CHEM 251 1st Edition Lecture 23 Outline of Last Lecture I. Clicker QuestionII. Hard-Soft AcidsIII. Hard-Soft BasesIV. Reactivity PredictionsV. Application: Hydrogen StorageOutline of Current Lecture VI. Clicker QuestionsVII. Group QuestionsCurrent LectureVIII. Clicker Questionsa. Predict the product: 6CaO + P4O10 →i. (A) 6CaO2 + P4O4ii. (B) 6CaO + P4 + 3O2iii. (C) (Ca3P2O8)2iv. (D) Ca6(PO4)4 b. Place the following in order of increasing pKa:i. 2<1<3<4c. Will the equilibrium favor reactants or products?2Fe(OCN)3(aq) + 3Fe(SCN)2(aq) 2Fe(SCN)3(aq) + 3Fe(OCN)2(aq)i. Reactants (hard-hard and soft-soft)IX. Group QuestionsThese 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.a. Consider: HCl(aq) + NaF(aq) D HF(aq) + NaCl(aq)i. Write reactions representing the pKa of HCl and HF, and show how they can be related the above balanced equation1.ii. The pKa of HCl is -8. The pKa of HF is 3.2. Based on these values, predict whether the equilibrium will lie towards the reactants or the products1. Towards products (HCl is stronger)iii. Will Keq be > 1 or < 1?1. >1b. Is the following carbocation a Lewis acid or a Lewis base?i. Lewis acidii. Draw the key acidic or basic orbitaliii. Predict the reactivity of Ph3C+ with H2O1. Reacts w/ water as a base (adds –OH)c. Consider: 1. Why is reaction 1 exothermic?a. Hard-soft mismatches in product2. Why is reaction 2 twice as favorable?a. Lattice enthalpy!d. Determine the free energy of:i. pKa (H3PO4) = 2.1410(-pKa) = Ka = Keq = 0.00725.ΔGº = –RT ln Keq = –(1.987x10–3 kcal·mol–1·K–1)·(298 K)·(ln 0.00725)ΔGº = +2.9
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