**Rate Equations from Experiment: I. Initial Rate Data.**

The rate of a chemical reaction slows during the process as the concentration of the reacting species falls. The change in rate is not a constant. However under certain circumstances the rate at the start of a reaction can be almost constant and the concentration changes with time follow almost a straight line so that the rate of the reaction can be calculated as D(conc)/D(time). Obtaining rates this way is termed the "initial rates" method and can provide a very quick way into a rate equation. Unfortunately it cannot be applied universally.

Here are data presented in the class on the reaction of persulphate ion with iodide ion:

S_{2}O_{8}^{2-} + 3 I^{-} 2 SO_{4}^{2-} + I_{3}^{-}

Run | Initial Concentrations (mol.L^{-1}) |
Initial Reaction Rate
(mol.L | |

[I^{-}] |
[S_{2}O_{8}^{2-}] |
||

1 | 0.077 | 0.077 | 3.6 x 10^{-5} |

2 | 0.077 | 0.038 | 1.8 x 10^{-5} |

3 | 0.038 | 0.077 | 1.8 x 10^{-5} |

4 | 0.019 | 0.077 | 9.0 x 10^{-6} |

^{}From this data, derive the values of k, m and n in the following rate equation:

rate = k[I^{-}]^{m}[S_{2}O_{8}^{2-}]^{n}.

In the simplest cases, visual inspection of the data can give the values of the exponents, m and n. Here, halving the concentrations of either species without changing the other halves the rate showing that both m and n are 1 for this reaction. Substitution of one data set into the rate equation can give k:

(from 1) 3.6 x 10^{-5} mol.L^{-1}.s^{-1} = k x 0.077 M x 0.077 M

So k = 6.07 x 10^{-3} L.mol^{-1}.s^{-1} (Note units for second order k)

Rate equation: rate = 6.07 x 10^{-3}L.mol^{-1}.s^{-1} [I^{-}][S_{2}O_{8}^{2-}].

Problem:

Determine the rate equation for the following process. Assume that it has the structure: rate =
k[I^{-}]^{n}[ClO^{-}]^{p}[OH^{-}]^{q}, and find values for each of k, n, p, and q.

Equation:

OH^{-} (catalyst)

I^{-}(aq) + ClO^{-}(aq) IO^{-}(aq) + Cl^{-}(aq)

Run | Initial Concentrations (mol.L^{-1}) |
Initial Reaction
Rate
(mol.L | ||

[I^{-}] |
[Cl_{}O_{}^{-}] |
[OH^{-}] |
||

1 | 0.010 | 0.010 | 0.010 | 6.1 x 10^{-4} |

2 | 0.020 | 0.010 | 0.010 | 12.2 x 10^{-4} |

3 | 0.010 | 0.020 | 0.010 | 12.2 x 10^{-4} |

4 | 0.010 | 0.010 | 0.020 | 3.0 x 10^{-4} |

Answer: Rate = 0.061 s^{-1} [I^{-}] [ClO^{-}] [OH^{-}]^{-1}

General approach:

1 Let the [product] = x at time = t.

rate = d[product]/dt = dx/dt

2. Let starting concentration of a chemical, A, be a M and assume that reaction coefficients are all 1:

A | products | ||

initial concentrations | a | 0 | |

concentrations at time, t | a - x | x |

3. Substitute these values into the rate equation and integrate without limits.

4. Find the constant by using the data: x = 0 at time, t, = 0

5. This gives an equation that relates x and t and which can be graphed.

Equations obtained this way are:

Finding the Rate Law.

The graphical method of finding a rate law involves graphing the LHS of each of the integrated rate equations against t. The correct integrated rate equation for the process will give a straight line graph of slope k, the rate constant.

**Example.**

Find the reaction order and the rate constant from the following data for the reaction: CH_{3}I +
(CH_{3})_{3}N (CH_{3})_{4}^{+} I^{-}.

**Given: **

Initial concentrations of CH_{3}I and (CH_{3})_{3}N are each 0.0200 M.

Time (sec) | 325 | 1295 | 1530 | 1957 |

% reacted | 31.4 | 64.9 | 68.8 | 73.7 |

**Method:** calculate the left hand side requirements and plot vs t.

To calculate x, = the concentration of product at time t, we need to take the initial concentration times the fraction reacted (since the stoichiometry is 1:1:1.

x | t | a - x | x/(a(a - x)) | ln(a/(a - x)) |

0.00628 | 325 | 0.01372 | 22.9 | 0.3769 |

0.01298 | 1295 | 0.00702 | 92.5 | 1.0470 |

0.01376 | 1530 | 0.00624 | 110.3 | 1.4648 |

0.01474 | 1957 | 0.00526 | 140.1 | 1.3356 |

Plotting each of the last two columns against t shows that the reaction follows second order
kinetics with a rate constant, k = 7 x 10^{-2} M^{-1}.s^{-1}.

