CHEM 102 1nd Edition Lecture 18 Outline of Last Lecture I. Heterogeneous CatalystsII. EnzymesIII. Chemical EquilibriumIV. Homo- vs Hetero- V. Properties of Kc vs k VI. Magnitude of KOutline of Current Lecture I. Gas-phase Homogeneous EquilibriaII. Variations of KcCurrent LectureI. Gas-phase Homogeneous Equilibriaa. The equilibrium constant can be written as Kp because P = (n/v)RT and (n/v) is theconcentrationb. When you have: 2H2S (g) + 3O2 (g) <-> 2H2O (g) + 2SO2 (g)c. Kp = H 2 O2SO 22H 2 S2O 23d. Calculate K using equilibrium concentrations and pressuresi. H2 + I2 <-> 2HIii. Equal amounts of H2 and I2 are injected into a 1.50 L flask, at equilibiurm, there is 1.80 mol of H2 and 1.80 mol of I2, and 3.6 mol of HI.iii. Calculate Kpiv. H2 = 1.8 mol/ 1.5 L = 1.2v. I2 = 1.8 mol/ 1.5L = 1.2vi. HI = 3.6/ 1.5L = 2.4vii.2.421.211.21 = 4 e. Calculate K using equilibrium concentrationsi. N2 + 3H2 -> 2NH3ii. Kc is 2.37e-3 at 1,000KThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.iii. The mixture has 1.5 mol N2 and 4.5 mol H2 in a 1.0 L container. What mustthe equilibrium concentration of NH3 be?iv. Kc = H 2¿¿[ N 2]1¿[NH 3]2¿= So, Kc[N2][H2]3 = [NH3]2v. (2.37e-4)(1.5)(4.5)3 = [NH3]2vi. 0.0324 = [NH3]2vii. 0.179 = [NH3]II. Variations of Kca. The value of Kc depends on the form of the balanced equation in the reactionb. If there is a reversed reaction, then the new Kc will be the reciprocal of the original Kcc. If the reaction is multiplied by a factor n, then the new Kc is obtained by raising the original Kc to the power of nd. The reactions are summed up, overall K is the product of the individual KsIII. Gas-phase Homogeneous EquilibriaIV. Variations of KcV. Reaction
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