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Chapter 16 Reaction Energy

Chapter 16 Reaction Energy. Chapter 16. Section 1 – Thermochemistry Section 2 – Driving Force of Reactions. Thermochemistry. Define temperature and state the units in which it is measured. Define heat and state its units. Perform specific-heat calculations.

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Chapter 16 Reaction Energy

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  1. Chapter 16Reaction Energy

  2. Chapter 16 • Section 1 – Thermochemistry • Section 2 – Driving Force of Reactions

  3. Thermochemistry • Define temperature and state the units in which it is measured. • Define heat and state its units. • Perform specific-heat calculations. • Explain enthalpy change, enthalpy of reaction, enthalpy of formation, and enthalpy of combustion. • Solve problems involving enthalpies of reaction, enthalpies of formation, and enthalpies of combustion.

  4. Thermochemistry • Virtually every chemical reaction is accompanied by a change in energy. • What two ways can energy be involved? • Thermochemistry: study of the transfer of energy as heat that accompanies chemical reactions and physical changes.

  5. Calorimeter • Calorimeter: used to measure heat absorbed or released. • Energy released from the reaction is measuredfrom the temp change ofthe water

  6. Calorimeter

  7. Temperature • measure of the average kinetic energy of the particles in a sample of matter. • The greater the kinetic energy of the particles in a sample, the hotter it feels. • Remember: How do we convert Celsius to Kelvin?

  8. Heat • Heat: the energy transferred between samples of matter because of a difference in their temperatures. • Joule: SI unit of heat and energy • Energy transferred as heat moves spontaneously  from matter at a higher temp  lower temp

  9. Specific Heat • Energy transferred depends on: • The material • The mass • Specific Heat: specific heat of a substance is the amount of energy required to raise the temperature of one gram by one Celsius degree (1°C) or one kelvin (1 K). • Units: J/(g•°C) or J/(g•K) • The temperature difference as measured in either Celsius degrees or kelvins is the same.

  10. Molar Heat Capacities of Elements and Compounds

  11. Specific Heat Calculation Heat released Specific heat: or absorbed: cp = specific heat at a given pressure q = energy lost or gained m = mass of the sample ∆T = difference between the initial and final temperatures.

  12. Combustion/Bomb Calorimeter Q = m x Cp x T

  13. Example Problem 1 • A 4.0 g sample of glass was heated from 274 K to 314 K, and was found to have absorbed 32 J of energy as heat.a. What is the specific heat of this type of glass? b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

  14. Example Problem 1 Solution Given:m = 4.0 g ∆T = 314 – 274 = 40. K q = 32 J Unknown: a. cp in J/(g•K) b. q for ∆T of 314 K → 344 K Solution: a.

  15. Example Problem 1 Solution b. **See Practice Worksheet for more examples

  16. Typical Heat Curve – Problem Solving • Assume we have 85 grams of ice at -10 °C . How much energy is needed to convert it to water vapor at 120oC? •  Step 1: -10°C to O°C (solid) • What is happening to the particles as heat is added? • General formula: q=(m)(cp)(∆T) • Answer = 1751 J  1.8kJ • Step 2: melting • Notice how there is no increase in temp. • Where is this energy going? • General formula: q=(∆H fusion)(mol) • Answer = 28 kJ

  17. Typical Heat Curve – Problem Solving; cont’d • Step 3: – 0 °C to 100 °C (liquid) • What is happening to the particles as we heat? • General formula: q=(m)(cp)(∆T) • Answer = 36,000  36 kJ • Step 4: Boiling • Notice how there is no increase in temp. • Where is this energy going? • General formula: q=(∆H vap)(mol) • Answer = 190 kJ • Step 5: 100 °C to 120 °C (gas) • What is happening to the particles as we heat? • General formula = (m)(cp)(∆T) • Answer = 3,200 J  3.2 kJ

  18. Typical Heat Curve – Problem Solving; cont’d • Now, add up all the steps for total heat. • Answer = 259 kJ Summary: • When the substance is heating • q=(m)(cp)(∆T) • During phase change • q=(mol)(∆H fus) OR • q=(mol)(∆H vap)

  19. Heating Curve of Water How much energy does it take heat water from ice to vapor?? (2 equations: 1 for change of temp./ 1 for phase change) Phase Change: • Heat of Fusion: Hfus • Heat energy needed to melt 1 mol (s to l) • Heat of Vaporation: Hvap • Heat energy need to boil 1 mol of a substance ( l – g)

  20. Enthalpy (Heat) of Reaction • Enthalpy (H): Heat • H is never directly determined, rather ∆H • Enthalpy(∆H): “change in enthalpy”energy absorbed as heat during a chemical reaction at constant pressure

  21. Enthalpy of a Reaction • When∆H is negative, the reaction is exothermic – releases heat • When ∆H is positive, the reaction is endothermic – absorbs heat

  22. Enthalpy change • The difference between the enthalpies of products and reactants. ∆H = Hproducts – Hreactants

  23. Exothermic Reaction Energy released -∆H

  24. Endothermic Energy is absorbed +∆H

  25. Stability • Compounds with a large negative enthalpy of formation are very stable. • Compounds with positive values of enthalpies of formation are typically unstable.

  26. Thermochemical Equation • Equation that includes the quantity of energy released or absorbed as heat 2H2(g) + O2(g) → 2H2O(g) ∆H = -483.6 kJ • States of matter are ALWAYS shown because it can influence the energy of released or absorbed. • Coefficients are always viewed as number of moles in the reaction.

