Module 4 Yr 11

Question 1

Why is energy required in chemical reactions
When is energy released in chemical reactions

Answer 1

Energy is required to break the bonds in reactant molecules
Energy is released upon forming the bonds in the product molecules

Question 2

Define exothermic and endothermic reaction

Answer 2

Exothermic: reaction that releases energy overall to the surroundings, usually in the form of heat. Occurs when energy released from bond formation exceeds the energy absorbed during bond breaking. Accompanied by increase in temperature.
Endothermic: reaction that absorbs energy overall from the surroundings usually in the form of heat. Occurs when energy absorbed from bond breaking is greater than the energy released from bond formation. Accompanied by decrease in temperature.

Question 3

What is dissolution

Answer 3

Process where solute in any phase dissolves in a solvent to form a solution

Question 4

What processes occur in dissolution

Answer 4

Solute-solute interactions in an ionic lattice are broken:

An ionic lattice is held together by ionic bonds i.e forces of electrostatic attraction between positively and negatively charged ions
Solvent-solvent interactions in the water solvent are broken:

In the liquid state, water molecules are held together by IMF i.ie hydrogen bonding, dipole dipole interactions and dispersion forces. Overcoming these forces require energy input.
Solute-solvent interactions between ions and water molecules are formed:

Called solvation or hydration (in case of water) by which solvent molecules form concentration rings called hydrogen spheres around individual ions. Ion-dipole forces are formed between the ions and the water molecules, which are stronger than the hydrogen bonding interactions between water molecules. Since new forces are formed, energy is released in this process.

Question 5

An example of dissolution

Answer 5

When NaCl dissolves in water, the slightly positive hydrogens in the water molecule orient themselves towards the Cl- anion, while the slightly negative oxygen orient itself toward the Na+ cation

Question 6

Name salts that dissolve exothermically and endothermically

Answer 6

Exothermically:NaOH, KOH, CaCl2, Na2CO3
Endothermically: NH4Cl, NaCl, NH4NO3, NaHCO3, KCl

Question 7

Define molar heat of solution (dissolution)
How to calculate molar heat

Answer 7

Refers to the quantity of heat released upon the dissolution of one mole of a particular substance usually in water. Its units are J mol^-1 or kJ mol^-1
Plug into equation:
molar heat=-Q/n
Q: quantity of heat released or absorbed
n: number of moles
Note: if salt dissolves exothermically, the molar heat is negative. If salt dissolves endothermically, molar heat is positive

Question 8

For experiment investigating temperature changes in dissolution of ionic substances:
Why are polystyrene cups used and why is one cup stacked inside the other?
Why is the thermometer suspended in the water by clamps and not left to rest at the bottom of the cup

Answer 8

Styrofoam cup won’t take much energy (good insulator) ensuring heat is absorbed by the water and not by the cup and then lost to the environment. Stacking cups introduces an air layer which improves insulation.
We want to be measuring the temperature changes in the water not the cup effectively improving the validity.

Question 9

For experiment investigating temperature changes in combustion reactions:
Why is an aluminium can used instead of a polystyrene cup
Why does the temperature of the water increase
Why is there different increase in water temperatures

Answer 9

Al is a good conductor of heat and we want to transfer heat from spirit burner to the water so a good conductor is required
Combustion is an exothermic reaction i.e energy released via bond formation>energy used for bond breaking
Larger alcohol means more carbon and hydrogen to combust per mole meaning more energy is used for bond formation than in bond breaking

Question 10

Define enthalpy
How to find enthalpy change of reaction
Where does missing energy come from in any chemical reaction

Answer 10

Amount of stored heat energy within a substance. The enthalpy change of a reaction ( ΔHreaction or ΔH) is the difference between the enthalpy of the products and that of the reactants.
ΔHreaction= ΔHproducts -ΔHreactants
ΔHreaction= ΔHbond-breaking + Δbond-forming
Note: ΔHproducts and ΔHreactants cannot be quantified, only ΔHreaction can be observed and measured
The missing energy is the heat released to or absorbed by the environment

Question 11

What is a transition state and activated complex
Conditions for converting reactants into products

Answer 11

State where original bonds are broken
The chemical structure with maximum enthalpy in the reaction pathway
They must go through a transition state or activated complex where one more more covalent bonds in the reactants are broken. It is the point in the reaction pathway in between the breaking of at least one covalent bond and the forming of new covalent bond (s).

