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Chemical Equilibrium

Chemical Equilibrium. PHYS 1090 Unit 15. Direction. A chemical reaction proceeds in a specific direction under specific conditions. A  B. Why? Entropy increases!. Increase Entropy. Disperse matter Dissolve a solid in a liquid Melt a solid Release a gas Disperse energy

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Chemical Equilibrium

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  1. Chemical Equilibrium PHYS 1090 Unit 15

  2. Direction • A chemical reaction proceeds in a specific direction under specific conditions. A  B • Why? Entropy increases!

  3. Increase Entropy • Disperse matter • Dissolve a solid in a liquid • Melt a solid • Release a gas • Disperse energy • Generate heat • Biggest effect at low temperature • When they conflict, one dominates. • Direction can change with temperature.

  4. Uncomplicated Examples • Dissolve CaCl2 in water • Gets hot: disperses energy • Solid dissolves: disperses matter • Burn wood: (C6H10O5)n(s) + 6n O2(g)  6nCO2(g) + 5nH2O(g) • Gets hot: disperses energy • Generates CO2 and H2O gas: disperses matter

  5. Complicated Examples • Dissolve NH4NO3 in water • Gets cold: constrains energy • Solid dissolves: disperses matter • Here, dispersing matter wins • Solidify super-cooled sodium acetate • Gets hot: disperses energy • Liquid solidifies: constrains matter • Here, dispersing energy wins

  6. Reaction Rates A  B • Rate is D[B]/Dt= –D[A]/Dt • Depends on • Intrinsic reactivity of A • Temperature • How much A there is • Proportional to [A]

  7. Reaction Rates A+ B  C • Rate is D[C]/Dt= –D[A]/Dt =–D[B]/Dt • Depends on • Intrinsic reactivity of A and B • Temperature • How much A and B there are • Proportional to [A][B]

  8. Rate constants A + B  C • Rate = k[A][B] • k = rate constant • Independent of [A], [B] • Depends on temperature

  9. Equilibrium  A  B • Two reactions: A  B and B  A • If both have the same rate, [A] and [B] don’t change • “Equilibrium” • kf[A] =kb[B] • [B]/[A] = kf /kb = K= equilibrium constant

  10. Equilibrium Constants • For A + B  C • For A + A  B • For A + B  C + D

  11. K = [HCl]2 K = [H2][Cl2] Examples NH3 + HCl  NH4Cl [NH4Cl] [NH3][HCl] H2 + Cl2 2 HCl

  12. Unchanging Concentrations • Solvent • Solid • Unmixed liquid Drop out of equilibrium constant

  13. [H2O][NaCl] K = [KOH][HCl] [NaCl] K = [KOH][HCl] Example KOH(aq) + HCl(aq)  H2O + NaCl(aq) • But H2O is the solvent: its concentration does not change. So:

  14. [H+][OH–] K = [H2O] Another Example H2O  H+(aq) + OH–(aq) • But again H2O is the solvent. So: K = [H+][OH–]

  15. [CaCO3][NaCl]2 K = [CaCl2][Na2CO3] [NaCl]2 K = [CaCl2][Na2CO3] Precipitation Example CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2 NaCl (aq) • CaCO3 is a solid precipitate, not dissolved. So:

  16. Le Chatelier’s Principle • If a chemical system at equilibrium is perturbed, the system will adjust to minimize the perturbation

  17. Instances • If reactant concentration is high: • More product will be generated • If products are removed: • More product will be generated • If products are insoluble: • Effectively removes them • More product will be generated • (precipitation reactions)

  18. Instances • If a reaction generates a gas: • Increasing the pressure consumes products • Decreasing the pressure consumes reactants • Letting gas escape generates more • If a reaction generates heat: • Increasing the temperature inhibits it • Decreasing the temperature facilitates it

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