We can represent the reaction using the reaction

Question

We can represent the reaction using the reaction mechanism below:

6I^-(aq)+BrO3^-(aq)+6H(aq)——> 3I2(aq)+Br^-(aq)+3H2O

(L). (Slow)

I2(aq)+2S2O3^2-(aq)——>2I^-(aq)+S4O6^2-(aq). (Fast)

I2(aq)+Starch——>starch-I2(aq) (Bluish-black). (Fast)

consider the reaction to answer these questions.

1. given the reaction mechanism, write the rate law expression for this reaction.
2. suppose that a chemist performed variations on the iodine clock experiment and obtained the experimental data, below. Calculate the value of K for your rate law expression, then write the equation including the value of K.

[I^-] in mol/L [BrO3^-] in mol/L [H^+] in mol/L

0.32 0.22 0.41

0.64 0.22 0.41

0.64 0.44 0.41

0.64 0.44 0.82

Rate of reaction mol/Ls

1.15*10^-3

2.3*10^-3

4.6*10^-3

9.2*10^-3

3- Discuss the factors affecting the rate of a chemical reaction (concentrations of reactants, surface area of reactants, temperature, presence of a catalyst) and how they might affect the rate of this reaction, using collision theory as the basis of your explanation.

i know it is a lot, sorry, but please help me.

Answer 1

Since the slow step is the first elementary step, the rate law will be governed by that step.

rate = k[I^-]⁶[BrO₃^-][H⁺]⁶ HOWEVER, looking at the rate data, it is 1st order in I-, BrO3- and H+ as indicated by the fact that doubling the [ ] of any one of them while holding the other 2 constant results in a doubling of the rate. Therefore, ..

rate =k[I^-][BrO³-]H⁺] and solving for k, we have...

k = rate / [I^-][BrO³-]H⁺] = 1.15x10^-3 M/s / (0.32 M )(0.22 M)(0.41 M)

k = 3.98x10^-2 M^-2s^-1

For #3, look up kinetic theory (collision theory) and you should be able to address this question. (Hint: increase in [reactants] increases rate unless it is zero order; increase surface area increases rate; increase in temperature increases rate if rx is endothermic; presence of a catalyst increases rate because it provides an alternate pathway with a low activation energy.)