(aq) + OCI-(aq) rightarrow 10-(aq) + Cl-(aq) The above reaction was studied (at a particular temperature) and the following data were obtained: Which of the following expressions for the rate law are completely consistent with the above experimental data. Calculate the rate constant (k).

Explanation / Answer

rate law is that when I- is doubled, the rate isdoubled. => rate is proportional to I- when I- and OCl- are both halved, the rate is reduced by 4 => rate is also propotional to OCl- rate = [I-][OCl-]*k rate = -d[I-]/dt = -d[OCl-]/dt = d[IO-]dt = d[Cl-]/dt bystoichiometry negative for d[I-]/dt because it is consumed/ So the first,second, and fourth are true statements -====== EDIT: sorry did not realize that there is a part 2) k = rate/{[I-][OCl-]} = 2.39e-2 / [0.1 *0.06] = 3.98L*mol-1*s

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