  27. Enthalpy of Formation • molar enthalpy of formation: enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25°C and 1 atm. • Enthalpies of formation are given for a standard temperature and pressure so that comparisons between compounds are meaningful. “Standard State” Heat (enthalpy) of FORMATION

  28. Where do I find ΔHf? • In your textbook! (or handout) • The values are given as the enthalpy of formation for one mole of the compound from its elements in their standard states. • **because ΔHf is reported for the formation of 1 mol of a substance, it may be necessary to write on ½ mol for one of the reactants • Ex: 1/2N2 (g) + 1/2O2 (g)  NO (g)

  29. Enthalpy of Combustion • ∆Hc : enthalpy change that occurs during the complete combustion of one mole of a substance **Enthalpy of combustion is defined in terms of one mole of reactant **Enthalpy of formation is defined in terms of one mole of product

  30. Determining Enthalpy of Reaction • Review: ∆Hrxn energy absorbed or released during a chemical reaction • Use Hess’s Law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. • This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

  31. Some Laws we should know *from enthalpy packet • 1. The sign ∆Hrxn of depends on the direction of the reaction • (If you flip the reaction around, the a positive enthalpy would become negative) • 2. The magnitude ∆Hrxn of is proportional to the amount of the substance in a reaction. • (If you change of the coefficients then multiply the ∆H, too)

  32. Example 1 • If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally. What is the heat of formation of methane? C(s) + 2H2(g) → CH4(g)

  33. Calculating using Hess’s Law • The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane: Goal Reaction: C(s) + 2H2(g) → CH4(g) C(s) + O2(g) → CO2(g) H2(g) + ½O2(g) → H2O(l) 2 [ H2(g) + ½O2(g) → H2O(l) ] CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆H0 = +890.8 kJ This reaction must x2! The ΔH is x2! This reaction must be reversed! The ΔH is changed to positive!

  34. ADD IT UP!! C(s) + O2(g) → CO2(g) 2H2(g) + O2(g) → 2H2O(l) CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) C(s) + 2H2(g) → CH4(g)

  35. Calculating Enthalpy from Heat of Formation ∆H0= Σ[(ΔH0fof products) × (mol of products)]– Σ[(ΔH0f of reactants) × (mol of reactants)] *Need to use reference sheet for Hf values **Keep in mind: elements in their standard state have an enthalpy of formation of zero • Ex: Ag(s) = 0 ; N2 (g) = 0 • See Practice WKST (Calculating Enthalpy from Heats of Formation)

  36. 16.2 Driving Force of Reactions • Explain the relationship between enthalpy change and the tendency of a reaction to occur. • Explain the relationship between entropy change and the tendency of a reaction to occur. • Discuss the concept of free energy, and explain how the value of this quantity is calculated and interpreted. • Describe the use of free energy change to determine the tendency of a reaction to occur.

  37. Spontaneous Reaction • Most chemical reactions are exothermic, liberating heat • Products have less energy than the reactants • Nature’s tendency is toward a lower energy state

  38. Spontaneous Reaction • But, some processes that are spontaneous are endothermic • Example: melting (particles have more energy in the liquid state) • So there are 2 factors that drive a reaction

  39. Entropy • 2 Factors: Enthalpy, but also Entropy • Entropy (S) : the degree of randomness or disorder of the particles in a system • Think about your bedroom, does it always stay neat and organized? Or does it get messier and messier? • Increase in disorder  Increase in entropy

  40. Entropy • Nature’s tendency is toward more disorder or greater entropy. • Also means, the entropy of the universe is ALWAYS increasing.

  41. Changes in Entropy • As you go from a solid  liquid  gasWhat happens to the entropy? • What about when you dissolve a solute? • Or what about when you dilute a substance? • How does entropy change with temperature?

  42. Standard Entropy Changes for Some Reactions

  43. Change in Entropy • Unit: kJ/mol K or J/mol K • Change in Entropy ΔS • Difference between the entropy of the products and the reactants • Increase in Entropy  +ΔS (more disorder) • Decrease in Entropy  -ΔS (less disorder)

  44. Change in Entropy Calculations • Calculate entropy like calculating enthalpy from heats of formation, but use S values instead • ∆S = Sproducts - Sreactants • See practice WKST, use thermodynamic data sheet for Sf values

  45. Gibb’s Free Energy • Gibb’s Free Energy (G): • Combined enthalpy-entropy function • Spontaneous Reactions: • Least enthalpy (very negative #) • Greatest entropy (large positive #)

  46. Change in Free Energy • Only the change in free energy can be measured (KJ/mol) • At a constant pressure and temperature… ∆G0= ∆H0– T∆S0 *T must be in Kelvin since ∆S will be in units of KJ/mol k *∆S should also be in KJ

  47. Relating Enthalpy and Entropy to Spontaneity If -∆G the reaction is spontaneous +∆G the reaction is not spontaneous See “Free Energy” worksheet to practice calculation See Practice Worksheet – “Pg. 78 Gibbs Free Energy” to review concepts

  48. Ice - Enthalpy vs Entropy

  49. Example Problem For the reaction NH4Cl(s) → NH3(g) + HCl(g) ∆H0 = 176 kJ/mol and ∆S0 = 0.285 kJ/(mol•K) at 298.15 K. Calculate ∆G0, and tell whether this reaction is spontaneous in the forward direction at 298.15 K.

  50. Example Problem Solution Given:∆H0 = 176 kJ/mol at 298.15 K ∆S0 = 0.285 kJ/(mol•K) at 298.15 K Unknown:∆G0 at 298.15 K Solution:The value of ∆G0 can be calculated according to the following equation: ∆G0= ∆H0– T∆S0 ∆G0= 176 kJ/mol – 298 K [0.285kJ/(mol•K)] ∆G0= 176 kJ/mol – 84.9 kJ/mol ∆G0= 91 kJ/mol

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