CHEM 104 1st Edition Lecture 19- Magnitude of kc gives information about amount of reactants/products at equilibriumo If kcis much larger than 1, the reaction is strongly product-favored at equilibrium o If kc is much smaller than 1, the reaction is strongly reactant-favored at equilibriumo If kc is approximately 1, there is a significant amount of both reactants and products present at equilibrium - Equilibrium constants depend on temperature o Example: 2NO2(g) N2O4(gas) Forward rate = k1[NO2]2 and reverse rate = k-1[N2O4]Forward rate = reverse rate at equilibriumKc = [N2O4]/[NO2]2= k1/k-1At 25 C kc = 3.5*108and at 450 C kc = 0.16there’s more [NO2] at higher temp and more [N2O4] at lower temp- At equilibrium, forward and reverse reactions occur at the same rate - For gases (only), equilibrium constants can be expressed using partial pressures o Example: CO(g) + 3H2(g) CH4(g) + H2O(g)Kp = P(CH4)P(H2O)/P(CO)P(H2)3Kc = [CH4][H2O]/[CO][H2]3o Relationship between partial pressure and concentration derived from PV = nRT P = (n/v)RT P = [x]RT o Kp = kc(RT)change in n Change in n = moles of gaseous product – moles of gaseous reactanto Example: CS2(g) + 4H2(g) CH4(g) + 2H2S(g) Kc = 0.28 at 900 K Kp = kc(RT)n = (0.28)(0.082*1173)-2 = 3.02*10-5o Kc = kp when there are equal number of moles on each side of equation- LeChatelier’s Principle: If a system is at equilibrium and a change in conditions is made so that the system isn’t at equilibrium, the system will react to the new equilibrium such as to partially counteract the change o Changes in concentration or partial pressure (of reactants or products)o Example: cisbutanetransbutanekc = [trans]/[cis] = 1.5 at 600 KIf we increase [cis], the equilibrium will shift rightIf we increase [trans], the equilibrium will shift lefto The Reaction Quotient (Q) aA + bBcC + dDKc = [C]c[D]d/[A]a[B]bQ = [C]c[D]d/[A]a[B]bDifference is Q is actual concentration when system isn’t at equilibrium, whereas kc is equilibrium concentration Q = kc at equilibriumQ <kc fewer products, equilibrium will shift right Q >kc fewer reactants, equilibrium will shift left o Example: Cr2O72-(aq) + 2HO-(aq) 2CrO42-(aq) + H2O(l) If we add NaOH to double [HO-]… Kc = [CrO42-]2/[Cr2O72-][HO-]2Q = [CrO42-]/[Cr2O72-]2[HO-]2 = kc/4 Q <kc, so there are fewer
View Full Document
Unlocking...