It is time to work through the final two big ideas in topic 6: reaction mechanisms and the effects of catalysts. As we approach this topic, it is increasingly important that we understand reaction pathways and how energy ties in to a specific elementary step.
- Predict a rate law for a reaction having a multistep mechanism given the individual steps in the mechanism. (Section 14.6)
- Explain the principles underlying catalysis. (Section 14.7)
To become familiar with the topics presented in this mission, view the slides below and take note of the key ideas. These are from sections 14.6-14.7 of your text.
Now work through the practice problems, and post your work to OneNote.
Work through these mastery problems and post your work to OneNote. The key is available on OneNote also.
14.66 Consider the following energy profile:
(a) How many elementary reactions are in the reaction mechanism?
(b) How many intermediates are formed in the reaction?
(c) Which step is rate limiting?
(d) For the overall reaction, is ΔE positive, negative, or zero?
14.68 The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:
H2O2(aq) + I−(aq)→H2O(l)+ IO−(aq)(slow)
IO−(aq) + H2O2(aq)→H2O(l)+O2(g)+ I−(aq)(fast)
(a) Write the chemical equation for the overall process.
(b) Identify the intermediate, if any, in the mechanism.
(c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.
14.70 You have studied the gas-phase oxidation of HBr by O2
4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g)
You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism:
HBr(g) + O2(g) → HOOBr(g)
HOOBr(g) + HBr(g) → 2 HOBr(g)
HOBr(g) + HBr(g) → H2O(g) + Br2(g)
(a) Confirm that the elementary reactions add to give the overall reaction.
(b) Based on the experimentally determined rate law, which step is rate determining?
(c) What are the intermediates in this mechanism?
(d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?
14.111 The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride:
Reaction 1: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
This reaction is very slow in the absence of light. However, Cl2(g) can absorb light to form Cl atoms:
Reaction 2: Cl2(g) + hv → 2 Cl(g)
Once the Cl atoms are generated, they can catalyze the reaction of CH4 and Cl2, according to the following proposed mechanism:
Reaction 3: CH4(g) + Cl(g) → CH3(g) + HCl(g)
Reaction 4: CH3(g) + Cl2(g) → CH3Cl(g) + Cl(g)
The enthalpy changes and activation energies for these two reactions are tabulated as follows:
(a) By using the bond enthalpy for Cl2 (242 kJ/mol), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found?
(b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4.
(c) By using bond enthalpies, estimate where the reactants, CH4(g) + Cl2(g), should be placed on your diagram in part (b). Use this result to estimate the value of Ea for the reaction CH4(g) + Cl2(g) → CH3(g) + HCl(g) + Cl(g).
(d) The species Cl(g) and CH3(g) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of CH3, and verify that it is a radical.
(e) The sequence of reactions 3 and 4 comprises a radical chain mechanism. Why do you think this is called a “chain reaction”? Propose a reaction that will terminate the chain reaction.
That's it for topic 6. Aren't you relieved???