Question

Chemical Kinetics Kit for AP Chemistry Student Guide d. Is it likely that this reaction occurs in one step? Explain. 2. A student would like to determine the orders of the reaction for an iodine clock reaction that occurs in three steps 3r(aq)+ H,O:(aq) + 2H'(aq )-,(aq) +2H 2O0) slow I(aq) + 25,0,(aq) 3r(aq) + S40(aq) 21, (aq) + starch-starch-I" complex + r(aq) fast fast Write the rate law for this reaction, using variables in place of the exponents a b. In this reaction, clear solutions are poured together. After a period of time, the solutions suddenly turn blue-black. Given the following chemicals, describe how you would conduct the iodine clock experiment and what data you would collect 0.5 M potassium iodide, KI 1% starch solution 3% hydrogen peroxide 0.10 M hydrochloric acid, HC water (deionized or distilled) 0.01 M sodium thiosulfate, Na S,O3

Answer 1

2a) The reaction mechanism consists of three elementary steps. The first step is the slowest step of the reaction and is hence, the rate-determining step.

The rate of a reaction depends only on the rate of the slow rate-determining step; hence, the rate law for the reaction can be written as

Rate = k[I^-]^x[H₂O₂]^y[H⁺]^z

where x, y and z are variables (exponents) (ans).

Note that the question asks to replace the stoichiometric co-efficients with variables.

Further, it must be noted that x, y and z are the orders of the reaction with respect to the concentrations of the reactants.

2b) In this reaction, I^- reacts with H₂O₂ to generate triiodide, I₃^-. This reaction is slow. There is another reaction which goes on in the iodine clock reaction, that between I₃^- and thiosulfate, S₂O₃^2-. The second reaction is fast. By controlling the quantities of the reactants (I^-, H₂O₂ and S₂O₃^2-), a situation can be obtained where the S₂O₃^2- is used up so that there is excess I₃^-. The excess I₃^- then reacts with starch to give a blue color which indicates the end point of the reaction.

In a test tube labelled as 1, a known volume of 0.5 M KI, a known volume of 3% hydrogen peroxide and a known volume of 0.10 M HCl are added. In a separate test tube labelled as B, a known volume of 0.01 M Na₂S₂O₃ containing 1% starch solution is taken. A stopwatch is kept handy to measure the time. The solution in test tube 2 is poured into test tube 1 and the stopwatch is started. The two solutions are mixed well and the time is noted till the blue color appears. The recorded time gives the rate of the reaction.

In another part of the experiment, the concentrations of the reactants in test tube 1 are varied one by one and the corresponding rates are recorded. The ratio of the rates is then used to determine the orders of the reaction with respect to the reactants.