3.7 entropy and feasibility of reactions

Question 1

apart from enthalpy, what other factor determines whether or not a reaction will take place?

Answer 1

the change in entropy

Question 2

• a reaction that is exothermic will result in products that are more thermodynamically stable than the reactants
• this is a driving force behind many reactions and causes them to be spontaneous
• however this cannot be reconciled with the fact that many endothermic reactions proceed readily under normal conditions
• it follows that there must be other factors other than the enthalpy change
• the other factor which determine whether or not a reaction takes place is the change in entropy

Question 3

what is entropy?

Answer 3

• measure of randomness of the particle
• measure of order/disorder in a system (chaos)

Question 4

what are the units of entropy?

Answer 4

JK^-1mol^-1

Question 5

the entropy of gases is (greater/less) than that of liquids which is (greater/less) than that of solids

Answer 5

greater
greater

S (gas) > S (liquid) > S (solid)

Question 6

what does the third law of thermodynamics state?

Answer 6

• at 0K, particles are in a maximum state of order and have zero entropy as they are stationary

Question 7

lots of order = low entropy

Question 8

if asked to predict whether a reaction is likely to have an increase or decrease in entropy, what do you do?

Answer 8

• look at the state each substance is in
• as well as how many moles you start and end with in the reaction
• an increase in disorder and entropy will lead to a positive entropy change ∆S = +ve

Question 9

NH4Cl (s) —> HCl (g) + NH3 (g)
is the entropy change positive or negative?

Answer 9

1 mol —> 2 mol
solid —> gas

∴ more disorder ∴ ∆S=+ve

Question 10

in general a significant increase in the entropy will occur if:

Answer 10

• there is a change of state from solid or liquid to gas
• there is a significant increase in the number of molecules between products and reactants

Question 11

systems will always seek to move towards a state of more disorder

Question 12

what factors could stop reactions occurring so quickly?

Answer 12

a large activation energy

Question 13

how to calculate enthalpy change (∆S°) quantitatively?

Answer 13

∆S° = ΣS° products - ΣS° reactants

Question 14

reactions that have a +ve ∆S end up in (more/less) stable products?

Answer 14

more

Question 15

diamond is described as metastable.
what does metastable mean?

Answer 15

has a high activation energy

Question 16

what do you calculate to see if a reaction will proceed?

Answer 16

the change in free energy (∆G)

• whether or not a reaction will proceed depends on a balance between entropy and entropy
• these two quantities combine to give a single term known as free energy, G

Question 17

what is the gibbs equation?

Answer 17

∆G = ∆H - T∆S

∆G = kJmol^-1
∆H = kJmol^-1
T = kelvin
∆S = kJK^-1mol^-1

Question 18

if ∆G is positive, the reaction (occurs/doesnt occur)?

Answer 18

doesnt occurs / not feasible

Question 19

if ∆G is negative, the reaction (occurs/doesnt occur)?

Answer 19

occurs
- feasible

Question 20

a reaction that has increasing entropy (+ ∆S) and is exothermic (- ∆H) will make ∆G (positive/negative) and so will always (be feasible/not be feasible)

Answer 20

negative
be feasible

Question 21

if ∆G is negative, is there still a possibility that the reaction will not occur?

Answer 21

• yes
• or may occur so slowly that effectively it doesnt happen
• if the reaction has a high activation energy the reaction will not occur

Question 22

• thermodynamics tells us information about whether a reaction can occur or whether it is possible for a reaction to occur
• can it tell us anything about the rate at which the reaction occurs?

Answer 22

no

Question 23

how can you predict at what temperature a reaction would become feasible?

Answer 23

• if ∆G tells us whether a reaction is feasible, so ∆G=0 can predict at what temperature a reaction would become feasible
  ∆G = ∆H - T∆S
  0 = ∆H - T∆S
  T∆S = ∆H
  T = ∆H / ∆S

Question 24

any phase changes for any reactions such as melting and boiling have a ∆G=?

Answer 24

∆G = 0

Question 25

calculate the temperature range that this reaction will be feasible N2 (g) + O2 (g) —> 2NO (g) ∆H = 180 kJmol^-1 ∆S = 25 JK^-1mol^-1

Answer 25

∆G = ∆H - T∆S ∆G = 0 T = ∆H / ∆S = 180 / (25/1000) = 7200K the reaction will be feasible when ∆G ≤ 0 so T must be >7200K * all units must be the same *

Question 26

how do you graphically determine ∆G?

Answer 26

y = mx + c ∆G = -T∆S + ∆H ∆H = c intercept -∆S = gradient of graph

Question 27

when graphically determining ∆G need to consider:

Answer 27

- a positive gradient means ∆S is negative - when ∆G < 0 then the reaction is spontaneous - the slope of the line may change as if the products start as a gas and change to liquid the entropy change would be different

Question 28

when working out the Ecell value for an electrochemical cell, if the value is positive then the reaction is (spontaneous/not spontaneous), with ∆G therefore being (positive/negative)

Answer 28

spontaneous negative Zn 2+ (aq) + 2e- —> Zn (s) E = -0.76V LHS Fe2+ (aq) + 2e- —> Fe (s) E = -0.44V RHS EMF = RHS - LHS = + 0.32 = +ve so feasible so ∆G is -ve

Question 29

if Kc > 1 is ∆G positive or negative?

Answer 29

negative

Question 30

if Kc < 1 is ∆G positive or negative?

Answer 30

positive

Question 31

what does the second law of thermodynamics state?

Answer 31

- in any spontaneous reaction the total entropy increases - so tells us that there us a general tendency towards disorder

Question 32

what is the standard entropy, S°, of a substance?

Answer 32

the entropy of one mole of that substance under standard conditions

Question 33

what are the units of gibbs free energy?

Answer 33

kJmol^-1 or Jmol^-1

Question 34

if: ∆H = +ve ∆S = -ve ∆G = always +ve what is the feasibility of the reaction?

Answer 34

- the reaction can never be feasible

Question 35

if: ∆H = -ve ∆S = +ve ∆G = always -ve what is the feasibility of the reaction?

Answer 35

- the reaction is always feasible - (it may not happen quickly as the activation energy needs to be considered)

Question 36

if: ∆H = +ve ∆S = +ve ∆G = -ve at high temperatures what is the feasibility of the reaction?

Answer 36

- this reaction is opposed by ∆H (endothermic) but favoured by ∆S - under standard conditions the reaction would not proceed, but if T is high enough to make -T∆S as large as possible (so T∆S > ∆H) then the reaction will proceed

Question 37

if: ∆H = -ve ∆S = -ve ∆G = -ve at low temperatures what is the feasibility of the reaction?

Answer 37

- this reaction is feasible whilst -T∆S is a small value - once T becomes significantly higher the reaction is less likely to occur spontaneously - the only way to find out is to plug numbers into the equation

Question 38

what does it mean if a reaction is ‘spontaneous’?

Answer 38

- a spontaneous process is capable of taking place without needing to be driven by an outside source of energy

Question 39

is a reaction with positive or negative entropy change more likely to be spontaneous?

Answer 39

- positive

Question 40

is a reaction with positive or negative enthalpy change more likely to be spontaneous?

Answer 40

negative (exothermic)

Question 41

what does it mean for a reaction to be ‘feasible’?

Answer 41

- for a reaction to be feasible at a given temperature, the reaction must occur spontaneously - this means no external energy input is required for the reaction to take place