Energetics
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Energetics. Exothermic and Endothermic reactions. Energy = force x distance (Joules). In chemical reactions, we need energy usually in the form of heat. Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products.

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Energetics

Energetics


Exothermic and endothermic reactions

Exothermic and Endothermic reactions

Energy = force x distance (Joules)

In chemical reactions, we need energy usually in the form of heat.

Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products.

Exothermic reactions

Endothermic reactions


Heat and temperature

Heat and Temperature

  • Heat is the energy transferred between objects that are at different temperatures.

  • The amount of heat transferred depends on the amount of the substance.

    • Energy is measured in units called joules (J).


Energetics

  • Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance.

  • It does not depend on the amount of the substance.

Do both beakers contain the same amount of heat?


Energy changes in chemical reactions

Energy Changes in Chemical Reactions

  • All chemical reactions are accompanied by some form of energy change

  • ExothermicEnergy is given out

  • Endothermic Energy is absorbed

  • Activity : observing exothermic and endothermic reactions


Enthalpy h and enthapy change h

Enthalpy (H)and Enthapy change(ΔH)

  • Enthalpy (H) is the heat content that is stored in a chemical system.

  • We measure the change in enthalpy ∆H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in

    kilojoules per mole (kJmol-1).

    ∆H = H(products) – H(reactants)


Enthalpy level diagram exothermic change

Enthalpy Level Diagram -Exothermic Change

  • For exothermic reactions, the reactants have more energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants)

  • ∆H is negative since H(products) < H(reactants)

  • There is an enthalpy decrease and heat is released to the surroundings

Enthalpy


Examples of exothermic reactions

Examples of Exothermic Reactions

  • Self-heating cans

    • CaO (s) + H₂O (l)  Ca(OH)₂ (aq)

  • Combustion reactions

    • CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)

  • neutralization (acid + base)

    • NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)

  • Respiration

    • C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)


Enthalpy level diagram endothermic change

Enthalpy Level Diagram -Endothermic Change

  • For endothermic reactions, the reactants have less energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants)

  • ∆H is positive since H(products) < H(reactants)

  • There is an enthalpy increase and heat is absorbed from the surroundings

Enthalpy


Examples of endothermic reactions

Examples of Endothermic Reactions

  • Self-cooling beer can

    • H ₂O (l)  H₂O (g)

  • Thermal decomposition

    • CaCO₃ (s)  CaO (s) + CO ₂ (g)

  • Photosynthesis

    • 6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)


  • Specific heat capacity

    Specific heat capacity

    • Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin.

    • Uint : Jg-1 0C-1

    The specific heat capacity of alminium is 0.90 Jg-1 0C-1 .

    If 0.90J of energy is put into 1g of aluminium, the temperature

    will be raised by 10C.

    Calculating heat absorbed and released

    q = c × m × ΔT

    q = heat absorbed or released

    c = specific heat capacity of substance

    m = mass of substance in grams

    ΔT = change in temperature in Celsius


    Calorimetry

    Calorimetry

    • Heat given off by a process is measured through the temperture change in another substance (usually water).

    • Due to the law of conservation of energy, any energy given off in a process must be absorbed by something else, we assume that

      the energy given out will be absorbed by the water and cause a temperature change.

    • calculate the heat through the equation Q = mcΔT


    Example

    Example

    How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg-1 0C-1) from 500C to 700C?


    Enthalpy change of combustion reactions

    Enthalpy change of combustion reactions

    • The standard enthalpy change of combustion for a substance is the heat released when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions.

    • Example,

      CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHƟc=-698 kJmol-1

    • The heat given out is used to heat another substance,e.g. water with a known specific heat capacity.

    • The experiment set-up can be used to determine the enthalpy change when 1 mole of a liquid is burnt.

    Example : refer to page 185


    Problems with calorimetry

    Problems with calorimetry

    • Loss of heat to the surroundings (exothermic reaction); absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter.

    • Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for.

    • Others include – e.g incomplete combustion. Some of the ethanol could be used to produce CO & soot & water (less heat is given out)

    Use bomb calorimeter – heavily insulated & substance is ignited

    electronically with good supply of oxygen


    Example1

    Example

    • If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change when 1 mole of methanol is burnt.

