1 / 30

Energy, Kinetics and Equilibrium in Chemical Reactions

Learn about concepts such as energy absorbed, enthalpy, system vs. surroundings enthalpy, and the significance of signs in endothermic and exothermic reactions. Explore thermodynamic equations, energy per mole, entropy, and free energy for predicting spontaneous reactions.

ehugh
Download Presentation

Energy, Kinetics and Equilibrium in Chemical Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Energy, Kinetics and Equilibrium in Chemical Reactions

  2. ENERGY

  3. ENTHALPY • energy absorbed as heat during a chemical reaction at constant pressure • cannot be measured directly, it is not the same as temperature • Unit • joules (J) • Symbol • H

  4. ENTHALPY • theamount of heat absorbed or lost by a system during a process • Enthalpy of reaction is the quantity of energy transferred as heat during a chemical reaction • the difference between the stored energy of the reactants and the products

  5. SYSTEM VS. SURROUNDINGS

  6. ENTHALPY • DH = H products – H reactants

  7. ENTHALPY EXAMPLE Ex: Given the equation: C2H4(g) + H2(g) C2H6(g) The product of the reaction has an enthalpy value of -285.8 kJ/mol and the reactants enthalpy values are -1410.9 kJ/mol and +1559.7 kJ/mol, respectively. What is the enthalpy of this reaction?

  8. ENTHAPY PROBLEM • H products= -285.8 kJ/mol • H reactants = (-1410.9kJ/mol + 1559.7 kJ/mol) = 148.8 kJ/mol DH = H products – H reactants DH = -285.8 kJ/mol - 148.8 kJ/mol = -434.6kJ/mol

  9. SIGNIFICANTS OF SIGN • DH = + • Endothermic reaction • heat is absorbed by the system • the energy of the products is greater than the energy of the reactants

  10. ENDOTHERMIC RXN DIAGRAM

  11. SIGNIFICANTS OF SIGN • DH = - • Exothermic reaction • heat is released to the surroundings • the energy of the reactants is greater than the energy of the products

  12. EXOTHERMIC RXN DIAGRAM

  13. THERMOCHEMICAL EQUATIONS • Shows the energy involved in the reaction • If the reaction is exothermic the DH value is on the product side • If the reaction is endothermic the DH value is on the reactant side

  14. EXAMPLES • 2 H2(g) + O2(g)  2 H2O + 483.6 kJ • Is this reaction exothermic or endothermic? • 2C(s) + 2H2(g)+ 52.3 kJ  C2H4(g) • Is this reaction exothermic or endothermic?

  15. Energy per mole • Because the enthalpy values represent kJ/mol we can calculate the energy released or absorbed in a chemical reaction. • Ex: Given the following equation find the energy released when 0.25 moles of C2H2 reacts in a complete combustion reaction. The enthalpy of combustion of C2H2 is -1301.1 kJ/mol.

  16. EXAMPLE • Step 1: write the balanced equation. (remember the DH value is per mol: this equation has 2 mols of C2H2(g) so we have to double the DH value in the equation) • 2 C2H2(g) + 5 O2(g)  2 CO2(g) + 2 H2O(l) + 2602.2 kJ • Step 2: solve the problem: in the reaction • 2 molsC2H2(g) have a DH = -2602.2 kJ (exothermic) • I want to know what is the DH value for 0.25 mol C2H2(g)

  17. EXAMPLE • 2 C2H2(g) + 5 O2(g)  2 CO2(g) + 2 H2O(l) + 2602.2 kJ • 0.25 mol X 2602.2 kJ/2 mol = -325.2 kJ • Remember in the balanced equation the energy value is for 2 moles of C2H2(g)

  18. ENTROPY • A measure of the degree of randomness of the particles, such as molecules in a system. • Reactions favor progress towards greater randomness

  19. ENTROPY • The entropy of a pure crystalline solid is zero at absolute zero • As energy is added molecular motion increases • The change is enthalpy can be calculated

  20. ENTROPY • Change in entropy is the difference between entropy of the products and the reactants • Increase in entropy is a positive value • Decrease in entropy is a negative value

  21. ENTROPY • Symbol for entropy S • Symbol for change in entropy DS • Unit kJ/mol * K • Entropy is temperature dependent

  22. ENTROPY • The process of forming a solution involves an increase in entropy • Solids are more ordered, they have low entropy • Liquids are less ordered than solids, the molecules have more movement, they have more entropy than solids

  23. ENTROPY • Gases have very little order, they move very fast and take the shape of their container. They have the most entropy. • (s) < (l) < (g)

  24. ENTROPY • CH4(g) + H2O(g)  CO(g) + 3H2(g) • Does this reaction become more random or less random? • Will the reaction have a positive DS or a negative DS?

  25. EXAMPLE • Determine if the entropy change will be positive or negative for the following reactions:(NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)2 H2(g) + O2(g) → 2 H2O(g)PCl5(g)→ PCl3(g)+ Cl2(g)

  26. FREE ENERGY (DG) • difference between the change in enthalpy and the product of the kelvin temperature and the entropy change • DG = DH – TDS • This can be used to make predictions as to the reactions ability to progress spontaneously.

  27. SPONTANEOUS REACTION PREDICTION

  28. SPONTANEOUS REACTION PREDICTION

  29. EXAMPLE • The gas-phase reaction of H2 and CO2 to produce H2O and CO2 has DH = +11 kJ and DS = +41 J/mol*K. Is the reaction spontaneous at 298.15K? What is the DG?

  30. ANSWER • DG = DH – TDS • DG = 11kJ – (298.15K * 0.041 kJ/mol *K) • DG = -1.2 kJ • The reaction will be spontaneous

More Related