Lecture 3: Gases, Kinetics, And Equilibrium

Question 1

Gases

Answer 1

1. STP:
   1a. Standard temperature: 0C
   1b. Standard Pressure: 1atm = 10^5Pa=1bar
   1c. Velocity of oxygen= faster than sound (343 m/s); 480m/s
2. Mean free path: average distance travelled by gas molecule between 1600A
   2a. For oxygen at STP=1600A
3. Gases will always mix despite polar differences, but at low temperatures the heavier gas/cold is below

Question 2

Kinetic molecular theory: ideal gases have

Answer 2

1. No volume
2. No force: except collisions
3. Elastic collisions: no energy loss
4. Kinetic energy is proportional to temperature: KE=(3/2)RT

Question 3

Ideal gas law and derivations

Answer 3

1. Ideal gas law: PV=nRT (R=0.08Latom/molK or 8 J/molK)
2. Avagadros law: V1/n1=V2/n2 (constant T and P)
3. Boyles law: P1V1=P2V2 (constant T and n)
4. Charles law: V1/T1=V2/T2 (constant P and n)

Question 4

STP conditions

Answer 4

1. 1 mol =22.4 L= 273K =1atm= 0.08Latm/kmol

Question 5

Partial pressure

Answer 5

1. Pa=XaPtotal
   1a. Xa=mole fraction
2. Dalton’s law says that if you have mixture of gasses, you can sum the partial pressures to get the total pressure
   2a. Explains why putting 3mol of any gas at STP=3atm

Question 6

Grahams law

Answer 6

1. KE=KE
   1a. V(gas1)/V(gas 2)=sqrt(m(gas 2)/m(gas 1))
   1b. Diffusion rate gas 1/diffusion rate gas 2=sqrt (mass gas 2/mass gas 1)
2. Tells us rate of diffusion of gases when mixed

Question 7

Real gases

Answer 7

1. Deviate from ideal gases when molecules are very close together
2. Shown in Van der Waals equation: [P+(an^2/V^2)]=nRT
3. Compression factor: Z=PV/nRT

Question 8

Collision theory

Answer 8

1. The energy to break a chemical bond comes from energy of collision between the reactants (usually 2 molecules)
2. In order for a collision to result in reaction: kinetic energies of colliding molecules must reach activation energy and colliding molecules must have a spatial orientation relative to each other
3. Shown in Arrhenius equation: k=Ae^(-Ea/RT) where A=zp

Question 9

Reaction rate equations

Answer 9

1. In aa+bB—>cC+dD
   1a. Change in reactant: -[A]/at=[C]/ct

Question 10

The rate law

Answer 10

1. 1. In aa+bB—>cC+dD
      1a. Rate forward=k[A]^a[B]^b

Question 11

Reaction orders

Answer 11

1. Zero: [A]
2. First: ln[A]
3. Second: 1/[A]

Question 12

Rate determining step

Answer 12

1. Slow step

Question 13

Equilibrium approximation

Question 14

Steady state approximation

Question 15

Catalysis

Answer 15

1. Increase rate ONLY by lowering Ea
   1a. Heterogenous catalyst: in diff phase than reactants and products (s vs g or l)
   1b. Homogenous catalysts: in same phase (usually gas or liquid)

Question 16

Effects of solvent on rate

Answer 16

1. Stirring/shaking increases reaction rate greatly

Question 17

Equilibrium conditions

Answer 17

1. Chemical equilibrium occurs: forward=reverse
2. Greatest point of entropy

Question 18

Nonequilibrium conditions

Answer 18

1. Reaction runs to completion here…product is removed as reaction proceeds
2. Steady state equilibrium: rate of forward doesn’t equal rate of reverse

Question 19

Law of mass action

Answer 19

1. Equilibrium constant (K)=used to relate the rate of the reaction to equilibrium
   1a. K=[C]^c[D]^d/[A]^a[B]^b

Question 20

Extra

Question 21

Reaction quotient

Answer 21

1. Q=K: reaction in equilibrium
2. Q>K: reverse is favoured
3. Q<K: forward is favoured

Question 22

Le chateliers principle: types of stresses

Answer 22

1. Addition or removal of a product or reactant
2. Changing the pressure of the system
3. Heating or cooling the system