Chemistry: Chemical Kinetics and Equilibrium

Question 1

Chemical Kinetics

Answer 1

The study of the rates (or speed) of reactions.

Question 2

Reaction Rate

Answer 2

The change of concentration of reactant or finished product with respect to time.

Question 3

Mechanism

Answer 3

The mechanism of a chemical reaction is the actual series of steps through which it occurs. Knowing the accepted mechanism of a reaction often helps to explain the reaction’s rate, position of equilibrium, and thermodynamic characteristics.

Consider this reaction: A2 + 2B –> 2AB

Suppose this equation actually takes place in two steps.

Step 1: A2 + B –> A2B (slow)
Step 2: A2B + B –> 2AB (fast)

Note that these two steps add up to the overall (net) reaction. A2B is the intermediate.

Question 4

Intermediate

Answer 4

Molecule that does not appear in the overall reaction because it’s neither a reactant or a product.

Reaction intermediates are often difficult to detect.

Question 5

Rate-Determining Step

Answer 5

The slowest step in a proposed mechanism. The overall reaction cannot proceed faster than this step.

Question 6

Rate

Answer 6

Consider a reaction 2A + B –> C, where one mole of C is produced from every two moles of A and one mole of B. The rate of this reaction may be described in terms of either the disappearance of reactants over time or the appearance of products over time.

Rate = decrease in concentration of reactants/time = increase in concentration of products/time

aA + bB –> cC + dD

Rate = (-1/a)(deltaA/deltat) = (-1/b)(deltaB/deltat) = (1/c)(deltaC/deltat) = (1/d)(deltaD/deltat)

Rate is expressed in units of moles per liter per second (mol/L x S) or molarity per second (molarity/s),

Question 7

Rate Law

Answer 7

For nearly all forward, irreversible reactions, the rate is proportional to the product of the concentrations of the reactants, each raised to some power. For the general reaction

aA + bB –> cC + dD

the rate is proportional to [A]^x x [B]^y, that is:

rate = k x [A]^x x [B]^y

This is the rate law expression where k is the rate constant. The exponents x and y are called the orders of reaction.

Question 8

Rate Constant

Answer 8

k in the rate law expression. It is defined as a constant of proportionality between the chemical reaction rate and the concentration of the reactants. Multiplying the units of k by the concentration factors raised to the appropriate powers gives the rate in units of concentration/time.

Question 9

Orders of Reaction

Answer 9

The exponents x and y in the rate law expression. X is the order with respect to A, and y is the order with respect to B. These exponents may be integers, fractions, or zero and must be determined experimentally.

Important to note that the exponents or the rate law are not necessarily equal to the stochiometric coefficients in the overall reaction equation. The exponents are equal to the stochiometric coefficients of the rate-determining step. If one of the reactants or products in this step is an intermediate not included in the overall reaction, then calculating the rate law in terms of the original reactants is more complex.

Question 10

Reaction Order

Answer 10

Or overall order of reaction. It’s the sum of the exponents.

Question 11

Zero-Order Reactions

Answer 11

Has a constant rate, which is independent of the reactant’s concentrations.

rate = k

WRT the administration of medication, a zero order reaction is one in which the amount of drug administered/eliminated per unit time remains constant. The concentration of Drug A can be calculated by

A = A0 - (k0)(t)

A0 = initial concentration of drug A in the body.
k0 = zero order rate constant
t = time

The zero order half life changes with time and is proportional to the initial drug concentration. It is inversely proportional to the zero order rate constant and can be represented with the following equation.

Half-Life = (1/2)(A0/k0)

Question 12

First-Order Reactions

Answer 12

Order=1. It has a rate proportional to the concentration of one reactant.

rate = k[A]

Have units of s^-1.

