Final Review

Final Review - Final Exam Review Dr Briggs 1 Chapter 1...

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1 Final Exam Review Dr. Briggs

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2 Chapter 1 Problems • Book (Ch. 1, prob. 6) Does the entropy increase or decrease in the following processes? (a) N 2 + 3H 2 Æ 2NH 3 Ammonia (b) H 2 N-C(=O)NH 2 + H 2 O Æ CO 2 + 2NH 3 Urea Ammonia (c) 1M NaCl 0.5M NaCl (d) COO- COO- HC-OH HC-OPO 3 2- H 2 C-OPO 3 2- H 2 C-OH Æ Æ (Decrease) (Increase) (Increase) (No Change)
3 Chapter 1 Problems • Book (Ch. 1, prob. 7) Consider a reaction with H=15 kJ/mol and S=50 J/mol/K. Is the reaction spontaneous (a) at 10 o C, (b) 80 o C? (a) G=15 kJ/mol - (273+10)K x 50 J/mol/K x 1 kJ/1000 J = 0.85 kJ/mol (b) G=15 kJ/mol - (273+80)K x 50 J/mol/K x 1 kJ/1000 J = -2.65 kJ/mol Recall, G= H-T S, and if G<0, reaction is spontaneous. NO YES

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4 Chapter 1 Problems •B r i g g s (1) How many angstroms are there in a nanometer? (2) How many kJ/mol are there in 1 kcal/mol? (3) What is the Gas Constant (R) in J/K/mol? (4) What is the definition of a spontaneous process ? (5) Draw a carboxyl group 10 (1Å = 10 -10 m, 1nm - 10 -9 m) 4.184 (1 kcal/mol = 4.184 kJ/mol) R = 8.3145 J/K/mol A spontaneous process is one that occurs without input of additional energy from the surroundings - the free energy is negative. -C(=O)OH
5 Chapter 1 Problems •B r i g g s (6) Given a G o of 2.5 kJ/mol and a temperature of 25 o C, what is the equilibrium constant? Recall, G o =-RTlnK eq lnK eq = -2.5 kJ/mol/(8.3144 J/mol/K x 298K / 1000 J/kJ) K eq = 0.36

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6 Chapter 2 Problems N H N N N O H N H 2 N N H NH 2 O H C H 2 COO- 3 + OH 6 1 9 1 3 Book: (Ch. 2, Prob. 1) Identify the potential hydrogen-bond donors and acceptors: 7 Donor Donor Donor Acceptor Acceptor Acceptor Donor Donor Acceptor Acceptor Donor Donor Acceptor Acceptor
7 Chapter 2 Prob. 2, Companion Which gas in each of the following pairs would you expect to be more soluble in water? a). Oxygen and carbon dioxide b). Nitrogen and ammonia c). Methane and hydrogen sulfide Carbon dioxide is more polar; O=O vs. O=C=O Ammonia is more polar; N N vs. NH 3 Hydrogen sulfide is more polar; CH 4 vs. H-S-H

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8 Chapter 2, Briggs problem 1). In the molecular structure below, match up the functional group type in the list to the left with the molecule on the right. • Hydroxyl • Amino • Sulfhydryl • Carboxyl •A c y l •E s t e r c y l t h e r • Amido Acyl Ether Amino Amido Sulfhydryl Hydroxyl OH NH 2 O CH 2 SH O C H 2 O CH 3 N H 2 H
9 Chapter 2 Problems • Book (Ch. 2, prob. 5) Draw the structures of the conjugate bases of the following acids CH CH COO- COOH CH 2 NH 3 + COOH CH 2 NH 3 + COO- C H NH 3 + COO- C H 2 COOH

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10 Chapter 2 Problems • Book (Ch. 2, prob. 5) Draw the structures of the conjugate bases of the following acids CH CH COO- COOH CH 2 NH 3 + COOH CH 2 NH 3 + COO- CH CH COO- COO- CH 2 NH 3 + COO- CH 2 NH 2 COO- C H NH 3 + COO- C H 2 COOH C H NH 3 + COO- C H 2 COO-
11 Chapter 2 Problems pH = -log[H+] [H+]=K w /[HO-]; M=moles/liter [H+]=1x10 -14 M 2 /(0.05M)=2x10 -13 M pH=-log(2x10 -13 )=12.7 [] 0.05M 1.0liter 0.01liter * liter 5moles ] [HO = = • Book (Ch. 2, prob. 7a) Calculate the pH of a 1L solution containing: (a) 10 mL of 5M NaOH

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12 Chapter 2 Problems The concentration of glycine is much, much lower (2 orders of magnitude) than that of HCl so we can ignore it.
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Final Review - Final Exam Review Dr Briggs 1 Chapter 1...

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