Chem 2 IB Unit 6 – Energy and Thermochemistry
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Chem 2 IB Unit 6 – Energy and Thermochemistry Energy – the capacity to do work Two major types - PowerPoint PPT Presentation

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Chem 2 IB Unit 6 – Energy and Thermochemistry Energy – the capacity to do work Two major types Kinetic – E of motion Potential – E stored-up (can do work, but not doing it yet ) . W=F x d. First Law of Thermodynamics Three Parts: E of universe is constant E cannot be created or destroyed

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W=F x d

  • First Law of Thermodynamics

    • Three Parts:

      • E of universe is constant

      • E cannot be created or destroyed

      • E can be transformed

    • Average KE of particles  temperature

    • So how are temperature and heat related?

  • Heat – how energy is transferred from one substance to another

    • Higher kinetic energy  lower kinetic energy

    • Hot  cold

    • When there are energy changes in chemical reactions there are two parts to the universe. The SYSTEM (part of the rxn i.e. reactants/products) and the SURROUNDINGS (everything else).

    • The E w/in a system is called the INTERNAL ENERGY and is = to the sum of KE and PE of all particles in the system.

    • It’s impossible to truly isolate system from surroundings during rxn; more important to measure Δ in E of a system when rxn happens. The ΔE will occur in form of work (W) or heat (q), so…


W = F x d = F x Δh

  • Since P=F/A and F=PA, then…

W = P x A x Δh = PΔV

  • Remember: the sign for Δs in E always represents the system’s point of view. A positive ΔV means system has done work on surroundings (lost E), so we represent W as –PΔV

ΔE = q-PΔV

  • This value for heat (q) is important, b/c rxns are usually in open containers where V can Δ; so, this is given a special symbol, ΔH, and called “standard enthalpy change” for the rxn

Exothermic -

Energy is given off (lost) by the system

Endothermic -

Energy is absorbed (gained) by the system

  • Here’s a table to help with endothermic/exothermic ideas:

  • IB ♥’s to ask questions about…

    • Endothermic process  Take in E → DECREASE in temperature

  • Exothermic process give out E  temperature INCREASE

  • What about relative stabilities? Show with an enthalpy level diagram.

  • Q=CΔT

    • Q=heat…in normal lab conditions (can Δ volume )  (not in a

    • bomb calorimeter), q≈ΔH  ΔH=CΔT

    • C=heat capacity (not Cp – specific heat capacity)

    C = m x Cp

    • If Δ temp easily (ex – metals ) lower Cp

    • If d/n/Δ temp easily (ex – plastic)  higher Cp


    • Some values to know…

      • Cp H2O = 4.18 J/gC

      • Cm H2O = 75.3 J/molC (molar heat capacity  J/molK)

    Ex: The Cp of Al is 0.900 J/g 0 C. How much energy is needed to raise the temperature of an 851 g block of Al from 22.8 to 94.6 0 C? What is the molar heat capacity of Al?

    Ex: 15.0 g of copper is heated to 75.0 0C and added to 50.0g of water at 25.0 0C. If the final temperature of both is 26.4 0C, what is the specific heat capacity of the metal?

    Average Bond Enthalpies

    (Same as Bond Dissociation Energy in Chem 1)

    • Break substance up into individual gaseous

    • diatomic molecules

    • Then, measure energy required to break them

    • apart

    • Luckily for us, values already established 

    • listed in IBDB and on Handout

    • These values can be used to determine ΔH for a

    • reaction

      • Let’s try…(drumroll)…combustion of

      • methane!

        • CH4 + 2O2 CO2 + 2H2O

    Heats of formation
    Heats of Formation

    ∆H for a reaction can be determined for a chemical change when you know the change in enthalpy for the formation (∆Hf) of each reactant and product.

    B/C different conditions will result in different values for ∆Hf a Thermodynamic Standard State has been defined as 298.15 K (250C), 1 atm of pressure for each gas, and 1M concentration for each solution (n.b. – this is not STP).

    Standard Heats of Formation - ∆Hfө:

    The enthalpy change for the hypothetical formation of 1 mol of a substance in its standard state from the most stable forms of its constituent elements in their standard states.

    To calculate the change in enthalpy for a reaction (∆Hrxn):

    ∆Hrxn = Σnp∆Hfө(products) - Σnr∆Hfө(reactants)

    Ex: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

    [1 mol(-394kj/mol) + 2 mol(-286kj/mol)] – [1 mol(-75 kj/mol) + 2(0 kj/mol)]

    = (-394 kJ – 572 kJ) – (-75 kJ) = -966 kJ + 75 kJ = - 891 kJ

    Now rewrite the equation as a Thermodynamic Equation for the formation of 1 mol of water:

    ½ CH4(g) + O2(g) → ½ CO2(g) + H2O(g) + 446 kJ

    What about for other amounts reacted or formed? For example:

    If only 5.83 g of water is produced, how much energy is released?

    5.83 g x 1 mol x - 891 kJ = - 144 kJ

    18.02g 2 mol H2O

    n. b. - ∆Hfө for elemental substances Al, O2, S8, P4, etc. is always 0.

    For reversible reactions – The sign for ∆H in the reverse rxn is opposite that for the forward rxn. So if the the forward rxn is exo, the reverse rxn is?


    Hess’s Law

    Since energy and therefore enthalpy change is a state function, it is possible to determine ΔHfor reactions in which energy changes are difficult to measure (anything come to mind?).

    This is because the value for ΔH is the same no matter what path is taken. Thus we can sum enthalpy changes for a set of reactions which together give us the overall reaction for which we are trying to determine the value of ΔH.

    Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.

    Example: Use the following information to calculate ΔHcombθ for the combustion of methane.

    CH4(g) + O2(g) → CH2O(g) + H2O(g) ΔHθ = -284 kJ

    CH2O(g) + O2(g) → CO2(g) + H2O(g) ΔHθ = -518 kJ

    H2O(l) → H2O(g) ΔHθ = -44 kJ

    Enthalpy cycles
    Enthalpy Cycles

    An enthalpy cycle is a way of representing energy changes that take place in a process or reaction. It is similar to the idea of Hess’s Law since ΔH values in parts of the cycle can be used to calculate an unknown ΔH value in another part of the cycle.

    Consider the enthalpy cycle below:


    3 C(s) + 4 H2(g) 5 O2(g)

    C3H8(g) + 5 O2(g)



    3 CO2(g) + 4 H2O(g)

    Notice that the reactants in equation 1 (ΔH1 above the arrow) could form the products in equation 3 (ΔH3). The products from equation 1 could also react to form the same products (ΔH2).

    We can write reactions 2 and 3 as we would if we were solving a Hess’s Law problem:

    C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH2

    3 C(s) + 4 H2(g) 5 O2(g) + 3 CO2(g) + 4 H2O(g) ΔH3

    Notice that if equation 2 is reversed then the CO2 and H2O are cancelled out and we get equation 1. (Work it out)

    So what this tells us is that to determine the value for ΔH1 the value for ΔH3 could be added to the opposite (negative) of ΔH2 (Since the reverse of reaction 2 would produce the products of equation 1).

    That is ΔH1 = ΔH3 - ΔH2 (just as we would expect using Hess’s Law)