      Note: Specific heat capacity of water is 4.18 Jg-1 0C-1

    Practice questions page 187 #1-4


    Enthalpy change in solutions

    Enthalpy change in solutions

    Enthalpy change of neutralisation (ΔHn)

    • The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions.

      Example,

      NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1

    • The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.

    • Reaction between strong acid and strong base involves

      H+(aq) + OH-(aq)  H2O(l) ΔHƟ=-57 kJmol-1

      For sulfuric acid, the enthalpy of neutralisation equation is

      ½ H2SO4(aq) + KOH(aq)  ½K2SO4(aq) + H2O(l) ΔHƟ=-57 kJmol-1

    Example : refer to page 188


    Enthalpy change in solutions1

    Enthalpy change in solutions

    Enthalpy change of neutralisation (ΔHn)

    • The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions.

    • Example,

      NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1

      The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.

      Enthalpy change of solution (ΔHsol)

    • The enthalpy change when 1 mol of solute is dissolved in excess solvent to form a solution of ‘infnite dilution’ under standard conditions.

      NH4 NO3(s) in excess water  NH4 + (aq)+ NO3 -(aq)

    Example : refer to page 188


    Energetics

    For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol-1 (less exothermic)

    CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)

    ΔHƟ=-55.2 kJmol-1

    • Some of the energy released is used to ionise the acid.


    Example2

    Example

    • 200.0cm3 of 0.150 M HCl is mixed with 100.0cm3 of 0.350 M NaOH. The temperature rose by 1.360C. If both solutions were originally at the same temp, calculate the enthalpy change of neutralisation.

      Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is 4.18J Jg-10C-1.

    -56.8kJmol-1


    Possible errors

    Possible errors

    The experimental change of neutralisation is

    -56.8 kJmol-1

    The accepted literature value is -57.2 kJmol-1

    (1) Heat loss to the environment.

    Assumptions that

    the denisty of NaOH and HCl solutions are the same as water.

    the specific heat capacity of the mixture are the same as that of water


    Example3

    Example

    When 3 g of sodium carbonate are added to 50 cm3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is

    4.18J Jg-10C-1.

    Example : refer to page 189 dissolving ammonium chloride


    Possible errors page 189

    Possible errors (page 189)

    The experimental change of solution is

    +13.8 kJmol-1

    The accepted literature value is 15.2 kJmol-1

    (1) Absorption of heat from the environment.

    Assumptions that the specific heat capacity of the solution is the same as that of water

    The mass of ammonium chloride is not taken into consideration when working out the heat energy released.


    Energetics

    Example

    100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

    First step

    Make sure you understand the graph.

    Extrapolate to determine the change in temperature.

    The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used?


    Energetics

    100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

    Determine the limiting reactant

    Calculate Q

    Calculate the enthalpy for the reaction.


    Enthalpy changes of combustion of fuels

    Enthalpy changes of combustion of fuels

    The following measurements are taken:

    • Mass of cold water (g)

    • Temperature rise of the water (0C)

    • The loss of mass of the fuel (g)

      We know that it takes 4.18J of energy to raise the temperature of 1g of water by 10C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg-1K-1.

      Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules)

    • If one mole of the fuel has a mass of M grams, then:

    • Enthalpy transfer = m x 4.18 x T x M/y

    • where y is mass loss of fuel.


    Example4

    Example

    Given that:

    Vol of water = 100 cm3

    Temp rise = 34.50C

    Mass of methanol burned = 0.75g

    Specific heat capacity of water = 4.18 Jg-10C-1

    Calculate the molar enthalpy change of the combustion of methanol.

    What is the big assumption made with this type of experiment?


    Hess s law

    Hess’s Law

    States that

    • If a reaction consists of a number of steps, the overall enthalpy change is equal to the sum of enthalpy of individual steps.

    • the overall enthalpy change in a reaction is constant, not dependent on the pathway take.


    Standard enthalpy changes h

    Standard enthalpy changes, ΔHƟ

    • measured under standard conditions: pressure of 1 atmosphere (1.013 x 105 Pa), temperature of 250C (298K) and concentration of 1 moldm-1.

      e.g.

