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Reaction Energy & Kinetics in Chemistry

Discover the study of energy change in chemical reactions, from enthalpy to heat of formation. Learn about stability and Hess's Law in determining reaction energetics and driving forces.

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Reaction Energy & Kinetics in Chemistry

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  1. Chapters 16, 17, 18

  2. Reaction Energy and Reaction Kinetics • Thermodynamics: the study of energy change in chemical reactions • Substances have energy stored in them. It is stored in bonds, stored in pressure, and any other form of potential energy. • Substances also have kinetic energy (motion of the molecules). • The sum of the energy is the heat content known as Enthalpy (H). • Heat of reaction (enthalpy change): heat absorbed or released during a chemical reaction

  3. Exothermic versus Endothermic ReactantsProducts + heatReactants + heat Products The reaction is exothermic The reaction is endothermic.

  4. Heat of Formation (Hf) • Heat released or absorbed when onemole of a compound is formed from its elements • 2 H2 + O2 2 H2O + 572 kJ( not Hf ) • H2 + ½ O2 H2O + 286 kJ ( yes! Hf) Hf = - 286 kJ

  5. The Heat of Formation • EX 1: The heat of formation of NaCl is -411.15 KJ. Express this in three ways. 1. 2. Na(s) + ½ Cl2(g)  NaCl(s) + 411.15 KJ 3. Na(s) + ½ Cl2(g)  NaCl(s) ΔH = -411.15 KJ Na(s) + ½ Cl2(g) 411.15 KJ NaCl(s)

  6. The Heat of Formation • The heat of formation of HI is 26.48 KJ. Express this in three ways. 1. 2. ½H 2(g) + ½ I2(g) + 26.48 KJ  HI (s) 3. ½H 2(g) + ½I 2(g) HI(s) ΔH = +26.48 KJ HI(s) ½ H2(g) + ½ l2(g) 26.48 KJ

  7. Stability of Compounds • A high negative Hf means that the compound is very stable. • Why? The compound releases a lot of energy during its formation so it takes just as much to break the compound apart. • A high positive Hf means that the compound is very unstable. • Why? The compound took quite a bit of energy to make the elements join and will probably require very little activation energy to make the compound break apart to go to lower energy.

  8. Which is more stable? • CS 2(g)Hf = +117.07 Kcal/mole • FeCl3(s)Hf = -399.49 Kcal/mole • SO3(g)Hf = -395.72 Kcal/mole • Solid Iron III Chloride is more stable because it has a heat of formation that is more negative (-399.49 Kcal/mole).

  9. Hess’ Law • The overall enthalpy change in a reaction does not depend on the number of steps. The enthalpy change is equal to the sum of the enthalpy changes for the individual steps in the process.

  10. Hess’s Law • Steps: • Write equations to form compounds • Multiply the equation by the number of moles needed for the equation • Change the sign of H if the equation is written in the reverse • Add up the H values to get the overall value for the equation • Add up the equations to check your answer

  11. Hess’s Law • Ex 1: Calculate the heat change using Hess’ Law: 2 CO (g) + O 2(g)  2CO2(g) • 2 equations must be written in order to include CO and CO2 • C + ½ O2 CO (switch and multiply) • C + O2 CO2 (multiply)

  12. Hess’ Law • Example 1: Calculate the heat change using Hess’ Law: 2 CO (g) + O 2(g)  2CO2(g) 2CO  2C + O2H = 2 (26.4) = 52.8kCal 2C + 2O2 2CO2H = 2 (-94) = -188kCal ______________________________________________ • 2CO + O2 2CO2H = -135.2 kCal To convert to kJ multiply by 4.184 = -566 kJ

