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Warmup

Warmup. 1. Solve for x and y using elimination. x + y = -1 -2x + -2y = 2 3x + 2y = -5. x = -3 y = 2. Enthalpy, Entropy, and Gibb’s Free Energy. exothermic. reaction pathway : shows ∆H of reaction and compares the total energy stored by the reactants vs. the products . endothermic.

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Warmup

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  1. Warmup 1. Solve for x and y using elimination. x + y = -1 -2x + -2y = 2 3x + 2y = -5 x = -3 y = 2

  2. Enthalpy, Entropy, and Gibb’s Free Energy

  3. exothermic reaction pathway: shows ∆H of reaction and compares the total energy stored by the reactants vs. the products endothermic

  4. What is the value of ∆H? ∆H = Hproducts – Hreactants= 500 – 150 = +350 kJ Endothermic! More energy stored by products than was stored by reactants (kJ) products ΔH reactants

  5. SubstanceHfo NO2(g) 34 kJ H2O(l) -286 kJ NH3(g) -46 kJ O2(g) 0 kJ Given the standard enthalpy of formation values, calculate the enthalpy change when ammonia is burned in air to form nitrogen dioxide and water : 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l) ΔH = [4(34kJ) + 6(-286kJ)] - [4(-46kJ) + 7(0kJ)] = 136 – 1716 + 184 = -1396 kJ / 4 moles = -349 kJ/mole

  6. Given the standard enthalpies of formation,calculate the enthalpy of combustion of propane in J/mole Compound ΔHf°(kcal/mole) H2O(l) -68.3 CO2(g) -94.1 O2(g) 0.0 C3H8(g) -25.0 C3H8 + 5O2 3CO2 + 4H2O ΔH = Hproducts – Hreactants = [3(-94.1) + 4(-68.3)] - [(-25.0) + 5(0.0)] = -530.5 kcal/mole • -530.5 kcal (1000 cal)(4.184 J) = • 1 mole (1 kcal) (1 cal) -2.2 x 106J/mole

  7. Hess’s Law intermediate ΔH is the same for a particular reaction, no matter if it occurs in one step or a series of steps. - 221 kJ = -787 kJ + 566 kJ

  8. H2CO + O2 H2O + CO2 Ex1) Calculate the ∆H for the reaction above given the following information: H2O + CO2 H2CO3 ∆H = -108.5 kJ H2CO3 H2O + CO2 ∆H = +108.5 kJ H2CO + O2 H2CO3 ∆ H = -196 kJ ∆H = -87.5 kJ

  9. C2H4O(l) + 5/2O2(g)  2CO2(g) + 2H2O(g) 2C2H4O(l) + 5O2(g)  4CO2(g) + 4H2O(g) Ex 2) Considering the information below, calculate the ∆H for the reaction above. 2C2H4O(l) +2H2O(g) 2C2H6O(l) +O2(g) ∆H= 915.7kJ C2H6O(l)+3O2(g) 2CO2(g) + 3H2O(g) ∆H= -3084.7 kJ 2C2H6O(l)+ 6O2(g)4CO2(g) +6H2O(g) ∆H=(-3084.7 kJ)2 ∆H = -5253.7 / 2 = -2626.9 kJ

  10. Ex 3) The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is -283.3 kJ/mol CO. Use these data to calculate ∆H for the reaction 2C(s) + O2(g)  2CO(g) C(s) + O2(g)  CO2(g) ∆H= -393.7 kJ/mol 2C(s) + 2O2(g)  2CO2(g) ∆H=2(-393.7 kJ/mol) 2CO(g) + O2(g)  2CO2(g) ∆H= -283.3 kJ/mol 2CO2(g)  2CO(g) + O2(g) ∆H=+283.3 kJ/mol ΔH = -504.1 kJ or 252.1 kJ per mole C

  11. Entropy The 2nd law of thermodynamics says that the disorder of the universe, meaning its entropy, or ΔS, is constantly increasing. Particles tend to move spontaneously from a state of low entropy (more order) to high entropy (less order) The third law says that the entropy of a perfect crystal at 0K is zero. Everything is more disorganized than a crystal, therefore everything has energy.

