Exam 4 - Version 155 Exam 4 Mccord (52385) 1 This print-out...

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Unformatted text preview: Version 155 Exam 4 Mccord (52385) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord CH301 This exam is only for McCords TTH CH301 class. 001 10.0 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. False correct 2. True Explanation: From S = q T since T is in the denomina- tor, S will be larger (more positive) when- ever T is smaller . 002 10.0 points Calculate the average H S bond enthalpy in H 2 S(g) given the standard enthalpies of for- mation for H 2 S(g), H(g), and S(g) as- 20 . 1, 218, and 223 kJ mol 1 , respectively. 1. 231 kJ mol 1 2. 679 kJ mol 1 3. 10 . 1 kJ mol 1 4. 340 kJ mol 1 correct 5. 461 kJ mol 1 Explanation: Write an equation which represents the bond energy (or a multiple of it): H 2 S(g) 2 H(g) + S(g) H r = 2 (BE H S ) Reactants: H f H 2 S(g) =- 20 . 15 kJ/mol Products: H f H(g) = 218 kJ/mol H f S = 223 kJ/mol H rxn = summationdisplay n H f prod- summationdisplay n H f rct = [2 (218 kJ / mol) + 223 kJ / mol]- (- 20 . 1 kJ / mol) = 679 . 1 kJ / mol rxn , so H rxn = 2 BE BE = 679 . 1 kJ mol rxn 2 = 339 . 55 kJ mol rxn . 003 10.0 points Using the provided bond energy data, cal- culate the change in enthalpy for the reaction below. 6HCl(g) + N 2 (g)- 2NH 3 (g) + 3Cl 2 (g) Bond Energy (kJ mol 1 ) H- Cl 432 N N 945 N- H 391 Cl- Cl 199 1.- 2 , 357 kJ mol 1 2. 787 kJ mol 1 3.- 594 kJ mol 1 4. 2 , 357 kJ mol 1 5. 594 kJ mol 1 correct Explanation: H = BE reactants- BE products = (6 432 + 945)- (6 391 + 3 199) = 594 kJ mol 1 Version 155 Exam 4 Mccord (52385) 2 004 10.0 points Which of the following statements concerning internal energy is/are true? I) If the expansion work is small, H and U are close in value. II) The internal energy of a system is equal to q at constant volume. III) Assuming no heat is exchanged, when pressure-volume work is done on the sys- tem, U is positive. 1. I, II 2. II, III 3. I only 4. III only 5. I, II, III correct 6. II only 7. I, III Explanation: Statement I follows from the identity H = U + p V , because p V is the expansion work. Statement II follows from U = q + w , because w = 0 for processes that occur at constant volume. Statement III also follows from U = q + w , because w is the pressure- volume work. (Note: for reversible processes, expansion work and pressure-volume work are identical.) 005 10.0 points Heat absorbed by a system at constant vol- ume is equal to 1. V 2. S 3. E correct 4. H 5. G Explanation: E = q + w q = H w =- P V When V = 0, w = 0....
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Exam 4 - Version 155 Exam 4 Mccord (52385) 1 This print-out...

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