Question 12

Charge of bond breaking and bond forming
What does a positive and negative ΔH mean

Answer 12

ΔHbond-breaking is always positive as it requires energy
ΔHbond-forming is always negative as it releases energy
Positive: Hrctnt<Hprod, more energy required in bond breaking meaning overall energy is required making it an endothermic reaction
Negative: Hrctnt>Hprod, more energy required in bond forming meaning overall energy is released making it an exothermic reaction

Question 13

Describe an energy profile diagram for exothermic and endothermic reactions

Answer 13

Exothermic reaction:
Reactants have high potential energy and continue straight and then elevate slightly (peak is activated complex) and then rapidly drop to the products which have a low potential energy and then continue straight
Endothermic reaction:
Reactants have low potential energy and then continue straight and then elevate rapidly (peak is activated complex) and then drop slightly to the products which have a high potential energy and then continue straight
X axis is reaction pathway
Y axis is potential energy (kJ)
Must label location of activated complex, reactants, products, change in enthalpy, activation energy (Ea)

Question 14

What is the activation energy

Answer 14

The energy difference between the reactants and the activated complex, a larger difference would mean more energy needs to be inputted to initiate the reaction

Question 15

How to write thermochemical equations

Answer 15

The fuel that is combusted is given a stoichiometric coefficient of 1. Moles of oxygen can be written in fraction form to ensure equation is balanced. Enthalpy of reaction must be written initially. If coefficients are doubled, then enthalpy is doubled.
E.g.
C4H10(g) +13/2 O2(g) –>4CO2(g) +5H20 (l). ΔH=-2886kJ mol^-1
2C4H10(g) +13 O2(g) –>8CO2(g) +10H20 (l). ΔH=-5772kJ mol^-1

Question 16

How to calculate heat released in calorimetry experiment

Answer 16

q=mcΔT
q: energy released in J
m: mass of water or other liquid in g
c: specific heat capacity in J g^-1 K^-1. Refers to the amount of heat energy required to increase the temperature of 1.00g of a substance by 1.00K.
ΔT: temperature change of the water in Kelvin
Note: remember in chemistry, specific heat capacity of water is 4.18 x 10^-3

Question 17

How to calculate molar enthalpy change (ΔH)

Answer 17

ΔH=-q/n
q: energy released in J
n: number of moles of fuel (combustion), salt (dissolution)
Negative sign provides the direction of enthalpy change where a negative enthalpy change means that the reaction is exothermic. If the calorimetry reaction is endothermic, then q will be negative so ΔH will be positive

Question 18

Assumptions made when calculating molar enthalpy change

Answer 18

1. In both reactions, all the heat generated from the reaction is transferred completely to the water, and not conducted/radiated to the apparatus or the surroundings
2. In the combustion experiment, there is sufficient oxygen such that the fuel is completely combusted
3. In the dissolution experiment, the final solution is sufficiently dilute such that the density of the solution increases negligibly that is isn’t exactly 1.000mg mL^-1
4. In the dissolution experiment, the final solution is sufficiently dilute such that the specific heat capacity of the solution approximates that of pure water meaning that isn’t exactly 4.18 J g^-1 K^-1

Question 19

Importance of percentage error

Answer 19

It is a superior method to quantitatively assess the accuracy where +/- 5% is considered accurate.
%error= 100% x [theoretical value-experimental value]/|[theoretical value]|