The classic example of first-order reaction is the process of radioactive decay. The concentration of radioactive substance A at any time t can be expressed by

[At] = [A0]e^-kt

[A0] = initial concentration of A
[At] = concentration of A at time t
k = rate constant
t = elapsed time

The half-life (t1/2) of a reaction is time needed for the concentration of the radioactive substance to decrease to half of its original value.

t1/2 = ln2/k = 0.693/k

Where k is the first order rate constant.

Question 13

Second-Order Reactions

Answer 13

Order=2. Has a rate proportional to the product of the concentration of two reactants or to the square of the concentration of a single reactant.

For example, rate = k[A]^2, rate = k[B]^2, or rate = k[A][B].

Units are (M^-1)/(s^-1)

Question 14

Higher-Order Reactions

Answer 14

Has an order greater than 2.

Question 15

Mixed-Order Reactions

Answer 15

Has a fractional order.

Question 16

Collision Theory of Chemical Kinetics

Answer 16

For a reaction to occur, molecules must collide with each other. The collision theory of chemical kinetics states that the rate of a reaction to the number of collisions per second between the reacting molecules. It is important to note that reaction rates almost always increase with increasing temperatures. On the other hand, reaction rates decrease with decreasing temperatures.

Not all collisions result in a chemical reaction. A effective collision occurs only if the molecules collide with correct orientation and sufficient force to break the existing bonds and form new ones. The minimum energy of collision necessary for a reaction to occur is called the activation energy, Ea, or the energy barrier. Only a fraction of colliding particles have enough kinetic energy to exceed the activation energy. This means that only a fraction of all collisions are effective.

rate = fZ

where Z is total number of collisions occurring per second and f is the fraction of collisions that are effective.

Question 17

Transition State Theory

Answer 17

WHen molecules collide with sufficient energy, they form a transition state, in which the old bonds are weakened and the new bonds are beginning to form. The transition state then dissociates into products, and the new bonds are fully formed.

Also called the activated complex, the transition state has greater energy than either the reactants or the products. The activation energy is required to bring the reactants to this particular energy level. Once an activated complex is formed, it can either dissociate into products or revert to reactants without any additional energy input. Transition states are distinguished from intermediates in that, existing as they do at energy maxima, transition states do not have a finite lifetime.

Question 18

Potential Energy Diagram

Answer 18

Illustrates the relationship between the activation energy, the heats of reaction, and the potential energy of the system. The most important factors in such diagrams are the relative energies of the products and reactants.

The activated complex exists at top of the energy barrier. The difference in potential energies between the activated complex and the reactants is the activation energy of the forward reaction; the difference in potential energies between the activated complex and the products is the activation energy of the reverse reaction.

Question 19

Enthalpy

Answer 19

Difference between the potential energy of the products and reactants.

A negative enthalpy change indicates an exothermic reaction (where heat is given off).

A positive enthalpy change indicates an endothermic reaction (where heat is absorbed).

Question 20

How Reactant Concentrations Effect Reaction Rate

Answer 20

Greater the concentrations of reactants, the greater will be the number of effective collisions per unit time, and therefore the reaction rate will increase for all but zero-order reactions. For reactions occurring in the gaseous state, the partial pressures of the reactants can serve as a measure of concentration.

Question 21

How Temperature Affects Reaction Rate

Answer 21

For nearly all reactions, reaction rate will increase as the temperature increases. Since the temperature of a substance is a measure of the particle’s kinetic energy, increasing temperature increases kinetic energy. Consequently, the proportion of molecules having energies greater than Ea increases with higher temperature.

Question 22

How Medium Affects Reaction Rate

Answer 22

The rate of a reaction may also be affected by the medium in which it takes place. Certain reactions proceed more rapidly in aqueous solution, whereas other reactions may proceed more rapidly in benzene. The state of the medium can also have an effect.

Question 23

Catalysts

Answer 23

Substances that increase reaction rate without themselves being consumed; they do this by lowering activation energy. Enzymes are biological catalysts.

