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Solutions Form 2

Solutions Form 2 - Solutions Part 1 FORM#9 Chem 1202 Final...

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Solutions Part 1 FORM #9 ————————————————————————————————————— Chem 1202 Final Exam Spring 2004 ————————————————————————————————————— 1 Solution: A very large equilibrium constant K { 1, means the equilibrium favors (produces more) products. At equilibrium the amounts of the reactants and products stop changing. The rates of the forward and reverse reactions are equal. 2 Solution: a. In CH 4 the C - H bond is not polar because H and C have very similar electronegativities. So that hydrogen is not acidic. b. Na ± is a negligibly weak acid. All alkali metal cations are negligibly weak acids. You should remember that. c. HC e O 4 , HC e O 3 , HBrO 3 are similar molecules. The acidic proton is bound to O in every case. Bond strengths are similar. What makes the difference here is how positive the H atom is. The molecule with the more electronegative atoms will have the most positive H atom and this will be the strongest acid. HC e O 4 is the strongest acid because it has the most O atoms and O is a very electronegative atom. 3 Solution: A positive enthalpy of reaction H rxn o means heat is absorbed by the reaction and this cools off the surroundings. Actually H rxn o ² heat only at constant pressure. 4 Solution: H rxn o ² 1 H f o C 2 H 4 g   ± 6 H f o F 2 " 2 H f o CF 4 ± 4 H f o HF ² ¡ ± 52.3  ¢ " 2 " 680.0   ± 4 " 268.5    ² ± 2486.3 kJ You have to know that the formation reaction for elements (like F 2 ) in standard states has H f o ² 0 5 No solution provided. 6 Solution:
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Remember the connection between K and G . K ² e " G / RT There are two things to remember when using this equation. (1) Make the units of G and R agree. R ² 8.314 J/mol ± K. so G must be in Joules. (2) Be careful to carry the sign of G into the exponent. In other words, if G ² " 10 you will have exp " " 10/ RT    . Don’t forget! K ² exp " 20. 1 1000 J/kJ / 8.314 J/ mol ± K 3. 19 10 2 K ² 5. 112 10 " 4 In this case the equilibrium we calculated was for a weak acid so the value ² K a . 7 Solution: The salt NaCHO 2 is NOT the reaction for which we need to solve the equilibrium. The salt completely dissociates: NaCHO 2 v Na ± aq   ± CHO 2 " aq   The ion CHO 2 " is a weak base. We need to solve the equilibrium for the weak base CHO 2 " . CHO 2 " aq ± H 2 O e   v OH " aq   ± HCHO 2 aq   Calculate the equilibrium concentrations for the weak base using an equilibrium table. CHO 2 " OH " HCHO 2 Initial 0.099 M 0 0 Equilibrium " x x x Change 0.099 " x x x think ahead so you eliminate mistakes. Weak bases require a K b and x in the table usually represents OH " not H ± . To solve for x put the equilibrium values into the expression for K b . K b ² K w K a ² OH " HCHO 2 CHO 2 " K w K a ² 1.0 10 " 14 1. 8 10 " 4 ² 5. 556 10 " 11 ² ¡ x ¢¡ x ¢ ¡ 0.099 " x ¢ Solving for x gives (we ignore the x in the denominator because the equilibrium constant is small.) x ² 5. 556 10 " 11 0.099 Always check the table to see what x is and how it is related to what we are trying to find.
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