1/4a. Define the overall order of a reaction, and the order with respect to one reagent.

The overall order of reaction is the sum of the powers to which the concentrations are raised in the rate equation.

The order with respect to one reagent is the power that its concentration is raised to in the rate equation.

1/4b. What does it mean when the order of reaction with respect to a particular reagent is zero?

That concentration of the particular reagent does not affect the rate of the reaction.

1/4c. The rate equation cannot be written down from the balanced chemical equation. Why not, do you think?

The balanced equation is a summary of what overall happens in a reaction. The rate equation reflects the overall rate of the reaction which is the rate only of the slowest step.

1/4d. The rate equation for a particular process is: rate = k[A]

What is the overall order of reaction, and the order with respect to A, to B, and to C?

Overall order = 4; order wrt A = 2; order wrt B = 1 order wrt C = 1

1/5a. What is entailed in the "Initial Rates" method of determining a rate equation?

Data to be gathered are the initial concentrations of the various reactants together with the initial rate of the reaction.

1/5b. Why do we not use this method all the time?

The method depends on the graph of concentration vs time to be a straight line over the first little while of the reaction, which it is not. If it is close to a straight line, this method works fairly well. Otherwise it doesn't work well.

1/5c. If the initial rate of a reaction increases by a factor of eight when a particular reagent concentration is doubled, what is the order of the reaction with respect to this particular reagent.

order is 3 {(2)^{3} = 8}

1/6a. What is meant by the term "Integrated Rate Equation"?

This is the equation which relates actual concentrations during a reaction to the time.

1/6b The first-order integrated rate equation can be quoted as:

ln(a/(a-x)) = kt.

What are the meanings of all symbols used in this equation?

a is the initial concentration of the reactant.

x is the concentration of product at time t

(a-x) is the concentration of reactant at time t.

1/6c. If a first order reaction is 75% complete in 100 minutes, how long will it be before it is 99% complete?

ln {a/(a-x)} = kt

ln {100/25} = k*100 min

k = 0.0139 min^{-1}.

ln {100/1} = 0.0139*t

t = 331 min

1/6d. How are integrated rate equations used to determine the rate equation for a reaction?

Integrated rate equations can be written in the form: LHS = kt. If a graph of the LHS against t is plotted, the correct rate law will show up as a straight line.

1/6e. If the function x/(a(a-x)) plotted against t gives a straight line, what is the rate equation, and how would you evaluate k?

This is the second order integrated rate equation, so the rate equation is second order:

rate = k [A]^{2}.

k can be obtained as the slope of the straight line.

1/7a. What is meant by the term "Half-Life" of a reaction?

The half life of a reaction is the amount of time for half of the initial material present to react.

1/7b. How does the half-life of zero-, first-, and second-order reaction depend on the initial amount of reactant?

order | dependence on initial amount of reactant |

0 | direct: decreasing initial amount increases the half life. |

1 | no dependence. |

2 | indirect: decreasing the initial amount increases the half life. |

1/7c. A reaction goes 50% in 10 minutes and 75% in 20 minutes. What is the order of the reaction?

First order (the half life is constant).

1/7d. Another reaction goes 50% in 5 hours and 75% in 8 hours. What is the order of this reaction?

Zero order (the half life decreases as amount decreases).

1/8. Nuclear decompositions follow first order kinetics. A certain radioactive product of a nuclear reaction must be stored until it is 99.9% decomposed. If its half life is 5 000 years, for how long must it be stored?

ln {a/(a-x)} = kt

ln 2 = kt_{½}

k = 0.0001386 years^{-1}

ln {100/0.1} = 0.0001386 * t

t = 49 800 years

How would you suggest that it be stored so that it remains undisturbed for the requisite length of time?

This is the current unsolved problem of nuclear waste. What do we do with it?