The two plots for first order and second order are shown below

First Order Plot

Second Order Plot

Half Lives

Definition:

The half life of a process is the time taken to reduce the amount of reactant(s) to half their original value.

The relationship between the half life and the starting reactant concentration can be obtained from the integrated rate equations by substituting in the conditions that x = a/2 at t = half life:

For zero order: t_{½} = a/2k. As the reaction proceeds successive half lives become shorter.

For first order: t_{½} = (ln2)/k. The half life is independent of the amount of material present.

For second order: t_{½} = 1/(ak). As the reaction proceeds, successive half lives become longer.

It may be possible to extract the order of a reaction from the kinetic data by examining what
happens to t_{½} during successive half lives.

Exercises:

1. The following data is from a classic 1933 kinetics paper. Examine the data from the half life perspective and suggest an order for the reaction.

Decomposition of N_{2}O_{5} in CCl_{4} at 45ºC.

Time (secs) | [N_{2}O_{5}] (M) |

184 | 2.08 |

319 | 1.91 |

526 | 1.67 |

867 | 1.36 |

1198 | 1.11 |

1877 | 0.72 |

2315 | 0.55 |

3144 | 0.34 |

2. The dimerization of butadiene was followed in the following kinetics experiment. Examine the data from a half life perspective and suggest an order for the reaction.

[C_{4}H_{6}] (M) |
Time (secs) |

0.01000 | 0 |

0.00625 | 1000 |

0.00476 | 1800 |

0.00370 | 2800 |

0.00313 | 3600 |

0.00270 | 4400 |

0.00241 | 5200 |

0.00208 | 6200 |

**Radioiosotopes.**

Nuclear process follow first-order kinetics and so have a half life which is a constant.

The radioactive isotopes that we encounter today may be classified into three or four catagories.

1. Long lived isotopes that are very slowly decomposing.

Examples are:

^{232}Th, t_{½ }= 1.41 x 10^{10} years

^{235}U, t_{½} = 7.1 x 10^{8} years

^{238}U, t_{½} = 4.51 x 10^{9} years.

2. Short-lived isotopes: products of the decay of these long-lived isotopes.

Examples are:

^{226}Ra, t_{½} = 1.60 x 10^{3} years (from ^{238}U)

^{222}Rn, t_{½} = 3.82 days (from ^{238}U)

^{219}Rn, t_{½} = 4.00 secs (from ^{235}U)

^{220}Rn, t_{½} = 55.3 secs (from ^{232}Th)

3. Short-lived isotopes: products of man-designed nuclear processes.

Examples are:

^{239}Pu, t_{½} = 24,390 years (transuraneum element,
nuclear power product)

^{255}No, t_{½} = 185 s (transuranium element,
laboratory synthesized)

^{90}Sr, t_{½} = 27.7 years (nuclear power product)

^{60}Co, t_{½} = 5.263 years (laboratory synthesized)

4. Short-lived isotopes: products of high-energy cosmic radiation in the earth's upper atmosphere.

Examples are:

^{3}H, t_{½} = 12.26 years

^{14}C, t_{½} = 5730 years

1. Very toxic and radioactive ^{90}Sr is produced in a nuclear power plant. If the material
containing this radioisotope must be stored until the amount has dropped by 99.9%, how long is
this?

t_{½} = 27.7 years. k = (ln2)/t_{½} = 0.02502 years^{-1}

ln(a/(a - x)) = kt. ln(100%/0.1%) = 0.02502 * t

t = 276 years

Note that this time is approximately 10 half lives. This figure of 10 half lives can be used to give a rough indication of how long a radioisotope must be stored.

2. A bottle of wine is claimed to be 80 years old. The wine in the bottle shows a tritium (3H) count of 13 counts per minute. A fresh sample of water shows a tritium count of 100 counts per minute. How old is the wine?

t_{½} = 12.26 years. k = (ln2)/t_{½} = 0.05654 years^{-1}

ln(a/(a - x)) = kt ln(100counts/13counts) = 0.05654 * t

t = 36 years.

So whilst fairly old, the wine is only half the age claimed for it.

Note that the ratio inside the ln function has no units, and the numeric values put in can be from any measurement that is directly proportional to the concentrations involved.

3. A wooden art object claimed to be found in an Egyptian pyramid is offered for sale to a museum. You are called in as an expert to determine the age of the artifact. Non-destructive radiocarbon dating revealed a disintegration rate of 12 disintegrations per minute per gram of carbon. A modern piece of wood gave 15.3 disintegrations per minute per gram of carbon. How old is the object?

t_{½} = 5730 years. k = (ln2)/t_{½} = 0.000121 years^{-1}

ln(a/(a - x)) = kt ln(15.3dis/12dis) = 0.000121 * t

t = 2000 years.