  12. Which phase of matter represents the highest entropy state?Based off positional probability; whichever phase has more positional options is the most likely

  13. Which of the following reactions results in the largest increase in entropy? • CO2(s)→CO2(g) • H2(g) + Cl2(g)→2HCl(g) • KNO3(s)→KNO3(l) • C(diamond) →C(graphite) All of the reactions result in an increase in disorder, but A, in which CO2 moves from a solid state to a gaseous one, represents the largest change in disorder. These changes to a system lead to an increase or larger entropy: • The greater the disorder or randomness in a system • Change of state from solid to liquid to gas. • When a pure solid or liquid dissolves in a solvent • When a gas molecule escapes from a solvent • The molecular complexity increases. • Reactions that increase the number of moles of particles

  14. Calculate the entropy change at 25̊C in J/K for • 2SO2(g) + O2(g) 2SO3(g)given the following data: • S° SO2(g): 248.1 J/mol·K • S° O2(g): 205.3 J/mol·K • S° SO3(g): 256.6 J/mol·K • ΔS = Sproducts– S reactants • [2(256.6)] - [2(248.1) + 1(205.3)] = -188.3 J/K The higher the S value, the more disordered the system. A positive (+) S value is more disordered, and a –S value is less disordered (“more ordered”).

  15. Gibb’s Free Energy The two driving forces for chemical reactions are enthalpy and entropy. Δ G = Δ H – TΔS The Gibb’s free energy equation is the total available energy in a system (minus an entropy term) and thus is the energy available for doing useful work.

  16. Consider a reaction at 25°C. If the enthalpy change is -230 J and the entropy change is 78 J/K, calculate the change in Gibbs free energy. Is this reaction spontaneous? G = H - TS = (-230J) – (298K)(78 J/K) = -23474 J • In thermodynamics, the term "spontaneous" means the process is "allowed“ or “is possible”, in contrast to "is not allowed" or "will not likely occur". • If G is negative, the reaction is spontaneous • If G is positive, then the reaction is nonspontaneous in the forward direction, but the reverse reaction will be spontaneous. • If G is equal to zero, the reaction is at equilibrium (more on this later)

  17. For the reaction CaSO4(s)  Ca2+(aq) + SO42-(aq)a. Calculate ∆H b. calculate ∆S c. calculate ∆G Hf°Ca2+ = -542.8 kJ S°Ca2+ = -53.1 J/K Hf° SO42- = -909.3 kJ S° SO42- = 20.1 J/K Hf° CaSO4= -1434.1 kJ S° CaSO4= 106.7 J/K a. ∆H = -542.9 kJ – 909.3 kJ –(-1434.1 kJ) = -18.0 kJ b. ∆S = -53.1 J/K + 20.1 J/K – 106.7 J/K = -139.7J/K (change units to kJ/K for part c) = -0.1397 kJ/K c. ∆G = -18.0 kJ – (298K)(-0.1397kJ/K) = 23.6 kJ

  18. We will use G = H - T S to find G a. Find Hf° for each component Hf° H2S = -20.1 kJ Hf° O2= 0 kJ Hf° H2O = -285.8 kJ Hf° S = 0 kJ H =[2(-285.8)-0] - [2(-20.1)+ 1(0)]= -531.4 kJ 2H2S(g) + O2(g) ® 2H2O(l) + 2S(rhombic) Determine if this reaction is spontaneous at 25ºC b. Find S for each component S° H2S = 205.6 J/K S° O2= 205.0 J/K S° H2O = 69.94 J/K S° S = 31.9 J/K S=[2(69.94)-2(31.9)] - [2(205.6)+ 205]= -412.5 J/K

  19. c. Now use G = H - T S  G = -531.4 kJ - 298K(-412.5 J/K)  G = -531.4 kJ - -123000 J  G = -531.4 kJ - -123 kJ  G = -408.4 kJ(spontaneous, releases free energy)

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