hw1solution_pdf

4 yes because δ g for the reaction is neg ative

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4. Yes, because Δ G 0 for the reaction is neg- ative. correct 5. Cannot tell unless we are given the stan- dard molar Gibbs free energies of formation at 25 C of Br 2 ( ) and Cl 2 (g). Explanation: Δ G 0 f HBr(g) = - 53 . 43 kJ/mol Δ G 0 f HCl(g) = - 95 . 30 kJ/mol Δ G 0 f Br 2 = 0 kJ/mol Δ G 0 f Cl 2 = 0 kJ/mol Δ G 0 rxn = summationdisplay n Δ G 0 fprod - summationdisplay n Δ G 0 frct = 2( - 95 . 30 kJ / mol) - 2( - 53 . 43 kJ / mol) = - 83 . 74 kJ / mol For spontaneous reactions, Δ G is negative. 021 3.8points Δ G vap for H 2 O( ) at 85 C is ( < 0, = 0, > 0) and at 100 C is ( < 0, = 0, > 0). 1. > 0; > 0 2. < 0; < 0 3. > 0; = 0 correct 4. < 0; > 0 5. < 0; = 0 Explanation: 022 3.8points Consider the following reaction: CO(g) + Cl 2 (g) COCl 2 (g) Species ΔH 0 f S 0 kJ/mol J/mol · K CO - 110 . 5 197 . 6 Cl 2 0 . 0 223 . 0 COCl 2 - 223 . 0 289 . 2 Calculate Δ G 0 for the reactions at 298 K. 1. - 7 . 34 J/mol 2. - 73 . 4 kJ/mol correct 3. - 500 . 0 kJ/mol 4. - 39 . 3 kJ/mol 5. 221 . 1 kJ/mol 6. - 151 . 6 kJ/mol Explanation: The information available to us suggests we should use the standard state Gibbs equation: Δ G 0 = Δ H 0 - T Δ S 0 , However, we have to calculate Δ H 0 and Δ S 0 first. Δ H 0 can be determined using Hess’s Law: Δ H 0 = summationdisplay n Δ H 0 f products - summationdisplay n Δ H 0 f reactants = (Δ H 0 f for COCl 2 ) - ( Δ H 0 f for Cl 2 ) - ( Δ H 0 f for CO ) = ( - 223 . 0 kJ / mol) - ( - 0 kJ / mol) - ( - 110 . 5 kJ / mol) = - 112 . 5 kJ / mol Then Δ S 0 can be determined from the equation similar to Hess’s Law:

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casey (rmc2555) – Homework 1 – holcombe – (51395) 8 Δ S 0 = summationdisplay n Δ S 0 products - summationdisplay n Δ S 0 reactants = ( S 0 for COCl 2 ) - bracketleftbig( S 0 for CO ) + ( S 0 for Cl 2 )bracketrightbig = (289 . 2 J / mol · K) - (197 . 6 J / mol · K) - (223 . 0 J / mol · K) = - 131 . 4 J / mol · K With Δ H 0 and Δ S 0 in hand, we can now use the Gibbs equation. Δ G 0 = ( - 112 . 5 kJ / mol) - (298 K) parenleftbigg - 131 . 4 J mol · K parenrightbigg · 1 kJ 1000 J = - 73 . 3428 kJ / mol 023 3.8points A 22 . 4 L vessel contains 0 . 02 mol H 2 gas, 0 . 02 mol N gas, and 0 . 1 mol NH 3 gas. The total pressure is 700 torr. What is the partial pressure of the H 2 gas? 1. 100 torr correct 2. 7 torr 3. 14 torr 4. 28 torr 5. None of these Explanation: n total = 0 . 14 mol P total = 700 torr n H 2 = 0 . 02 mol X H 2 = n H 2 n total = 0 . 02 mol 0 . 14 mol = 0 . 142857 P H 2 = X H 2 P total = (0 . 142857) (700 torr) = 100 torr 024 3.8points A 5.0 L flask containing O 2 at 2.00 atm is connected to a 3.0 L flask containing H 2 at 4.00 atm and the gases are allowed to mix. What is the mole fraction of H 2 ? 1. Cannot be determined 2. 0.33 3. 0.55 correct 4. 0.45 5. 0.67 6. 0.25 Explanation: Applying the ideal gas law P V = n R T = P V R T N O 2 = (2 atm) (5 L) R T N H 2 = (4 atm) (3 L) R T Assume the temperature of the two gases remains the same before and after the mixing occurs. The mole fraction of H 2 is χ H 2 = n H 2 n total = n H 2 n H 2 + n O 2 = (4 atm) (3 L) R T (4 atm) (3 L) R T + (2 atm) (5 L) R T = (4 atm) (3 L) (4 atm) (3 L) + (2 atm) (5 L) = 0 . 545455 025 3.8points The total barometric pressure on a certain day was 755.1 torr. The humidity was such that water vapor in the air contributed 20.4 torr. The other components of air that day contributed a pressure of
casey (rmc2555) – Homework 1 – holcombe – (51395) 9 1. 760 torr.
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