  13. Hess’ Law • Example 2: Calculate the heat change using Hess’ Law: NH3 + O2 NO + H2

  14. Bond Energy • EX 1: Calculate the Heat of Reaction using Bond Energies 2 H 2(g) + O2(g)  2 H2O (g) • Must draw correct Lewis pictures: 2[H-H] + 1[O=O]  2[H-O-H] H-H  104.2 x 2 = 208.4 H-O  110.6 x 4 = 442.4 O=O  119.1 x 1 = 119.1 327.5 kcal • The energy difference is 114.9 Kcal. Because more energy is released than absorbed, the reaction is exothermic and the answer is: H -114.9 Kcal

  15. Bond Energy • EX 2: Calculate the Heat of Reaction using Bond Energies HCl  H2 + Cl2

  16. The Driving Force of Chemical Reactions • In regard to enthalpy, lower is more favorable (- H ) • For example: • C8H18 + O2 CO2 + H2O + heat • However, this is not the only driving force. • Entropy (S): the measure of disorder in a system. The higher disorder (more + S), the more likely the reaction is to occur (messy room, leaves on trees). • Systems tend to go towards minimum energy (-H) and maximum randomness (+S).

  17. Entropy • General trends in entropy: • s  l  g = + S • g  g + g + g = + S •  temperature = + S • Spontaneous reactions take place without outside influence (they can be fast or slow).

  18. Entropy • Ex1: • 2 C8H18(l) + 25 O2(g) 16 CO 2(g) + 18H2O (g) + heat • More pieces and less heat • Entropy (S) is positive and enthalpy (H) is negative. • Thus, the reaction is spontaneous.

  19. Entropy • Ex2: • CO2(g) + N2(g) + H2O(g) + heat  C3H5(ONO2)3(l) • Less pieces and more heat • Entropy (S) is negative and enthalpy (H) is positive. • Thus, the reaction is not spontaneous.

  20. Entropy • Ex3: H2O (s) H2O (l) • Does it happen? • Entropy (S) is positive and enthalpy (H) is positive. • Thus, we can’t determine the spontaneity based on the information given. • GibbsFree energy formula is used to determine the spontaneity. • Free energy of a system: G = Gibbs Free Energy (combined enthalpy/entropy function) • G = H - (TS) • If G is negative, the reaction is spontaneous

  21. Possible combinations of entropy and enthalpy: What determines a maybe? Temperature

  22. Gibb’s Free Energy • Is this reaction spontaneous? • H2O + C  CO + H2 • H = + 131.3 kJ/mole • S = + 0.134 kJ/mole.K at 25 Co G = 131.3 - [298(0.134)] G = +91.4 kJ/mole The reaction is not spontaneous.

  23. How about at 900 Co? • G = 131.3 - [1173(0.134)] • G = -26.0 kJ/mole • The reaction is spontaneous .

  24. The Reaction Process • Reaction Mechanism: The series of steps in a reaction. • We have learned the following: • H2 + I2  2 HI H = + 26.5 kJ • However, we now know that it is really: Step 1 I2  2 I Step 2 I + H2 H2I Step 3 H2I + I  2 HI H2 + I2 2 HI

  25. Reaction Intermediate Step 1 I2  2 I Step 2 I + H2 H2I Step 3 H2I + I  2 HI H2 + I2 2 HI • Reaction Intermediate: a species that appears in some steps but not in the overall reaction. It is relatively short lived. So, in the above example, H2I is the reaction intermediate.

  26. Collision Theory • Collision Theory: In order for molecules to react, they must collide, but collision doesn’t guarantee reaction. • For a reaction to take place, 1. The collisions need enough energy to break bonds 2. the particles need proper orientation

  27. Collision Theory

  28. Activated complex: a short lived structure existing when old bonds are broken and new bonds are being formed. • Energy of activation: energy needed to transform reactants into an activated complex. • Reactants must have sufficient kinetic energy and proper orientation.

  29. Energy Diagrams:

  30. Energy Diagrams

  31. Reaction Rate – reaction rates are affected by: • nature of reactants (type of bond – ionics react faster than covalents) • Temperature • Concentration of reactants • Pressure (gases only) • Catalyst (A substance that is used in a reaction, but is not used up in the reaction. Catalysts work by changing the mechanism of a reaction and lowering the activation energy). • Surface area • All of these factors can be explained in relation to the collision theory.

