Rate of Reaction
For a chemical reaction to be feasible it must take place at a reasonable rate. Chemical kinetics is the study of reaction rates.
Reaction rate is a positive quantity that expresses how the concentrations of a reactant or product change with time.
aA + bB ----- cC + dD
Substances are in gas phase or aqueous solution.
rate = -Δ[A] = - Δ[B] = Δ[C] = Δ[D]
aΔt bΔt cΔt dΔt
The coefficients from the balanced equation cause concentrations to change at different rates.
Example
N2O5 --> 2 NO2 + ½ O2
From the balanced equation the rate that N2O5 reacts is 1/2 the rate that NO2 is produced (or NO2 is produced at twice the rate N2O5 is consumed). Along the same lines, O2 is produced at 1/2 the rate that N2O5 is consumed. Or mathematically:
rateN2O5 = 0.5 rateNO2
rateN2O5 = 2 rateO2
Example
In the formation of ammonia, molecular nitrogen disappears at a rate of 0.10 mole per liter.
N2(g) + 3 H2(g) --- 2 NH3(g)
rate = -Δ[N2] = - Δ[H2] = Δ[NH3] = 0.10 mol
Δt 3Δt 2Δt L . min
By defining rate this way it is independent of what species you focus on.
Measuring Rate
Rate can be measured by absorption of visible light as color changes occur or changes in pressure resulting from changes in the number of moles.
To determine instantaneous rate at a particular concentration you can use the plot of concentration versus time and draw a tangent to the curve at the desired point. Rate will equal the negative slope of tangent.
This tells us that the rate at [N2O5] = 0.080 M is 0.028 mole/L . min.
Reaction Rate and Concentration
Reaction rate is directly related to reactant concentration. Reactions occur as the result of collisions between reactant molecules.
A higher number of molecules produce a greater number of collisions and a faster reaction.
This plot shows that rate is directly proportional to concentration.
rate = k [N2O5] (rate expression, where k is the rate constant)
Order of Reaction (Single Reactant)
A --- products
Rate = k[A]m
m is the power to which A is raised and is called the order of reaction. m = 0 (zero order), m = 1 (first order), m = 2 (second order), etc.
Most reaction orders are integral but you can have fractional orders.
The order cannot be determined from the coefficients. It must be determined experimentally.
One way to do this is to measure the initial rate (rate at t = 0) as a function of the concentration of reactant.
Example
Make two reactant mixtures varying only in [A].
rate 2 = k[A]2m rate 1 = k[A]1m
rate2/ rate 1 = ([A]2/[A]1)m then solve for m
Example
The initial rate of decomposition of acetaldehyde, CH3CHO, at 600o C
CH3CHO(g) ---- CH4(g) + CO2(g)
was measured at a series of concentrations with the following results:
[CH3CHO] 0.20 M 0.30M 0.40 M 0.50M
Rate (mol/ L . s) 0.34 0.76 1.4 2.1
Using these data, determine the reaction order (the value of m) in the equation:
rate = k [CH3CHO]m
Order of Reaction (More Than One Reactant)
aA + bB ----- products
rate = k[A]m x [B]n
m = order with respect to A
n = order with respect to B
The overall order for a reaction is the sum of the exponents m + n.
If m =1 and n =2 then overall the reaction is third order.
When more than one reactant is involved, the order can be determined by holding the initial concentration of one reactant constant while varying that of the other reactant.
Example
[A]1 and [A]2 differ but [B] is constant
rate 1 = k[A]1m x [B]n rate 2 = k[A]2m x [B]n
rate 2/ rate1 = ([A2]/[A1])m then solve for m
Example
Consider the reaction at 55 0C:
(CH3)3CBr(aq) + OH-(aq) -- (CH3)3COH(aq)+ Br-(aq)
A series of experiments is carried out with the following results:
Exp. 1 / Exp. 2 / Exp. 3 / Exp. 4 / Exp. 5[(CH3)3CBr] / 0.50 / 1.0 / 1.5 / 1.0 / 1.0
[OH-] / 0.050 / 0.050 / 0.050 / 0.10 / 0.20
Rate (mol/L.s) / 0.0050 / 0.010 / 0.015 / 0.010 / 0.010
Find the order of the reaction with respect to both (CH3)3CBr and OH-. Write the rate expression for the reaction.