      N2(g) + 3H2(g) 2NH3(g) ΔHƟ = -92 kJmol-1

      The enthalpy change of reaction is -92 kJmol-1

      92 kJ of heat energy are given out when 1 mol of nitrogen reacs with 3 mols of hydrogen to form 2 mols of ammonia.


    Reaction in aqueous soln

    Reaction in aqueous soln

    Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide.

    NaOH(aq)

    NaOH(s) NaCl(s) + H2O(l)

    + HCl(aq)

    1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔHƟ1=-43kJmol-1

    2. NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ1=-57kJmol-1

    3. NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l).

    Indirect path

    + HCl(aq)

    + H2O(l)

    ΔH2

    ΔH1

    Direct path


    Combustion reaction using cycles

    Combustion reaction (using cycles)

    Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide.

    C(s) + O2(g)  CO2(g) ΔHƟ=-394 kJmol-1

    2C(s) + O2(g)  2CO(g) ΔHƟ= -222kJmol-1

    2CO(s) + O2(g)  2CO2(g)

    2CO(g) + O2(g) 2CO2(g)

    ΔHƟ

    ΔHƟ = -(-222)+2(-394) = -566kJmol-1


    Combustion reaction manipulating equations

    Combustion reaction (manipulating equations)

    Example : refer to page 196 evaporation of water & 197 formation of ethanol from ethene


    Example decomposition reaction

    *Example : Decomposition reaction

    Calculate the enthalpy change for the thermal decomposition of calcium carbonate.

    CaCO3(s)  CaO(s) + CO2(g)

    CaCO3(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-17 kJmol-1

    CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-195kJmol-1

    CaCO3(s) CaO(s) +CO2(g)

    CaCl2(aq) + H2O(l) +CO2(g)

    ΔH

    Direct path

    + 2HCl(aq)

    + 2HCl(aq)

    -17 kJmol-1

    -195kJmol-1

    Indirect path


    Example enthalpy of hydration of an anhydrous salt

    *Example : Enthalpy of hydration of an anhydrous salt.

    Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change.

    CuSO4(s) +5H2O(l)  CuSO4.5H2O (s)

    CuSO4(s) +5H2O(l) CuSO4.5H2O (s)

    Cu2+(aq) + SO42-(aq)

    ΔH

    Direct pathway

    ΔH1

    ΔH2

    Indirect pathway


    Example5

    *Example

    From the following data at 250C and 1 atmosphere pressure:

    Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1

    Eqn 2: 3CO(g) + O3(g) 3CO2(g)ΔHƟ=-992 kJmol-1

    Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O2(g) O3 (g)


    Example6

    *Example

    Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions.

    C (s,graphite) + O2(g) CO2 (g) ΔHƟ=-393 kJmol-1

    C (s, diamond) + O2(g) CO2(g)ΔHƟ=-395 kJmol-1


    Energetics

    Practice questions page 199 #7-9


    Bond enthalpies bond energies

    Bond enthalpies (Bond energies)

    • Enthalpy changes can also be calculated directly from bond enthalpies.

    • The bond enthalpy is the amount of energy required to break one mole of a specified covalent bond in the gaseous state.

    • For diatomic molecule the bond enthalpy is defined as the enthalpy change for the process X-Y(g) X(g) + Y(g) [gaseous state]


    Bond enthalpies

    Bond Enthalpies

    • Bond enthalpy can only be calculated for substances in the gaseous state.

      Br2(l)  2Br(g) ΔHƟ= 224 kJmol-1

    atomisation

    2 x ΔH Ɵat

    Br2(l) 2Br(g)

    Br2(g)

    ΔH Ɵvap

    enthalpy change

    of vaporisation

    Br-Br

    bond enthalpy

    Energy must be supplied to break the van der Waals’ forces between the Bromine

    molecules and to break the Br-Br bonds. Endothermic process


    Average bond enthalpies

    Average bond enthalpies

    • Ave bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH4 , alkanes and other hydrocarbons.


    Some average bond enthalpies

    Some average bond enthalpies

    Refer to page 201


    Bond breaking and forming

    Bond breaking and Forming

    When a hydrocarbon e.g. methane (CH4) burns,

    CH4 + O2 CO2 + H2O

    What happens?