1/9. Carbon-14 dating depends on the observation that living matter equilibrates with the carbon-14 in the atmosphere, dead matter does not. If living matter has a count of 15 decompositions per second, and a count of 3 decompositions per second is the minimum your apparatus can detect, what is the maximum age of a specimen that you can determine?

ln {a/(a-x)} = kt

ln 2 = kt_{½}

k = (ln 2)/5730 years

k = 0.000121 years^{-1}

ln {15/3} = 0.000121 * t

t = 13 300 years.

1/12. When the temperature of a reaction is increased, the rate of the reaction increases. How does the kinetic theory of gases explain why this happens? (Note that the observation that the number of collisions increases as the average kinetic energy increases, although true, is by no means a large enough factor to explain the rate increase.)

The rate increases with the number of effect collisions. Raising the temperature significantly increases the number of molecules
which have a certain amount of energy (such as the E_{a} required for a reaction to occur).

(You should illustrate this answer with a diagram.)

1/14. Why are forward reaction rates and backward reaction rates different from each other?

Reactants and products usually do not have the same energy, yet they are connected via a reaction which has the same transition
state. Consequently, the energy needed to attain the transition state, E_{a}, will differ for reactants and products. The rate of the
reaction depends on the activation energy, and so the two rates will differ.

(You should illustrate this answer with a diagram.)

1/20. What is meant by the term "Catalyst"? Does a catalyst take part in a reaction?

A catalyst increases the rate of a reaction without itself being consumed. The initial and final states of the system are not changed, but the path of the reaction is changed to give a lower activation energy by the involvement of the catalyst.

The catalyst must take part in the reaction, though it is regenerated before the end of the process.

1/14.94

The time will be the same. The half life of the process is constant, independent of the initial concentrations.

2. A common and serious mistake is to assume that the rate equation for a reaction can be derived from the balanced chemical equation for the reaction using the coefficient of the chemical equation as exponents in the rate expression. Why cannot rate equations be derived this way?

See answer to 1/4c above. (No additional marks were assigned to this question.)

3. A. The following data were collected for the gas phase reaction between nitric oxide and bromine at 273ºC:

2 NO(g) + Br_{2}(g) ---> 2 NOBr(g)

Experiment # | [NO] mol/L | [Br_{2}] mol/L |
Initial rate(mol/L.s) |

1 | 0.10 | 0.10 | 0.012 |

2 | 0.10 | 0.20 | 0.024 |

3 | 0.20 | 0.10 | 0.048 |

4 | 0.10 | 0.10 | 0.108 |

3Aa) Determine the rate law.

Doubling [Br_{2}] doubles the rate; rate varies as [Br_{2}]^{1}

Doubling [NO] quadruples the rate; rate varies as [NO]^{2}.

The rate law is: rate = k [NO]^{2} [Br_{2}]

3Ab) Calculate the value of the rate constant.

From run 1: 0.012 mol/L.s = k (0.10 M)^{2} (0.10 M)

So: k = 12 L^{2}/mol^{2}.s

3Ac) How is the rate of appearance of NOBr related to the rate of disappearance of Br

By stoichiometry, 2 NOBr appear for each 1 Br_{2} that disappears so rate of appearance of NOBr is twice the rate of disappearance of
Br_{2}

3. B. The mechanism of this reaction might be:

NO(g) + Br_{2}(g) ---> NOBr_{2}(g)

NOBr_{2}(g) + NO(g) ---> 2 NOBr(g)

3Ba) Describe the molecularity for each step in the mechanism.

Step 1 is bimolecular

Step 2 is bimolecular

3Bb) Write the rate law for each elementary step in the mechanism.

Step 1: rate = k [NO] [Br_{2}]

Step 2: rate = k [NOBr_{2}] [NO]

3Bc) Identify the intermediate (if any).

An intermediate is a compound which appears during the course of a reaction but is not present at the end. NOBr_{2} is an
intermediate.

3Bd) If the rate law obtained in part A pertains and this mechanism is correct, what can be said about the relative rates of steps 1 and 2. What assumption is also being made concerning step 1?