  32. Writing Rate Laws •  For: A + B  C • rate = k [ A ]x [ B ]y(determined experimentally) •  If a reaction is one step, then the coefficients equal the exponents. • If a reaction is multi step, then the coefficients do not equal exponents.

  33. Ex. 2 H2 + 2 NO  N2 + 2 H2O •  If the reaction is one step, the reaction rate formula is … • R = k [ H2 ]2 [ NO ]2 • If the reaction had multiple steps, the reaction rate formula would be something else. Perhaps something like … •  R = k[ H2 ] [ NO ]2 • Notice that the coefficients and exponents are not the same. An experiment would have to be performed to show this.

  34. Ex. X + 2Y  XY2(a single step reaction) • Write the rate law: R = k [ X ] [ Y ]2 • If you double X, the rate doubles. • If you double Y, the rate quadruples. • If Y is reduced to 1/3, the rate is 1/9 of the original. • If X is cut in half and Y is doubled, the rate doubles.

  35. Reversible Reactions _Fe3O4(s) + 4 H2(g) 3 Fe(s) +4 H2O(g) This reaction is also possible in reverse. If water is removed from the vessel, the reaction shifts to the right, but if in a closed container… _Fe3O4(s) +4 H2(g) 3 Fe(s) +4H2O(g) The double arrows represent equilibrium.

  36. Reversible Reactions Equilibrium: when the rate of the forward reaction equals rate of the reverse reaction.

  37. Equilibrium Constant Do NOT include liquids or solids Note: If K1, products are favored. If K1, reactants are favored • For… H2+ I2 2 HI Keq = [HI]2 [H2][I2]   • Ex1: At the beginning of a reaction, the Hydrogen concentration is 1.00 M and so is the Iodine concentration. At equilibrium, [H2] = 0.228 M, [I2] = 0.228 M, and [HI] = 1.544 M. Calculate Keq. • Keq =(1.544)2 (.228)(.228) Keq= 45.9

  38. Applications of Keq • Ex2: If the concentration of HI at equilibrium is 0.158 M, what are the concentrations of Hydrogen and Iodine? (Keq = 45.9)  • H2 + I2 2 HI 45.9 = (0.158)2 (x)(x) 45.9(x)2 = (0.158)2 x2 = 5.44 x 10-4 x = .0233 M

  39. Solubility Equilibrium • Equilibrium is also reached when solids dissolve in water. • AgCl(s) + H2O(l) Ag+1(aq) + Cl-1(aq) • We don’t count solids or liquid water in K expressions, so… Ksp= [Ag+1][Cl-1] Ksp= solubility product constant K = moresoluble

  40. Solubility Equilibrium • EX 1: • Ag2SO4(s) + H2O(l)2 Ag+1(aq) +SO4-2(aq) Ksp = [Ag+1]2[SO4-2]

  41. Solubility Equilibrium • If Ksp for CdS = 1.0 x 10 –28, what is the concentration of each ion? • CdS(s)+ H2O(l) Cd+2(aq) + S-2(aq) Ksp= [Cd+2][S-2] 1.0 x 10-28 = [x][x] 1.0 x 10-28 = [x]2 x = 1 x 10-14 M

  42. Solubility Equilibrium • EX 2: If the Ksp for Ag2SO4 = 1.10 x 10-12, calculate the [ ] of each ion. Ag2SO4(s) + H2O(l)  2 Ag+1(aq) + SO4-2(aq) Ksp = [Ag+1]2[SO4-2] 1.10 x 10-12 = [2x]2[x] 1.10 x 10-12 = 4 x3 x3 = 2.75 x 10-13 x = 6.5 x 10 –5 M The [SO4-2] is 6.5 x 10 –5 [Ag+1] two times [SO4-2], thus the [Ag+1] is 1.3 x 10 –4

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