Relating Reactant Concentration and Time
First-Order Reactions
A -- products
rate = k[A]
ln [A]o = kt
[A]
Where [A]o = the original concentration of reactant and [A] is the concentration at time, t. k is the first-order rate constant.
ln a/b = ln a - ln b, so the first-order equation is:
ln [A]0 – ln [A] = kt
ln [A] = ln [A]0 –kt
Compare to y = mx +b
The plot of ln [A] vs. t should be a straight line with a y-intercept of ln [A]0.
Example
Example
For the first-order decomposition of N2O5 at 67 0C, where k = 0.35/min, calculate (a) the concentration after six minutes, starting at 0.200M (b) the time required for the concentration to drop to 0.150 M (c) the time required for half a sample of N2O5 to decompose.
The time required for one half of a reactant to decompose (half-life) via a first-order reaction has a fixed value, independent of concentration.
t1/2 = ln 2/ k = 0.693 / k
For radioactive decay (where an unstable nucleus decomposes) if X is the amount of radioactive isotope present (in mol, g, # of atoms) at time t,
rate = kX
ln Xo / X = kt where Xo is the initial amount
Example
Plutonium-240, produced in nuclear reactors, has a half life of 6.58 x 103 years. Calculate (a) the first-order rate constant for the decay of Plutonium-240 (b) the fraction of a sample that will remain after a thousand years.
Zero and Second-Order Reactions
Zero-Order
A -- products
rate = k[A]0 = k
[A] = [A]0 – kt
If the plot of [A] vs. t is linear, the reaction must be zero-order and k must equal - m.
Half-life = [A]0/2k
Second-Order
A -- products
rate = k[A]2
1/[A] – 1/[A]0 = kt
For a second-order reaction the plot of 1/[A] vs. t should be linear.
Half-life = 1/k[A]0
Example
The following data were obtained for the gas-phase decomposition of hydrogen iodide:
Time (h) = 0 2 4 6
[HI] = 1.00 0.50 0.33 0.25
Is this reaction zero, first, or second-order in HI?
Example
A certain reaction is first-order in A and second –order in B. In the box shown below, which is supposed to have a volume of one liter, a mole of A is presented by “a” and a mole of B by “b”.
(1)
In which of the three boxes shown below is the rate of the reaction the same as that in the box shown above?
(2) (3) (4)
Models For Reaction Rate
Collision Model
Molecules need to collide in order to react. Not every collision produces a reaction.
Molecules must be properly oriented.
There is a minimum energy (activation energy, Ea,) the molecules must possess for a collision to be effective. The larger the Ea the slower the rate of reaction.
Ea is measured in kilojoules.
r = p x Z x f
rate = steric factor (proper orientation) x collision frequency x fraction of collisions where energy of colliding particles > Ea.
Transition –State Model
CO(g) + NO2(g) --- CO2(g) + NO(g)
The activated complex (transition state species) is an intermediate, unstable high-energy species that must be formed before a reaction can occur.
The intermediate state between reactants and products is the transition state.
The transition state model assumes the activated complex is in equilibrium (at low concentrations) with reactants and may go to product (by climbing over the energy barrier) or revert back to reactant.
Reaction Rate And Temperature
The rates of most reactions increase as temperature rises. As a very general rule as T increases by 10 0C, reaction rate doubles.
The explanation for this lies in kinetic theory. Higher T means ability to meet Ea requirements and increases the number of collisions.
The Arrhenius Equation:
k = Ae-Ea/RT(where A is a constant)
ln k = ln A – Ea/RT
A plot of ln k vs. 1/T is a straight line.
Ea = -R x m
The Arrhenius equation can be expressed as a two point equation relating k and T.
ln k1 = ln A – Ea/RT1 (R = 8.31 J/mol . K)
ln k2 = ln A – Ea/RT2
ln k2/k1 = Ea/R [1/T1-1/T2]
Example
For a certain reaction, the rate constant doubles when the temperature increases from 15 0C to 25 0C. Calculate (a) Ea (b) the rate constant at 1000C, taking k at 25 0C to be 1.2 x 10-2 L/mol . s. (c) the percentage by which the rate constant increases going from 25 0C to 35 0 C.
