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For OCl- + I- giving OI- + Cl-, trials give: 0.0040 M OCl- with 0.0020 M I- reacts at 0.00184; 0.0020 M with 0.0040 M at 0.00092; 0.0020 M with 0.0020 M at 0.00046 mol/L/s. What is the overall order?

ASecond order overall
BThird order overall
CFirst order overall
DFourth order overall
Answer & Solution
Correct answer: B. Third order overall
1. Compare the trials where only iodide changes: 0.0020 M OCl- with iodide doubling from 0.0020 to 0.0040 M. 2. The rate doubles from 0.00046 to 0.00092 mol per litre per second, so the order in I- is 1. 3. Now compare trial one against trial two: OCl- doubles while I- halves. 4. The rate doubles from 0.00092 to 0.00184; halving I- alone would have halved it, so doubling OCl- must have supplied a factor of four. 5. A factor of four from doubling means the order in OCl- is 2. 6. The overall order is 2 + 1 = 3, third order. 7. Answering second order counts only the OCl- dependence and drops the iodide term. 8. Fourth order would need the OCl- comparison to yield a factor of eight, which the data do not show. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.3 Rate Laws_
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