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Thermodynamics, image, image - Coggle Diagram

- - - - Enthalpy change when water is changed from solid to liquid to gas
        
        Enthalpy change of fusion and vaporization are positive value
        
        Thus transition of both are endothermic
        
        If enthalpy change value is negative, then transition is exothermic
  - - - Example
        
        A Cl2(g) molecule has a bond energy of 244 kJ/mol.
        
        It absorbs 244 kJ of energy to break the Cl-Cl bond of one mole of chlorine in its gaseous state.
        
        It releases 244 kJ of energy to form the Cl-Cl bond of one
        mole of chlorine in its gaseous state.
      - Number of bonds broken:
        5(C-H) + 1(C-C) + 2(N-H) + (C-N)
        Energy is absorbed in the bond breaking process.
        ∆Hbond breaking is positive.
        ∆Hbond breaking
        = [5(C-H) + 1(C-C) + 2(N-H) + (C-N)] = [5(412) + (348) + 2(388) + 305] = + 3489kJwing reaction to produce ethene (CH2=CH2) and ammonia (NH3) in the presence of heat.
      - Number of bonds formed: 4(C-H) + 1(C=C) + 3(N-H)
        Energy is released in the bond forming process.
        ∆Hbond forming is negative.
        ∆Hbond forming
        = [4(C-H) + 1(C=C) + 3(N-H)] = [4(412) + 1(612) + 3(388)] = - 3424k
      - The enthalpy change of a reaction is equal to the sum of the amount of energy absorbed for the bond breaking process and the sum of energy released during the bond forming process
        
        ∆ rxn H = ∆Hbond breaking + ∆Hbond forming = + 3489kJ + (- 3424kJ) = +65kJ
  - - - Element A + Element B = One more compound AB
        
        The reference state of an element is its most stable state at at the specified temperature and 1 bar
        
        Standard enthalpy of formation is expressed as enthalpy per
        mole of the compound formed.
        
        Example
        
        the standard enthalpy of formation of liquid benzene
        at 298K, is + 49.0 kJ/mol.
        
        6C (s, graphite) + 3 H2(g)  C6H6 (l) Δf
        H ° =+ 49.0 kJ/mol
    - - Monoatomic elements (e.g. C, Na, Zn)
        seven diatomic elements – H2, N2, O2, F2, Cl2, Br2 and I2
        
        standard enthalpy of formation = Zero
        
         Example: ∆fH
        
        H°(Cl2) = 0 kJ/mol.
  - - - Exothermic reaction
        
        Products have lesser
        energy than reactant
        
        Heat released to the
        surroundings
        
        ∆H negative
      - Endothermic reaction
        
        Products have more energy
        than reactants
        
        Heat absorbed from
        surroundings
        
        ∆H positive
    - - For a reaction or physical process (ΔH°)
        Is enthalpy difference between the products in standard state and the reactants in standard state at same temperature
        
        Example:
        
        The standard enthalpy of vaporization (ΔvapH°) of water is the enthalpy
        change per mole when a pure water at 1 bar vaporizes to a gas at 1 bar
        
        H2O (l)  H2O (g) ΔvapH° (373K) = +40.66kJ/mol
      - Standard State
        
        standard state of a substance at a specified
        temperature is its pure form at 1 bar
        
        1 bar = 0. 0.986923 atm
        
        Example:
        
        Standard state of ethanol at 298K, 1 bar is liquid.
        
        298 K = 25 °C
        
        ° means standard state
  - - - calculated from the standard enthalpies change of formation of reactants and products in equation below.
        
        Formula could only be used with the standard enthalpies change of formation of reactants (∆fH° (reactants)) and products (∆f(products))
        
        Elements in their standard states are not included in ∆rxnH° calculations, as ∆f H° for an element in its standard form is zero.
        
        ΔrxnH °
        
        Calculated Unit is kJ
        Thermodynamics table is kJ/mol
      - Methane undergoes combustion with oxygen to form water and
        carbon dioxide
        
        CH4 (g) + 2O2 (g)  2H2O (l) + CO2 (g)
        
        Using the standard enthalpies change of formation (∆f H°) of compounds given in the table below, calculate the enthalpy change of this combustion reaction.
- - - - A Hess Cycle can be used to work out Enthalpy Changes
        
        Hess Cycle
        
        Enthalpy change of the reaction is Sum of Products(H2) + Sum of Reactants(H1)
        
        Positive reaction is endothermic
        
        positive change of H means energy is stored
        
        Negative reaction is Exothermic
        
        Negative change of H means energy is released
        
        Δ H2 is between the reactants and the elemental
        
        Δ H3 is between the elemental and the product
      - Route 1: Reactant to product
      - Route 2: Reactant to element to product
    - - By Hess' Law: A + B + C = D
        
        Form the first equation to get the first of Δ H2
        
        Form the second equation to get the second of ΔH3
        
        Enthaply change = Δ H2 + Δ H3
- - - - cant be created nor destroyed
        
        total energy in an isolated system is a constant
      - is conserved
        
        determine the amount of energy in the system
        
        equation of total energy in the system
        
        equation of initial internal energy of the system
        
        equation of work done on the system
        
        equation of heat added or removed from the system
        
        Pravin
      - can be converted to other form of energy
        
        for example for cycling activity
        
        kinetic energy
        
        potential energy
        
        all are equivalent to constant
        
        cellular energy
        
        heat