Question 20

Discussion of calorimetry experiments

Answer 20

In most calorimetry experiments, the magnitude of the enthalpy change will be significantly less than the theoretical value.
The key assumption of this investigation is that all of the heat generated by the reaction is absorbed by the water, raising its temperature in proportion to the amount of heat absorbed
However, some energy will not reach the water and instead be lost to the surrounding apparatus and external environment via radiation and conduction, especially to glass or metal.
When combustion reactions are involved, incomplete combustion may occur which releases less energy per mole than complete combustion
Standard values for enthalpy changes in combustion assume complete combustion which cannot be achieved in a school laboratory
Values achieved in school-based experiments for heat of combustion will generally have a lower magnitude than the theoretical values

Question 21

Factors that may increase the magnitude of ΔH of combustion. List the problem and the solution

Answer 21

Factor 1: Problem is not stirring, so water at the base of can is warmer. Stir vigorously with stirring rod to solve issue
Factor 2: Problem is thermometer is resting at the base of the can which may be warmer. Suspend thermometer in the water with a clamp from a retort stand to solve issue.

Question 22

Factors that may decrease the magnitude of ΔH of combustion. List the problem and the solution

Answer 22

Factor 1: Problem is heat radiated/conducted to apparatus and surroundings. Turn of any fans, use stacked styrofoam cups and use a draught shield/lid.
Factor 2: Problem is can is too close to the flame so there is insufficient O2 and incomplete combustion occurs. Move the can slightly further away from the flame to solve issue.
Factor 3: Problem is can is too far from flame or has sooty base so there is insufficient heat transfer. Cleaning the can and moving it slightly closer to the flame can solve issue.
Factor 4: Problem is impurities (water) in the spirit burner that don’t combust well. Cannot be solved without changing the spirit burner.
Factor 5: Problem is salt may not be completely dissolved. Stir vigorously with a stirring rod to solve issue.

Question 23

What do chemical bonds store

Answer 23

They don’t store energy.

Question 24

Amount of energy that unbonded atoms have in comparison to two atoms bonded together
Why is energy required to break bonds

Answer 24

Two atoms which are bonded have less energy (occupy a lower energy state) than those which are unbonded.
Essentially, bonds constitute negative potential energy within a molecule meaning energy is required to break bonds. Energy needs to be inputted to raise the atoms to a higher energy lvl and counteract the - potential energy of the bond. By atoms forming bonds, they move to a lower energy state where this reduction in energy lvl is associated with a release of energy ensuring that the total amount of energy remains constant.

Question 25

What is Hess’s law

Answer 25

A law that states that the enthalpy change accompanying a chemical change is independent of the route by which the chemical change occurs. Hess’s Law is an implication of the Law of Conservation of Energy.
To summarise, it means that regardless of the number of steps involved in a multi-step reaction, the enthalpy change of the overall reaction is the sum of the enthalpy change of all the steps.
Note: all steps must proceed at the same temperature and absolute enthalpy cannot be calculated rather enthalpy changes only.

Question 26

What is the standard enthalpy of formation

Answer 26

ΔH°f is the change in enthalpy associated with the formation of 1 mol of a substance in its standard state (25°C and 100kPa) from its constituent elements in their standard state. The equations for the standard formation is written such that the stoichiometric coefficient of the formed substance is given the value 1.

Question 27

Standard enthalpy of formation of SO3(g)
Do these substances have ΔH°f=0
O2(g)
CO2(g)
Br2(l)
Na(l)
O3(g)
Au(s)

Answer 27

Standard formation of SO3(g) is written as follow:
S(s)+3/2 O2(g) –> SO3(g) ΔH=-45.9 kJ mol^-1
Hence ΔH°f of SO3(g) is 45.9
Note: an element in its standard state is defined to have ΔH°f=0
O2(g): yes as it is an element in its standard state
CO2(g): No, as it isn’t an element
Br2(l): yes as it is an element in its standard state
Na(l): No, as it is a solid at standard state not liquid
O3(g): Standard form of oxygen is O2, O3 is variant
Au(s): yes as it is an element in its standard state

Question 28

How to calculate the enthalpy change of a chemical reaction at standard conditions

