Energetics
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Energetics. First Law of Thermodynamics. The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed. Heat and Temperature.

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Energetics

Energetics


First law of thermodynamics

First Law of Thermodynamics

Thelaw of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.


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.

    • Though energy has many different forms, all 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 and enthapy change

Enthalpy and Enthapy change

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

  • We cannot measure enthalpy directly, only 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)

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


Using a calorimeter

Using a Calorimeter

The particles are dissolved in the solution

The heat is exchanged from the particles into the solution

The heat exchange is measured using the thermometer

There are limitations in using a simple calorimeter which need to be known.


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.


Enthalpy diagram exothermic change

Enthalpy 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 diagram endothermic change

Enthalpy 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

      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


    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?


    Example1

    Example

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


    Example2

    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.


    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.


    Example3

    Example

    • 50 cm3 of 0.100 moldm-1 silver nitrate solution was put in a calorimeter and 0.200g of zinc powder is added. The temperature of th solution rose to 4.30C. Deduce which reagent was in excess and calculate the the enthalpy change for the reaction (per mole o Zn that reacts). Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is 4.18J Jg-10C-1. Ignore the heat capacity of metals and dissolved ions.


    Enthalpy change of combustion

    Enthalpy change of combustion

    • 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


    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.


    Energetics

    Example

    To determine the enthalpy of combustion for ethanol (see reaction), a calorimeter setup (below) was used. The burner was lit and allowed to heat the water for 60 s. The change in mass of the burner was 0.518 g and the temperature increase was measured to be 9.90 oC.

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


    Energetics

    The burner was lit and allowed to heat the water for 60 s. The change in mass of the burner was 0.518 g and the temperature increase was measured to be 9.90 oC.

    First step – calculate heat evolved using calorimetry

    Last step – determine ΔH for the reaction


    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.


    Enthalpy change of neutralisation

    Enthalpy change of neutralisation

    • 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.


    Energetics

    • 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


    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.


    Energetics

    Example

    50.0 cm3 of a 1.00 mol dm-3 HCl solution is mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7oC. After stirring and accounting for heat loss, the highest temperature reached was 33.5 oC. Calculate the enthalpy change for this reaction.

    NaOHHCl

    both 26.7o

    .

    26.7o 33.5o


    Energetics

    50.0 cm3 of a 1.00 mol dm-3 HCl solution is mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7oC. After stirring and accounting for heat loss, the highest temperature reached was 33.5 oC. Calculate the enthalpy change for this reaction.

    After writing a balanced equation, the molar quantities and limiting reactant needs to be determined.

    Note that in this example there is exactly the right amount of each reactant. If one reactant is present in excess, the heat evolved will associated with the mole amount of limiting reactant.


    Energetics

    50.0 cm3 of a 1.00 mol dm-3 HCl solution is mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7oC. After stirring and accounting for heat loss, the highest temperature reached was 33.5 oC. Calculate the enthalpy change for this reaction.

    • Next step – determine how much heat was released.

    • There are some assumptions in this calculation

    • Density of reaction mixture (to determine mass)

    • SHC of reaction mixture (to calculate Q)


    Energetics

    50.0 cm3 of a 1.00 mol dm-3 HCl solution is mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH. A neutralization reaction occurs. The initial temperature of both solutions was 26.7oC. After stirring and accounting for heat loss, the highest temperature reached was 33.5 oC. Calculate the enthalpy change for this reaction.

    Final step – calculate ΔH for the reaction


    Example5

    Example

    50.00 cm3 of 1.0 moldm-1 hydrochloric acid was added to 50.00 cm3 of 1.0 moldm-1 sodium hydroxide solution. The temperatre rose by 6.80C.

    Calculate the enthalpy change f neutralisation for the reaction. Assume that the density of the solution is 1.00gcm-1 and the specific heat capacity of the solutio is 4.18 Jg-10C-1.


    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.


    Example combustion reaction

    Example : Combustion reaction

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

    Route 1:C(s) + O2(g)  CO2(g) ΔHƟ1=-394 kJmol-1

    Route 2:C(s) + ½O2(g)  CO(g) ΔHƟ2= ?

    CO(s) + ½O2(g)  CO2(g) ΔHƟ1=-283 kJmol-1

    C(s) + O2(g) CO2(g)

    CO(s) + ½O2(g)

    ΔHƟ1

    Route 1

    ΔHƟ3

    ΔHƟ2

    Route 2


    Example reaction in aqueous soln

    Example : 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


    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

    -17 kJmol-1

    + 2HCl(aq)

    -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


    Example6

    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)


    Example7

    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


    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]

    • Average 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


    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

    = ∑(energy absorbed to break bonds) - ∑(energy released to form bonds)

    = a + (-b)

    = 2640 – 3338

    = -698 kJ/mol

    Why is this an exothermic reaction (produces heat)?


    Example8

    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


    Example9

    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


    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 carbo, hydrogen and methane are -393, -286 and -890 kJmol-1 resepctively.


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