    Energetics

    H

    O

    O

    +

    C

    O

    O

    H

    H

    H

    O

    O

    C

    O

    H

    H

    O

    H

    H

    Enthalpy Level (KJ)

    Bond Breaking

    O

    O

    C

    H

    ENERGY

    O

    H

    ENERGY

    O

    H

    CH4 + 2O2 CO2 + 2H2O

    H

    Bond Forming

    4 C-H

    2 O=O

    4 H-O

    2 C=O

    Progress of Reaction

    Energy Level Diagram


    Bond breaking and forming1

    H

    O

    O

    +

    C

    O

    O

    H

    H

    H

    O

    O

    C

    O

    H

    H

    O

    H

    H

    Bond breaking and Forming

    CH4 + 2O2 CO2 + 2H2O

    Why is this an exothermic reaction (produces heat)?


    Energetics

    H

    O

    O

    +

    C

    O

    O

    H

    H

    H

    O

    O

    C

    O

    H

    H

    O

    H

    H

    Break  Form

    CH4 + 2O2 CO2 + 2H2O

    Energy absorbed when bonds are broken

    = (4 x C-H + 2 x O=O)

    Energy given out when bonds are formed

    = ( 2 x C=O + 4 x H-O)

    = 4 x 412 + 2 x 496

    = 2640 kJ/mol

    = 2 x 803 + 4 x 464

    = 3338 kJ/mol


    Energetics

    Energy absorbed when bonds are broken (a)

    = 2640 kJ/mol

    Energy released when bonds are formed (b)

    = 3338 kJ/mol

    Enthalpy change, ΔH

    = ∑(bonds broken) - ∑(bonds made)

    = a + (-b)

    = 2640 – 3338

    = -698 kJ/mol

    Why is this an exothermic reaction (produces heat)?


    Example7

    Example

    What can be said about the hydrogenation reaction of ethene?

    H H

    C=C (g) + H-H (g)  H-C-C-H (g)

    H H

    H

    H

    H

    H


    Example8

    Example

    What can be said about the combustion of hydrazine in oxygen?

    H H

    N-N (g) + O=O (g)  NΞN (g) + 2 O (g)

    H H

    H

    H


    The bond enthalpy from an enthalpy change of reaction

    The bond enthalpy from an enthalpy change of reaction

    Example

    Calculate the mean Cl-F bond enthalpy given that

    Cl2(g)+ 3F2(g) 2ClF3(g)ΔHƟ= -164 kJmol-1

    Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1


    Using bond enthalpies enthalpies of atomisation

    Using bond enthalpies & enthalpies of atomisation

    Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions.

    Example C(s)  C(g)

    Calculate the enthalpy change for the process

    3 C(s)+ 4H2(g) C3H8(g)ΔHƟ= -164 kJmol-1

    Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1

    (ΔH Ɵat )

    ΔH Ɵat = 715 kJmol-1

    Practice questions page 206 #10,12,13


    Hess s law example

    Hess’s Law - example

    The combustion of both C and CO to form CO2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle.

    ΔHx = -393 – (-283)

    = - 110 kJmol-1

    ΔHx

    C(s)+ ½O2(g) CO(g)

    CO2(g)

    ½O2(g)

    -393kJmol-1

    ½O2(g)

    -283kJmol-1


    Hess s law example1

    Hess’s Law - example

    Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbon, hydrogen and methane are -393, -286 and -890kJmol-1 respectively.


    Entropy

    Entropy

    Dissolving sugar .

    Sugar molecules are dispersed

    throughout the solution and

    are moving around.

    More disordered or random.