Answer 28

ΔHrxn=ΔHreaction
ΔHrxn=ΔHproducts-ΔHreactants

Question 29

Calculate the ΔH of HCl(g)+C2H4(g)–>C2H5Cl(g) with the following information
Standard enthalpies are
C2H5Cl(g): 52.2 kJ mol^-1
C2H4(g): -92.3 kJ mol^-1
HCl(g): -109 kJ mol^-1

Answer 29

2C(s)+2H2(g) –>C2H4(g) ΔH=+52.2 kJ mol^-1
1/2 H2(g)+1/2 Cl2(g) –>HCl(g) ΔH=-92.3 kJ mol^-1
2C(s)+5/2 H2(g)+1/2 Cl2(g) –>C2H5Cl(g) ΔH=-109kJ mol^-1
Reverse 1 and 2=
1. C2H4(s) –>2C(s)+2H2(g). ΔH=-52.2 kJ mol^-1
2. HCl (g) –>1/2 H2(g) + 1/2 Cl2(g). ΔH=92.3 kJ mol^-1
=C2H4(s)+HCl(g) –>5/2 H2(g) + 1/2 Cl2(g) +2C(s) ΔH=40.1
3. 2C(s) +5/2 H2(g) +1/2 Cl2(g) –> C2H5Cl(s) ΔH=-109kJ mol^-1
=C2H4(s)+HCl(g) –>C2H5(g) ΔH=-68.9 kJ mol^-1

OR
ΔHrxn=ΔHproducts-ΔHreactants
=-109-(55.2-92.3)
-68.9kJ mol^-1

Question 30

Calculate the standard enthalpy of combustion of one mole of sulfur where sulfur dioxide is the final product.
1. The ΔH°f of SO3(g) is -395
2. SO2(g)+1/2 O2(g) –>SO3(g) has ΔH =-98kJ mol^-1

Answer 30

S(s) +O2(g) –>SO2(g) ΔH=?
S(s) +3/2 O2(g) –>SO3(g) ΔH=-395
SO2(g)+1/2 O2(g) –>SO3(g) has ΔH =-98
Flip 2
SO3(g)–>SO2(g)+1/2 O2(g) ΔH =98
S(s) + O2(g) –>SO2(g). ΔH =-297kJ mol^-1

Question 31

What is the equation for photosynthesis

Answer 31

6CO2(g) + 6H20(l) –>C6H12O6(s) +6O2(g)
Note: glucose in biological reaction is aqueous and photosynthesis is endothermic thus having a positive enthalpy change. Respiration is thermodynamic reverse reaction of this.

Question 32

Calculate the enthalpy change associated with photosynthesis at standard conditions
C6H12O6(s): -1274.5
O2(g): 0
CO2(g): -393.5
H2O(l): -285.8

Answer 32

ΔHrxn=ΔHproducts-ΔHreactants
=-1274.5 -(6x-393.5+6x-285.8)
=2801.3 kJ mol^-1

Question 33

It is given that the molar enthalpy of dissolution of glucose is +11.kJ mol^-1. Find the enthalpy change associated with photosynthesis under biological conditions.

Answer 33

6CO2(g) + 6H20(l) –>C6H12O6(s) +6O2(g). ΔH=2801.3
C6H12O6(s) –>C6H12O6(aq). ΔH=11
6CO2(g) + 6H20(l) –>C6H12O6(aq) +6O2(g). ΔH=2801.3+11=2812.3

Question 34

What is enthalpy of fusion and solidification

Answer 34

Latent heat of fusion is the enthalpy change associated with melting 1 mol of a solid to its liquid form, measured at 1 atm. Fusion, or melting is endothermic where in water, it represents the breaking of some intermolecular bonding. 6.01 kJ of energy is required to melt 1 mol of water at its mp of 0 degrees celsius.
H20 (s) –> H20(l). ΔH=+6.01kJ mol^-1
Conversely, solidification or freezing releases energy so it is an exothermic process where it is the negative of the enthalpy of fusion.
H20 (l) –> H20(s). ΔH=-6.01kJ mol^-1