If the first step is slow, the rate equation is that of the first step:

rate = k [NO] [Br_{2}].

As this rate law is not that observed the first step cannot be slow.

If the second step is slow, the rate equation is that of the second step:

rate = k [NOBr_{2}] [NO]

This rate law cannot be compared with the experimental rate law as it stands.

Use the equilibrium approximation: K = [NOBr_{2}] / ([NO][Br_{2}])

Giving [NOBr_{2}] = K [NO] [Br_{2}]

Substituting for [NOBr_{2}]: rate = k K [NO] [Br_{2}] [NO] = k' [NO]^{2} [Br_{2}].

This rate law is the same as that found experimentally and so this mechanism with the second step slow is acceptable.

4. A sample of polluted water was oxidized with O

Time (days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 10 | 20 |

Organic matter oxidized (%) | 21 | 37 | 50 | 60 | 68 | 75 | 80 | 90 | 99 |

a. Is the oxidation process first or second order?

50% of the reaction occurs in 3 days.

75% of the reaction occurs in 6 days.

The half life is constant at 3 days, therefore the reaction is first order.

b. What is the rate constant for this reaction?

For a first order reaction: ln 2 = kt_{½}

Thus k = 0.231 days^{-1}.

5. For the second order reaction

2 NO(g) ---> N_{2}(g) + O_{2}(g)

the rate constant is 0.143 L/mol.s at 1400 K and 0.659 L/mol.s at 1500 K. What is the activation energy for the reaction?

Using the Arrhenius equation: ln k = ln A - E_{a} / RT

ln {k_{2}/k_{1}} = {E_{a}/R}{1/T_{1} - 1/T_{2}}

ln {0.659/0.143} = {E_{a}/R} {0.000714 - 0.000667} = {E_{a}/R}*0.0000476

E_{a} = 267 kJ.

6. The rate of the reaction

CH_{3}COOC_{2}H_{5}(aq) + OH^{-}(aq) ---> CH_{3}COO^{-}(aq) + C_{2}H_{5}OH(aq)

was measured at several temperatures and the following data collected:

Temperature (ºC) | 15 | 25 | 35 | 45 |

k (M^{-1}.s^{-1}) | 0.0521 | 0.101 | 0.184 | 0.332 |

Using these data, construct a graph of log k vs 1/T and determine the value of E_{a}.

Slope = -E_{a}/R

-5654= -E_{a}/8.314

E_{a} = 47 000 J

E_{a} = 47.0 kJ

7a) The nuclide

ln {a/(a-x)} = kt

ln{2} = kt_{½}

k = 0.0420 hours^{-1}

ln {0.0100/y} = 0.0420 * 24.0

0.0100/y = 2.740

y = 0.00365

Answer: After 1.00 day there will be 0.00365 g of the isotope left.

7b) The rate of decay of

ln {a/(a-x)} = kt

ln {100/10} = k*369 min

k = 0.00624 min^{-1}

ln 2 = kt_{½}

t_{½} = 111 min

7c) Strontium-90 has a half life of 27.6 years. For how long would a sample of earth contaminated initially with 1.0 g of strontium-90 have to be stored so that the residual amount of strontium-90 was less than 1.0 mg ? Starting at our current date and looking backwards in time for this long, what was the state of science and technology then?

ln {a/(a-x)} = kt

ln 2 = kt_{½}

k = 0.0251 years^{-1}

ln {1.0/0.001} = 0.0251 * t

t = 275 years

1723! Even Mozart hadn't been born!

7d. The carbon from the heart wood of a giant sequoia tree gives 10.8

ln {a/(a-x)} = kt

ln 2 = kt_{½}

t_{½} = 5730 years.

k = 0.000121 years^{-1}

ln {15.3/10.8} = 0.000121 * t

t = 2879 years

The tree is about 2880 years old.

The approximation is that the equilibrium [^{14}C] was the same 2880 years ago as it is now.