Catalysis
A catalyst is a substance that increases the rate of reaction without being consumed by it.
A catalyst works by changing the reaction path to one with a lower activation energy. The catalyzed path may have 2 or more steps (but Ea is still lower).
A heterogeneous catalyst is one in a different phase from the reaction mixture. It is often a solid that increases the rate of a gas or liquid phase rxn.
Examples
N2O(g) ---Ag--- N2(g) + ½ O2(g)
2 CO(g) + O2(g) ---Pt-- 2 CO2(g)
A homogeneous catalyst is one that is present in the same phase as the reactants. It speeds up the reaction by forming a reactive intermediate that decomposed to give products.
Example
Adding NaI (aq) to cause the catalyzed decomposition of hydrogen peroxide.
Step 1 : H2O2(aq) + I-(aq) ---- H2O(l) + IO-(aq)
Step 2 : H2O2(aq) + IO-(aq) -- H2O(l)+ O2(g) + I-(aq)
2 H2O2(aq) ----- 2 H2O(l) + O2(g)
Enzymes protein molecules of high molar mass that allow reactions that are normally slow to occur readily in living organisms.
Enzymes are hugely effective and can increase the rate constant by a factor of 1012 or more. However, they often only work over a small temperature range.
Example
C12H22O11(aq) + H2O(l) –-maltase-- 2 C6H12O6(aq)
maltose glucose
Reaction Mechanisms
A reaction mechanism is a description of a path, or a sequence of steps, by which a reaction occurs at the molecular level.
The nature of the rate expression and the reaction order depend on the mechanism by which the reaction takes place.
Two reactions can have different mechanisms and different rate expressions, yet have the same final equation.
Elementary Steps are the individual steps that constitute a reaction mechanism.
They may be:
A -- B + C, rate = k[A], unimolecular
A + B -- C + D, rate = k [A] x [B], bimolecular
A + B + C - D + E, rate = k[A] x [B] x [C], termolecular
The rate of an elementary step is equal to k multiplied by the concentrations of each reactant molecule.
Often, one step in a reaction in a mechanism is much slower than any other. The slow step is the rate-determining step.
The rate of the overall reaction is the rate of the slow step.
Example
Step 1: A - B (fast)
Step 2 : B - C (slow)
Step 3 : C - D (fast)
A -- D
The rate at which A is covert to D is approx. the same as the rate of conversion of B to C.
Deducing a Rate Expression From A Proposed Mechanism
(1)Find the slowest step = overall rate of the reaction
(2)Find rate expression for the slowest step
Example
Step 1 : NO2(g) + NO2(g)--- NO3(g) + NO(g) (slow)
Step 2: CO(g) + NO3(g) --- CO2(g) + NO2(g) (fast)
CO(g) + NO2(g) -- CO2(g) + NO(g)
rate = rate step 1 = k[NO2] 2
If the reaction involves a reactive intermediate (a species produced in one step and later consumed in another) its concentration is usually too small to be measured and it must be eliminated from the rate expression.
Example
Step 1: NO(g) + Cl2(g) NOCl2(g) (fast)
Step 2 : NOCl2(g) + NO(g) 2 NOCl(g) (slow)
2 NO(g) + Cl2(g) --- 2 NOCl(g)
You would expect r (step 2 ) = k2[NOCl2] x [NO].
This is incorrect because NOCl2 is a reactive intermediate and conc. cannot be measured accurately.
Step one forward and reverse reaction rates are equal so k1[NO] x [Cl2] = k -1[NOCl2]
[NOCl2] = k1[NO] x [Cl2]
k -1
rate = k2[NOCl2] x [NO] = k2k1[NO]2 x [Cl2]
k -1
Example
The decomposition of ozone, O3, to diatomic oxygen, O2, is believed to occur by a two-step mechanism:
Step 1: O3(g) O2(g) + O(g) (fast)
Step 2: O3(g) + O(g) ----- 2 O2(g) (slow)
2 O3(g) --- 3 O2(g)
Obtain the rate expression corresponding to this mechanism.