Question 35

What is enthalpy of vaporisation and condensation

Answer 35

Latent heat of vaporisation is the enthalpy change associated with heating 1 mol of liquid to its gaseous form measured at 1 atm.
Vaporisation requires energy so its an endothermic process where in water, it represents the breaking of all intermolecular bonds in liquid water. 40.6 kJ of energy is required to evaporate 1 mol of water at its bp of 100 degrees celsius.
H20 (l) –> H20(g). ΔH=+40.7kJ mol^-1
Conversely, condensation releases energy so this is an exothermic process where it is the negative of the enthalpy of vaporisation
H20 (g) –> H20(l). ΔH=-40.7kJ mol^-1

Question 36

What processes are endothermic and exothermic

Answer 36

Exothermic:Deposition (gas to solid), condensation, solidification
Endothermic: Sublimation (solid to gas), vaporisation, fusion

Question 37

State collision theory
Define intermediate

Answer 37

A successful reaction occurs when particles collide with sufficient kinetic energy greater than activation energy at the correct orientation
A temporary species in a chemical reaction formed from a sub-reaction. Not a reactant or product

Question 38

Define catalyst
What are the types of catalysts

Answer 38

A non-reagent that lowers the activation energy by offering alternate reaction pathway, thereby increasing the rate of reaction.
Homogenous: same phase
Heterogeneous: different phase

Question 39

Describe the mechanism of action of homogenous catalysts

Answer 39

The catalyst reacts with one or more of the reactants to form a temporary intermediate. The intermediate then reacts with other reactants or decomposes to form the products of the reaction. This process also regenerates the catalyst so the catalyst isn’t consumed overall during the reaction

Question 40

Describe the mechanism of action of heterogenous catalysts

Answer 40

Usually solid catalysts in liquid/aqueous or gaseous reaction:
The reactants first adsorb to the catalyst at specific active site. An interaction between the surface of the catalyst and the reactant molecules causes the reactants to be more active. The chemical reaction then takes place on the surface of the heterogenous catalyst. An intermediate may or may not be formed at the surface of the catalyst. The product then desorbs from the surface.

Question 41

Provide example of mechanism of action of heterogenous catalyst

Answer 41

Hydrogenation of ethene
Hydrogen gas molecules adsorb onto the surface of the catalyst, where they are broken down into hydrogen atoms. The ethene molecule adsorbs onto the surface of the catalyst, causing the double bond to open up, forming two new bonding sites. Two adsorbed Hydrogen atoms react with the adsorbed ethene molecule on the surface of the catalyst. The product ethane (ethene when double bond is removed) from the catalyst surface.
Note: the state ad means adsorbed to the surface of the catalyst such is 1/2 N2(g) –>1/2N2 (ad)

Question 42

Do bonds store energy
When a bond is broken, energy comes out from where the bond used to be
Bond breaking is exothermic
Breaking a bond uses energy
Forming a bond releases energy
Do catalysts affect enthalpy

Answer 42

False
False
False
True
True
No, they decrease activation energies since initial and final values aren’t changed

Question 43

Define bond energy

Answer 43

Bond enthalpy refers to the amount of energy required to break 1 mole of a certain type of bond into its constituent gaseous atoms under standard state conditions (25°C, 100kPa). The values are positive since bond-breaking is exothermic so ΔH>0. Bond energy can be measured by breaking the bonds in 1 mole
e.g. calculate the bond energy of H-Cl
HCl(g)–> H(g)+Cl(g). ΔH=+433kj mol^-1
Note: bond energies are usually quoted as an average

Question 44

Define entropy
What is the standard molar entropy

Answer 44

The amount of randomness or lack of ordered structure in a substance is entropy (S). S° is the entropy of one mole of a substance in its standard state at 25°C and 100kPa. Units are JK^-1mol^-1
More chaotic, more entropy