The object is too young to have been placed in the pyramid when it was originally made, however it is quite old and may be of interest to the museum.

4. The half life for the first-order decomposition of nitramide is 123 minutes at 15ºC:

NH_{2}NO_{2}(aq) N_{2}O(g) + H_{2}O(l)

If 0.100 g of nitramide in 1.00L of water is allowed to decompose for exactly 5 hours at 15ºC, what mass of nitramide remains?

t_{½} = 123 minutes. k = (ln2)/t_{½} = 0.005635 minutes^{-1}

ln(a/(a - x)) = kt ln(0.01gL^{-1}/y gL^{-1}) = 0.005635*300

y = 0.0184 gL^{-1}

Answer: mass of nitramide remaining = 0.0184 g

**Approaches to the Mechanism of a Reaction****.**

We have examined several methods for the experimental determination of a rate equation. We have a theory of how a reaction may occur. What we need is the ability to compare the experimental observations with the theoretical prediction. To be an acceptable (though not necessarily correct) mechanism, the experimental and predicted rate equations must be the same.

__Obtaining a rate equation from a proposed step wise mechanism__.

Note that for each step only those reactants which collide and react are included. Thus for a step changes in the concentration of all species involved will change the number of collisions and hence the rate of the reaction. Thus:

**for a step in a step-wise mechanism, the rate equation can be written down directly.**

The nucleophilic substitution of iodomethane by hydroxide ion is considered to be a one-step process:

step 1: CH_{3}I + OH^{-} CH_{3}OH + I^{-}

The rate law for this step is: rate = k [CH_{3}I] [OH^{-}]

This is the theoretical rate equation for this one step mechanism.

But what is found experimentally? rate = k [CH_{3}I] [OH^{-}].

Thus the experimental and the theoretical rate equations are the same and the one-step mechanism is a possible one.

This mechanism is symbolized S_{N}2:

substitution, nucleophilic, bimolecular.

Measurement of the kinetics of the reaction of 2-chloro-2-methylpropane with iodide ion:

(CH_{3})_{3}C-Cl + I^{-} (CH_{3})_{3}C-I + Cl^{-}

gave a different rate equation than that for iodomethane and hydroxide ion:

rate = k [(CH_{3})_{3}-Cl] experimental.

This rate equation is not compatible with a one-step mechanism since the iodide ion concentration does not affect the reaction rate.

The simplest approach to this is to assume that the iodide ion is not involved in the rate-determining step of a multistep mechanism.

One suggested mechanism is:

step 1 (CH_{3})_{3}C-Cl (CH_{3})_{3}C^{+} + Cl^{-} slow

step 2 (CH_{3})_{3}C^{+} + I^{-} (CH_{3})_{3}C-I fast

Here the first step is rate determining and the theoretical rate law from that step is: rate = k
[(CH_{3})_{3}C-Cl]

which compares exactly with the experimental making this mechanism a possibility.

This mechanism is symbolized: S_{N}1,

substitution, nucleophilic, unimolecular.

Mechanisms where the rate-determining step is not the first step.

The rate equation for the reaction:

2 NO(g) + O_{2}(g) 2 NO_{2}(g)

is found to be: rate = k [NO]^{2}[O_{2}].

This is the rate equation that you would obtain theoretically if the reaction had only one step. But this would involve a termolecular collision, which, although not unknown, is rare.

An alternative mechanism to this one-step idea is:

step 1 2 NO(g) N_{2}O_{2}(g) fast

step 2 N_{2}O_{2}(g) + O_{2}(g) 2 NO(g) slow

The theoretical rate equation for this two step mechanism is

rate = k [N_{2}O_{2}] [O_{2}].

which cannot be compared with the experimental rate equation since N_{2}O_{2} is a postulated
intermediate not actually "seen" in the reaction, so its concentration cannot be measured
experimentally.

An approximation that can be made to resolve this difficulty is to assume that the first step (and
all fast steps) are fast in both directions and that an equilibrium exists. This enables the writing
of an equilibrium expression to evaluate the [N_{2}O_{2}]:

from which [N_{2}O_{2}] = K_{c} [NO]^{2}

Substituting into the theoretical rate equation gives:

rate = k K_{c} [NO]^{2} [O_{2}]

which now agrees with the experimental rate equation: the two step mechanism is acceptable.

Summary.

To check whether a suggested step-wise mechanism is feasible the theoretical rate equation must be identical to the experimental one.

The experimental rate equation is obtained using experimental techniques and analysis of data by

initial rates

integrated rate equations

half lives

The theoretical rate equation is obtained from the rate determining step in the proposed mechanism.

