Chemistry (Matter)

Thermo Chemistry

1. Enthalpy ( H) is the amount of heat content.
   1. Heat content is accounted for by a change in "heat flow" or enthalpy of the reaction system.
      1. Endothermic reaction: H > 0
         (i.e., H products >H reactants).
         Heat absorbed goes to increase the enthalpy of the reaction system.
      2. Exothermic reaction: H < 0
         (i.e., H products < H reactants).
         Heat is evolved at the expense of the reaction system.
   2. Thermochemical Equation: specify H in kilojoules/mole.
      1. CH_4(g) + 2O_2(g) " width="15" align="absmiddle" border="0" height="10"> CO_2(g) + 2H₂O_(l) + 890.3 kJ
         H = -890 kJ

         6.00kJ + H₂O_(s) " width="15" align="absmiddle" border="0" height="10"> H₂O_(l)
         H = +6.00kJ

         ! In some textbooks H is written as a product or reactant !

         The preceding is based upon the Law of Conservation of Energy (James Joule, 1818-1889, Joule also developed the First Law of Thermodynamics): energy is neither created nor destroyed in ordinary chemical or physical changes.

      2. Quantitative H
         H = q_{reaction mixture} (at constant temperature only)

         q = (m)(t)(C_p)

         q = heat absorbed by the water in joules (J)
         m = mass of substance
         t = t_final - t_initial
         C_p = specific heat of water = 4.184

         When using moles, molar heat capacity is used. The units are

         1 cal = 4.184 J

2. Calorimetry
   1. Coffee-cup calorimeter (only used for reactions in solution, must be at constant pressure)
      q_reaction=-q_water
   2. Bomb calorimeter (reaction gases, and must have constant volume)
      q_reaction=-(q_water+q_bomb)
      q_bomb=Ct, where C is the calorimeter constant (C_v of bomb x mass of bomb, really same equation)
   3. H vs. E for chemical reactions
      H=qp since E=qp-PV
      substituting gives H=E+PV
      where P will usually be in atmospheric pressure, and V is volume change at that pressure.
3. Laws of Thermochemistry
   1. The magnitude of H is directly proportional to the amount of reactant or product.
      -Thus H can be used as a conversion factor in a balanced equation to obtain amounts of reactant/product or H itself. (mole to mole ratio's).
   2. H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction.

      Problems 1: H Calculation

      1. When 1 mol of methane is burned at constant pressure, 890.3kJ of energy is released as heat. Calculate H for a process in which a 5.8 gram sample of methane is burned at constant pressure.
         CH_4(g) + 2O_2(g) " width="15" align="absmiddle" border="0" height="10"> CO_2(g) + 2H₂O_(l) + 890.3 kJ
         = 320 kJ
         H = -320 kJ

      2. For the reaction of methane with oxygen given in the notes, calculate the H in kJ if 5.8 grams of oxygen are consumed in the process.
         = 81 kJ
         H= -81kJ

      3. Ammonium nitrate, NH₄NO₃, is commonly used as an explosive. It decomposes by the following reaction:

         NH₄NO₃ " width="15" align="absmiddle" border="0" height="10"> N₂O_(g) + 2 H₂O_(g) + 37.0kJ

         H = -37.0 kJ

         If 72.0 grams of H₂O are formed from the reaction, how much heat was released?
         = 73.9 kJ

   3. Hess' Law: The value of H for a reaction is the same whether it occurs directly or in a series of steps (state function).

      H_total = H₁ + H₂

      often used to calculate H for one step, knowing H for all steps and for the overall reaction.
      **All of the laws of thermochemistry follow from the fact that the enthalpy H of a substance is one of its properties.**

4. Heats of Formation
   Molar heat of formation (H_f) is equal to the enthalpy change, H when one mole of the compound is formed from the elements in their stable forms at 25^oC and 1 atm is H^o (pronounced 'delta h naught'). H^o of a solution is of a 1M solution, at 1 atm and 25 ^oC.

   Heats of formation are usually negative quantities.

   H = H_{f products} - H_{f reactants} To apply the above relation, use the following rules:

   1. The contribution for each compound is found by multiplying the heat of formation in kJ per mole by the number of moles of compound, given by its coefficient in the balanced equation.

      Heats of formation can be found in appendix A-4

   2. Any element in its stable form is omitted.
      
   Can also apply to heats of formation to ions.

   Arbitrarily assign H⁺ ion to be zero. H_f H⁺_(aq) = 0

   Having established the above, a scale can be established with Hydrogen ion as the base.

   1. Calculate H⁰_rxn for the following reaction.

      2 C₃H_6(g) + 9 O_2(g) " width="15" align="absmiddle" border="0" height="10"> 6 CO_2(g) + 6 H₂O_(l)
      *Appendix 4*
      H_rxn = H_{f products} - H_{f reactants}
      H_rxn = [ 6 H₂O_(l) + 6 CO_2(g) ] - [ 2 C₃H_6(g) + 9 O_2(g) ]
      H_rxn = [ 6(-286 kJ) + 6 (-393.2 kJ) ] - [ 2(20.9 kJ) + 9(0)]
      H_rxn = [ -1716 kJ + -2361 kJ ] - [41.8 kJ ]
      H_rxn = = 2059
      

   2. 1. Calculate H⁰ for 2Al_(s) + Cr₂O_3(s) " width="15" align="absmiddle" border="0" height="10"> Al₂O_3(s) + 2Cr_(s).
      2. Compare this reaction to sample exercise 6.10, "thermite" reaction.

         Which reaction yields more energy per gram of metal formed?

      1. H_rxn = [ Al₂O_3(s) + 2 Cr_(s) ] - [ 2 Al_(s) + Cr₂O_3(s) ]
         H_rxn = [ (-1676 kJ) + 2(0 kJ) ] - [ 2(0 kJ) + -1128 kJ ]
         H_rxn = [ -1676 kJ ] - [ -1128 kJ ]
         H_rxn = = -10.1
      2. H_rxn = = -15.75
         "Thermite" reaction releases more energy per gram of metal formed.