Question 45

Theoretical entropy of perfect crystal
Change in entropy formula
Define spontaneous process

Answer 45

Absolute entropy can be measured 0 JK^-1mol^-1
ΔSrn=ΣS°products-ΣS°reactants
A process that will naturally occur without any ongoing addition of external energy

Question 46

Express 2nd law of thermodynamics mathematically
When will a process occur?
What values is ΔSsurroundings in exothermic and endothermic reactions and reason why

Answer 46

ΔSsystem +ΔSsurroundings=ΔSuniverse >0
A process only occurs if it leads to an increase in the entropy of the universe
Exothermic: ΔSsurroundings is positive. This is because the heat released to the surroundings will increase the KE of the surrounding particles and increase their entropy.
Endothermic: ΔSsurroundings is negative. This is because the heat absorbed will decrease the KE of the surrounding particles and decrease their entropy.

Question 47

State the entropy trends

Answer 47

Different solids have similar molar entropy. Likewise, different liquids and gases also only have slight variations in entropy.
Entropy increases with molar mass e.g. Entropy of iodine gas > fluorine gas entropy
Entropy increases with melting, vaporisation or sublimation e.g. water vapor entropy > liquid water entropy
Entropy increases when solids or liquids dissolve in water. E.g. NaCl(s) entropy <dissolved Na+(aq) and Cl-(aq) ions entropy
Entropy decreases when a gas is dissolved in water. E.g. O2(g) in the air entropy >O2(g) dissolved in H20 entropy
Entropy is lower in hard, brittle substances compared to malleable solids. E.g. Entropy of most metals > diamond entropy
Entropy increases with chemical complexity
Entropy of MgCl2 > NaCl entropy

Question 48

Structure to answer questions predicting higher entropy values

Answer 48

1. Identify no of moles on LHS and RHS
2. Find similarities and differences
3. How differences contribute to either increase or decrease in entropy

Question 49

Define forward and backward entropy drive
Define forward and backward enthalpy drive
What must a reaction do in order to occur?

Answer 49

The entropy of the system increases (positive ΔS). A reverse/backward entropy drive is one where the entropy of the system decreases (negative ΔS)
The enthalpy of the system decreases (negative ΔH). A reverse/backward enthalpy drive is one where the enthalpy of the system increases (positive ΔH)
Both enthalpy and entropy drives are forward
Enthalpy drive is forward and larger than a reverse entropy drive
Entropy drive is forward and larger than a reverse enthalpy drive.

Essentially: Forward entropy and enthalpy drive is always spontaneous
Reverse enthalpy drive and forward entropy drive is sometimes spontaneous
Reverse entropy drive and forward enthalpy drive is sometimes spontaneous
Reverse entropy and enthalpy drive is never spontaneous

Question 50

Compare entropy of water in solid, liquid and gas form

Answer 50

Solid: water molecules in ice are arranged in an ordered crystalline lattice –> low entropy
Liquid: water molecules in liquid water can tumble over each other –>moderate entropy
H2O(s) –> H2O(l)
Gas: water molecules in water vapour, water molecules can move about randomly –> highest entropy
H2O(l) –> H2O(g)
Generally, gaseous substances have the highest entropy

Question 51

Compare entropy between solid NaCl and aqueous NaCl

Answer 51

Solid: the ions are arranged in an ordered crystalline lattice
Aqueous: the ions can move about freely in hydration spheres
NaCl(s) –>Na+(aq) + Cl-(aq) ΔS>0
Therefore, entropy of aqueous NaCl is greater than of solid NaCl

Question 52

Similarities and differences between entropy and enthalpy

Answer 52

Similarities: both drivers of reactions, both changes can be measured, change in both doesn’t depend on its intermediate steps and both are labelled as forward drives
Differences (entropy): measures disorder, absolute entropy can be measured, ΔSrn=ΣS°products-ΣS°reactants, A system favours maximum entropy, symbol is S, units is JK^-1mol^-1, changes significantly with temperature
Differences (enthalpy): measures internal heat, absolute enthalpy cannot be measured only changes, ΔHrxn=ΔHproducts-ΔHreactants, a system favours minimum enthalpy, symbol is H, units of kJ mol^-1, doesn’t change significantly with temperature

Question 53

Why do gases have higher entropies than their respective solids

Answer 53

In a solid, each molecule has very few energy states it can take. If the energy of any molecule was too high, it would cease to be a solid. In a gas, each molecule can take a wider range of energy states. More energy states means more possible microscopic energy configurations and hence higher entropy.