Catalysts may increase frequency of collision between the reactants, change the relative orientation of the reactants to make a higher percentage of collisions effective, donate electron density to the reactants, or reduce intramolecular bonding with reactant molecules.

Question 24

Equilibrium Constant (Kc)

Answer 24

2A B+C

Since the reaction occurs in one step, the rates of the forward and reverse reaction are given by

ratef = kf[A]^2 and rater = kr[B][C]

When ratef = rater, equilibrium is achieved.

kf[A]^2 = kr[B][C] = kf/kr = [B][C]/[A]^2

Since both ks are constant,

Kc = [B][C]/[A]^2

Kc is the equilibrium constant. And the subscript c indicates it is in terms of concentration. For dilute solutions, Kc and Keq are used interchangeably.

While the forward and reverse reactions are equal at equilibrium, the molar concentrations of the reactants and products are usually not equal. This means that the forward and reverse rate constants are also usually unequal.

kf[A]^2 = kr[B][C]

kf = kr{[B][C]/[A]^2}

In a reaction with more than one step, the equilibrium constant for the overall reaction is found by multiplying together the equilibrium constants for each step of the reaction. When this is done, the equilibrium constant for the overall reaction is equal to the concentrations of products divided by reactants in the overall reaction, each raised to its stochiometric coefficient. The forward and reverse rate constants for any step n are designated kn and k-n respectively. If

aA + bB cC + dD

occurs in 3 steps, then

Kc = k1k2k3/k-1k-2k-3 = {([C]^c}{([D]^d)}/{([A]^a)([B]^b)}

This expression is known as the law of mass action.

Question 25

Reaction Quotient (Q)

Answer 25

A measure of the degree to which a reaction has gone to completion. Only a constant at equilibrium, when it’s equal to Kc.

Question 26

Properties of the Equlibrium Constant

Answer 26

• Pure solids and liquids don’t appear in the equilibrium constant expression.
• Keq is characteristic of a given system at a given temperature.
• If the value of Keq is very large compared to 1, an equilibrium mixture of reactants and products will contain very little of the reactants compared to products.
• If the value of Keq is very small compared to 1, an equilibrium mixture of reactants and products will contain very little of the products compared to reactants.
• If the value of Keq is close to 1, an equilibrium mixture of products and reactants will contain approximately equal amounts of both.

Question 27

Le Chatelier’s Principle

Answer 27

French chemist Henry Louis Le Chatelier indicated that if an external stress is applied to a system at equilibrium, the system will attempt to adjust itself to offset partially the stress. This rule is used to determine with direction in which a reaction at equilibrium will proceed when subjected to a stress.

Question 28

Changes in Concentration

Answer 28

Increasing the concentration of a species will tend to shift the equilibrium away from the species that is added to reestablish its equilibrium concentration, and vice versa.

Decreasing the concentration of a species will shift the equilibrium toward the production of that species.

Question 29

Change in Pressure or Volume

Answer 29

In a system at constant temperature, a change in pressure causes a change in volume, and vice versa. Since liquids and solids are incompressible, a change in pressure/volume of systems with only these phases has little or no effect on their equilibrium. Reactions involving gases however, may be greatly effected.

Pressure and volume are inversely related. An increase in pressure of a system will shift the equilibrium so as to decrease the number of moles of gas present. This reduces the volume of the system and relieves the stress of the increased pressure.

N2(g) + 3H2(g) 2NH3(g)

The left side has four moles, whereas right side only 2. Pressure increases, equilibrium shifts so that side with fewer moles is favored. So right. If volume is increased, pressure decreases, leading to shift to the left.

Question 30

Change in Temperature

Answer 30

Heat may be considered a product in an exothermic reaction and a reactant in an endothermic reaction.

A B + heat

If this system’s temperature decreased, the reaction would shift to right, replace heat loss. If temperature increased, reaction would shift left due to increased concentration of heat.

Temperature actually alters numerical value of Keq. Pressure, volume, or concentration of species do not have such an effect.