topic seven/seventeen

Question 1

in homogenous equilibria, the reactants and products are present in the same

Answer 1

phase

Question 2

four key characteristics of a reactiona t equilbrium

Answer 2

1. The forward and reverse reactions occur at the same rate.
2. The concentrations of reactants and products at equilibrium are constant.
3. Equilibrium requires a closed system.
4. There is no change in macroscopic properties at equilibrium.

Question 3

The same characteristics apply for physical equilibrium as for chemical equilibrium:

Answer 3

Equilibrium is dynamic – the liquid is continuing to evaporate and condense.

The rate of the forward process (evaporation) is equal to the rate of the reverse process (condensation).

The volume of liquid at equilibrium remains constant.

Physical equilibrium can only occur in a closed system.

At equilibrium, there are no changes in macroscopic properties such as colour.

Question 4

what is Kc

Answer 4

a constant for a given temperature, therefore, it is temperature-dependent. A change in temperature would result in a change in the value of Kc for a reaction (this is discussed in more detail in a later section).

Question 5

equilibrium constant expression

Answer 5

aA + bB ⇌ cC + dD

Kc=[C]c[D]d / [A]a[B]b

Question 6

If the Kc value at a given temperature is high, this indicates that

Answer 6

the reaction has almost gone to completion and a high proportion of reactants have been converted to products.

A high value of Kc means that there is a higher concentration of products than reactants in the equilibrium mixture

Question 7

If the Kc value at a given temperature is low, this indicates that

Answer 7

the forward reaction has barely proceeded and only a small proportion of the reactants have reacted to form products.

A low value of Kc means that there is a higher concentration of reactants than products in the equilibrium mixture

Question 8

value of Kc when Forward reaction hardly proceeds

Answer 8

Kc < 10^–10

Question 9

value of Kc ‘Reactants’ predominate at equilibrium

Answer 9

Kc = 0.01

Question 10

value of Kc when Equal amounts of reactants and products

Answer 10

Kc = 1

Question 11

value of Kc when ‘Products’ predominate at equilibrium

Answer 11

Kc = 100

Question 12

value of Kc when Reaction almost goes to completion

Answer 12

Kc > 10^10

Question 13

what is the value of Kc when the reaction has gone to completion

Answer 13

Kc is v much over 1

Question 14

what is the value of Kc when the reaction hasnt gone to completion

Answer 14

Kc is v much less than 1

Question 15

if a reaction is reversed, what happens to the kc

Answer 15

it will be the reciprocal of that value

Question 16

if the reaction coefficents are doubled, the Kc is

Answer 16

squared

Question 17

if the reaction coefficients are halved, the kc is

Answer 17

square rooted

Question 18

what is the reaction quotient Q

Answer 18

a measure of the relative amounts of reactants and products for a reaction that is not yet at equilibrium. The reaction quotient is useful for predicting in which direction a reaction will proceed to reach equilibrium.

Question 19

what is the different in reaction quotient expressiona dn equilibrum constant expression (q and kc)

Answer 19

kc uses equilibrium values

Question 20

when Q is less than Kc

Answer 20

the reaction is not yet at equilibriujm

forwards reaction favoured

Question 21

when Q = Kc

Answer 21

the reaction is at equolibrium

Question 22

when Q is bigger than Kc

Answer 22

the reacton has exceeded equilibrium

reverse reaction favoured

Question 23

Le Châtelier’s principle states that:

Answer 23

‘When a system at equilibrium is subjected to a change, the system will respond to minimise the effect of the change.’

Question 24

The effect of changes in conditions on the position of equilibrium

Answer 24

Question 25

The chemical process that produces ammonia gas involves the following reaction:

Answer 25

N2 (g) + 3H2 (g) ⇌ 2NH3 (g) ΔH = –92 kJ mol–1

Question 26

the haber process: This means that increasing the pressure will shift the equilibrium position to the right, yielding more what

Answer 26

ammonia

Question 27

which reaction in the haber process is exothermic

Answer 27

forward (-92kH mol^-1)

As a result the production of ammonia is favoured by lower temperatures. however uually temp of 450 is used for quick reaction time

Question 28

what catalyst is used in the haber process

Answer 28

iron

Question 29

what does contact process create

Answer 29

sulfuric acid

Question 30

main uses of contact process

Answer 30

Fertilisers.
Paints and pigments.
Detergents and soaps.
Dyestuffs.

Question 31

temperature in the contact process

Answer 31

Applying Le Châtelier’s principle suggests the highest yield of sulfur trioxide would be obtained using low temperatures and high pressures. However, in actuality, very low temperatures cannot be used, as the rate of reaching equilibrium would be too slow, and therefore uneconomic. A further factor to be considered here is that the vanadium(V) oxide catalyst becomes effective only at temperatures above 400°C. The percentage conversion is improved by passing the reacting gases through a sequence of four separate catalyst beds. This repetitive technique produces a conversion of sulfur dioxide to sulfur trioxide that is 99.5% efficient.

Question 32

pressure in the contact process

Answer 32

Given this high rate of conversion, the use of high pressure is unnecessary, even though it would increase the yield of sulfur trioxide, SO3. The cost of using high pressures is uneconomic and introduces dangers of explosion. A pressure of 1-2 atmospheres, slightly above atmospheric pressure, is used to ensure the gases circulate freely through the catalyst beds.

Question 33

how to work out Kc from equilibrium concentrations

Answer 33

use equilibrium cosntant equation

Question 34

how to work out Kc from initial concentrations

Answer 34

the initial concentration, I

the change in concentration, C

the equilibrium concentration, E.

Using the balanced equation for the reaction, we can calculate the equilibrium concentrations of each reactant and product and then find the value of Kc.

Question 35

the form of the equilibrium expression for Kc,

Answer 35

Kc = kf / kr

Question 36

If the rate constant for the forward reaction (kf1) is greater than that of the reverse reaction (kr1), the equilibrium constant will be

Answer 36

large, and the forward reaction is favoured

Question 37

if the rate constant for the backward reaction (kr2) is greater than that of the forward reaction (kf2), the equalibrium constant will be

Answer 37

small and the reverse reaction is favoured

Question 38

Gibbs free energy change must be what for a spontaneous reaction

Answer 38

negative

Question 39

The equilibrium constant, K, is related to the standard Gibbs free energy change, ΔG⦵, for a reaction by the equation below:

Answer 39

ΔG⦵ = −RT ln K

R is the universal gas constant (8.31 J K−1 mol−1)

T is the absolute temperature in kelvin (K)

ln K is the natural log of the equilibrium constant, K.

Question 40

You can see that the more negative the value of the standard Gibbs free energy change, the

Answer 40

larger the value of the equilibrium constant, K. This means that the more spontaneous a reaction is, the further the position of equilibrium lies to the right

Question 41

if gibbs is negative, ln k is positive and Kc is less than 1, waht is the pos of e?

Answer 41

Lies to the right – favouring the products

Question 42

if gibbs is zerio, ln k is zerio and kc is 1, what is the pos of e

Answer 42

Equal proportions of reactants and products

Question 43

if gibbs is positive, ln k is negative and Kc is over 1, waht is the pos of e

Answer 43

Lies to the left – favouring the reactants

Question 44

Change in Gibbs free energy for a spontaneous reaction.

Answer 44

Question 45

at equilibrium, gibbs is at a min or max and entropy is at a min or max

Answer 45

Gibbs free energy is at a minimum and entropy is at a maximum.