Question 54

What is the change in gibbs free energy

Answer 54

The maximum amount of work performed by a thermodynamic system or during a chemical process
ΔG=ΔH-TΔS
Units are J mol^-1 or kJ mol^-1
If enthalpy and entropy changes are measured under standard state conditions of 25°C, 100kPa and solutions of 1M, then
ΔG°=ΔH°-TΔS°

Question 55

How is enthalpy and entropy drive represented in gibbs energy equation

Answer 55

Enthalpy drive: qualitatively represented by ΔH°
Entropy drive: qualitatively represented by -TΔS°

Question 56

Properties of ΔG

Answer 56

Cannot be measured directly. Only ΔG can be measured.
Is independent of the path of transformation and depends only on difference in free energy of products and reactants.
Size of ΔG doesn’t determine rate of reaction
Sign determines whether reaction is spontaneous or not at certain temperatures

Question 57

Define spontaneous energy

Answer 57

One that occurs without a continual input of energy. Often a spark or flame is required to start a spontaneous reaction. This provides sufficient energy to overcome the activation energy barrier. Spontaneity like ΔG isn’t related to the rate of reaction.

Question 58

If ΔG<0, ΔG=0 or ΔG>0, what happens

Answer 58

If ΔG<0, the reaction is spontaneous and will go to completion without continuous external energy input. Called an exergonic reaction.
If ΔG=0, reaction is neither spontaneous or non-spontaneous. Both forward and reverse processes may occur simultaneously or not at all and are at equilibrium
If ΔG>0, the reaction is non-spontaneous so a continuous energy input is required to force the reaction to occur. Called an endergonic reaction.

Question 59

When are reactions spontaneous or not, four mark response

Answer 59

Reaction types: ΔH>0. ΔH<0
ΔS<0 ΔS>0 ΔS<0 ΔS>0
ΔG=ΔH-TΔS Low T Low T. ΔG=ΔH-TΔS
+. + ΔG=ΔH-TΔS ΔG=ΔH-TΔS - -
ΔG>0 + - +. - ΔG<0
non-spontaneous ΔG>0 ΔG<0 spontaneous
non-spontaneous spontaneous
High T High T
ΔG=ΔH-TΔS ΔG=ΔH-TΔS
+ - - +
ΔG<0 ΔG>0
spontaneous non-spontaneous
Forward enthalpy (ΔH<0) and entropy drive (ΔS>0) always spontaneous
Forward entropy drive (ΔS>0) and reverse enthalpy drive (ΔH>0) is spontaneous at high T
Reverse entropy drive (ΔS<0) and forward enthalpy drive (ΔH>0) is spontaneous at low T
Reverse entropy (ΔS<0) and enthalpy drive (ΔH<0) is never spontaneous

Question 60

Impact of temperature on ΔG

Answer 60

If ΔS is positive, then ΔG will decrease as temperature increases. By increasing temperature, the reaction with ΔS>0 are more likely to become spontaneous.

If ΔS is negative, then ΔG will decrease as temperature decreases. By decreasing temperature, the reaction with S<0 are more likely to become spontaneous.

Question 61

Why are reverse entropy and enthalpy drives never going to produce a spontaneous reaction

Answer 61

This is because the ΔG will be always positive meaning a reaction with this will not proceed naturally.

Question 62

Why do forward entropy and enthalpy drives produce a spontaneous reaction

Answer 62

This is because the ΔG will always be negative meaning a reaction with this will proceed naturally.