[12%] The air oxidation of an organoiridium compound (C9H11)3Ir takes place according to the followi — Kinetics Chemistry Question
Problem Context
[12%] The air oxidation of an organoiridium compound (C9H11)3Ir takes place according to the following equation:
(C9H11)3Ir + 0.5 O2 → (C9H11)3IrO
The reaction was studied with the initial [(C9H11)3Ir] = 1.3 × 10–4 M under two different O2 concentrations (run 1, [O2] = 1.0 × 10–2 M; run 2, [O2] = 2.0 × 10–3 M). The concentration of (C9H11)3Ir was measured as a function of time; below are plotted [(C9H11)3Ir], ln([(C9H11)3Ir]), and 1/[(C9H11)3Ir] vs. time.
Is the order of the reaction in (C9H11)3Ir 0, 1, or 2? Justify your answer. (Even if the data are not exactly consistent with an integer order, pick the closest integer order.)
Model Answer
In each run, [O2] >> [(C9H11)3Ir], so [O2] is effectively constant during each run. So the order in Ir can be determined by seeing whether the [Ir] changes linearly with time (zeroth-order), whether ln([Ir]) changes linearly with time (first-order), or whether 1/[Ir] changes linearly with time (second-order). Clearly it is only the last of these that is true; the reaction must be second-order in (C9H11)3Ir.
Is the order of the reaction in O2 0, 1, or 2? Justify your answer. (Even if the data are not exactly consistent with an integer order, pick the closest integer order.)
Model Answer
Since rate = k[O2]^n[(C9H11)3Ir]^2, we can see how the slope of the second-order plots (which are equal to k[O2]^n) change with [O2]. The slope of run 1 ([O2] = 1.0 × 10^-2 M) is 280 M^-1s^-1, while the slope for run 2 ([O2] = 2.0 × 10^-3 M) is 31 M^-1s^-1. A decrease of [O2] by a factor of 5 results in a 9-fold decrease in rate. This is not particularly close to either n = 1 (should be a 5-fold decrease) or n = 2 (should be a 25-fold decrease), but it is closer to n = 1. (Numerically, n = log(9)/log(5) = 1.37; as directed by the instructions, this should be considered as the closest integer, n = 1.)
Calculate the rate constant for the reaction.
Model Answer
If we use the data from run 1, k[O2] = k[1.0 × 10^-2 M] = 280 M^-1s^-1. So k for this overall third-order reaction is 2.8 × 10^4 M^-2s^-1. (Using run 2 gives k = 1.6 × 10^4 M^-2s^-1.)
The following mechanism has been proposed. Is the mechanism consistent with the observed rate law? Explain your reasoning.
(C9H11)3Ir + O2 ⇌ (C9H11)3Ir(O2) (forward: k1, reverse: k-1)
(C9H11)3Ir(O2) + (C9H11)3Ir → 2 (C9H11)3IrO (rate: k2)
Model Answer
If step 1 is rate-determining, then Rate = k[O2][(C9H11)3Ir], which gives the wrong order in iridium. If step 2 is rate-determining, then Rate = k[(C9H11)3Ir]^2[O2], which is consistent with the experimental data. So this mechanism is consistent with the data if the second step is rate-limiting (i.e., k2[(C9H11)3Ir] << k-1).