If the rds is the first step then the rate equation can be obtained directly from that step.

If the rds is not the first step then the rate equation for the rds is written and any unknown concentrations are obtained by writing equilibrium expressions for all fast steps preceding the rds.

If the theoretical rate equation differs from the experimental rate equation then the proposed step-wise mechanism is wrong.

When first considering the collision model for chemical processes, the rate of the reaction was dependent on a number of factors, only one of which was the concentrations of the species involved.

Two other factors mentioned were:

orientation of the collision

energy requirements

In the rate equations that we have been writing,

rate = k [A][B] for example,

these other two factors are not apparent. They must be within the rate constant, which seems likely since for a given process these factors will remain the same during the process.

Investigation of these other two factors must involve an investigation of how k varies.

In fact k is found experimentally to vary with temperature.

(There is one major exception to this, the k value for radioisotope decomposition is independent of temperature.)

Experimentally, the relationship between k and T was determined by Arrhenius 100 years ago as:

or (taking logarithms of both sides):

ln k = lnA - (E_{a} / RT)

This is known as the Arrhenius equation.

**The Arrhenius Equation**

This is an **experimental** relationship.

The connection with the theoretical interpretation of mechanisms by step wise collisions is that:

A (termed the pre-exponential factor) is a factor which depends on the number of collisions which are correctly orientated for reaction to occur. It is an entropy term.

Ea (termed the activation energy) is interpreted as the minimum energy needed by the process for it to occur. If a collision takes place between reacting species with less than this minimum, then reaction cannot occur.

**Determination of the Activation Energy for a Reaction**.

The experimental Arrhenius relationship in its logarithmic form is;

ln k = lnA - E_{a} / RT.

This equation is of the type y = mx + c, so that a graphical evaluation can be made by plotting ln
k against 1/T with the slope of the line being -E_{a} / RT.

Evaluation of much data has shown that E_{a} is approximately constant with temperature. A simple
approach to obtaining a value for E_{a} from this observation is to use just two points, giving two
equations with two unknowns. Usually the pre-exponential factor, A, is not calculated.

At temperature 1: ln k_{1} = ln A - E_{a} / RT_{1}

At temperature 2: ln k_{2} = ln A - E_{a} / RT_{2}

Knowledge of T_{1}, T_{2}, k_{1}, and k_{2} allows the calculation of E_{a}.

Note: as with any equation with unknowns, given all but one, that one unknown can be found.
The unknown does not have to be E_{a}:

given E_{a}, T_{1}, T_{2}, and k_{1}, find k_{2}.

given E_{a}, T_{1}, k_{1}, and k_{2} , find T_{2}.

Problem:

The rate of chirping of tree crickets and the flashing of fireflies both roughly double for a 10ºC rise in temperature.

Calculate E_{a} for these physiological processes.

At temperature 1: ln k_{1} = ln A - E_{a} / RT_{1}

At temperature 2: ln k_{2} = ln A - E_{a} / RT_{2}

Subtracting:

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

(This equation is a very useful one,)

From the data k_{2} = 2*k_{1}.

Make assumption that T_{1} = 15ºC and T_{2} = 25ºC.

ln (k_{1}/ 2k_{1}) = (E_{a}/R)*(1/298 - 1/288)

ln (½) = (E_{a}/R)*(-1.165 x 10-4)

-0.6931 = (E_{a}/R)*(-1.165 x 10-4)

(E_{a}/R) = 5950 K^{}

Substitute R = 8.314 J/K.mol :

E_{a} = 49 500 J/mol = 49.5 kJ/mol

**Collision Theory in Summary.**

- A chemical reaction can take place:
- when a collision between the reacting molecules occurs
- with the right orientation
- with at least the minimum energy, E
_{a}.

NOTE THE FOLLOWING:

The increase in the rate of a reaction with temperature is NOT attributable to the increased speed of the molecules giving more collisions.

A quick calculation tells you that this is so:

temperature rise: 288 K to 298 K (10ºC)

a factor of 1.03 (3%)

rate rise: doubles

a factor of 2 (100%)

The increase in the rate of a reaction is almost entirely due to an increase in the number of
collisions that occur with the right amount of energy, E_{a}.

Note also that the rate increase also depends on the size of E_{a}. If E_{a} is extremely large, then
temperature changes will have little or no effect on the rate of the reaction. Such seems to be the
case for radioisotope process. The activation energy needed for a nuclear process is extremely
high, nuclear process are effectively temperature independent.

**Catalysis.**

**A catalyst is a substance which speeds up a reaction without itself being consumed**.

From the collision theory, the only way this can happen is if the catalyst becomes involved in the
process and changes the course of the reaction in such a way as to give a** different step wise
mechanism with a lower activation energy**.