    • Other examples:

    • melting ice

    • H2O(s)  H2O(l)

    • evaporating water

    • H2O(l)  H2O(g)


    The sign of entropy s

    The sign of entropy, S

    Increasing entropy

    Entropy (S) : amount of disorder Unit : JK-1mol-1

    SƟ : standard entropy Δ SƟ : entropy change

    If Δ SƟ > 0 => increase in entropy => increase in disorder

    E.g. H2O(l )  H2O(g) Δ SƟ =+119JK-1mol-1

    If Δ SƟ < 0 => decrease in entropy => decrease in disorder

    E.g. NH3(g ) + HCl(g)  NH4Cl(s) Δ SƟ = - 285JK-1mol-1


    Factors affecting entropy

    Factors affecting entropy

    (1) State of matter

    • Gas (most) particle motion is more random in a gas

    • Liquid (middle) particle motion is less random than a gas but more than a solid

    • Solid (least) particle motion is restricted. Possible positions for molecules are restricted

    • Examples

      • (changing state) H2O(l)  H2O(g)

      • (changing state) H2O(s)  H2O(l)

    ΔS(gas) > ΔS(liquid) > ΔS(solid)


    Energetics

    (2) Temperature

    • Comparing two gasses, one at 20 C and one at 80 C

    • Molecules in the 80 C gas have more kinetic energy, they are moving more and colliding more


    Energetics

    (3) The number of molecules

    • More molecules means more possible positions relative to the other molecules

      • (more moles and change of state)

        Li2CO3(s)  Li2O(s) + CO2(g)

      • (more moles)

        MgSO48H2O  Mg2+(aq) + SO42-(aq) + 8H2O(l)

        (4) More complex molecules have higher entropy values


    Predict the sign of s

    Predict the sign of ΔSƟ

    Is there an increase or decrease in disorder of the system?

    Is there an increase or decrease in the no. of moles of gas?

    For reaction where the no. of moles of gas is the same on both sides,

    the ΔS = 0.

    E.g. F2(g) + Cl2(g)  2ClF(g)

    Practice Qn 24from pg 230 (textbook


    Example9

    Example

    Which of the following reactions has the largest ΔS value?

    • CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)

    • 2Al(s) + 3S(s)  Al2S3(s)

    • CH4(g) + H2O(l)  3H2(g) + CO(g)

    • 2S(s) + 3O2(g)  2SO3(g)


    Calculate s

    Calculate ΔSƟ

    Entropy change =

    total entropy of products – total entropy of reactants

    ΔSƟ =∑ ΔSƟproducts - ∑ ΔSƟreactants

    E.g. Calculate the standard entropy change for the reaction CH4 (g) + 2O2(g) CO2(g) + 2H2O(l)

    Use standard entropy from pg 230 (textbook)


    Spontaneity

    Spontaneity

    • Spontaneous reaction – one that occurs without any outside influence (no input of energy)

    • A spontaneous reaction does not have to happen quickly.

    • E.g.

      4Na(s) + O2(g) 2Na2O(s) can happen by itself.


    Gibbs free energy g

    Gibbs free energy, G

    • When heat is released in a chemical reaction, the surrounding is hotter and particles move around more => entropy increases.

      The entropy change ,ΔS is related to

      the enthalpy change, ΔH of the system.

      ΔG= ΔH – TΔS

      ΔG : free energy change

      ΔGƟ : standard free energy change

    For a reaction to be spontaneous, ΔG < 0


    Energetics

    • Temperature, T should be in Kelvin, K

      0°C = 273K (273.15K)

    • Check the units of ΔS, entropy is often given in JK–1mol–1 but must be converted to

      kJK–1mol–1

      ΔGө = ΔHө – TΔSө

    • ө = standard conditions, 25°C (298K) and 1 atm (101.3 kPa)


    Calculate g u sing g h t s

    Calculate ΔGƟusing ΔGƟ = ΔHƟ – TΔSƟ

    Given that the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then state whether the reaction is spontaneous at 25 C

    C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)


    Calculate g u sing standard free energy of formation

    Calculate ΔGƟusing standard free energy of formation

    ΔGƟ =∑ ΔGfƟproducts - ∑ ΔGfƟreactants

    standard free energy of formation : free energy change for the formation of 1 mole of substance from its elements in their standard states & under standard conditions.

    CalculateΔGƟ for the reaction

    CaCO3(s)  CaO(s) + CO2(g) given that

    Practice Qn from pg 235 (textbook


    Spontaneity of reactions gibbs free energy g

    Spontaneity of reactions – Gibbs Free Energy, ΔG

    ΔH : Enthalpy Change ΔS : Entropy Change

    For a spontaneous reaction ΔG is negative (–)

    For a non–spontaneous reaction ΔG is positive (+)


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