Exam 4 - Version 155 – Exam 4 – Mccord –(52385 1 This...

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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 McCord’s 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 Δ S = q T since T is in the denomina- tor, Δ 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) Δ H ◦ r = 2 (BE H − S ) Reactants: Δ H f H 2 S(g) =- 20 . 15 kJ/mol Products: Δ H f H(g) = 218 kJ/mol Δ H f S = 223 kJ/mol Δ H rxn = summationdisplay n Δ H f prod- summationdisplay n Δ 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: Δ H = Σ BE reactants- Σ 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, Δ H and Δ 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, Δ 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 Δ H = Δ U + p Δ V , because p Δ V is the expansion work. Statement II follows from Δ U = q + w , because w = 0 for processes that occur at constant volume. Statement III also follows from Δ 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. ΔV 2. Δ S 3. Δ E correct 4. Δ H 5. Δ G Explanation: Δ E = q + w q = Δ H w =- P Δ V When Δ V = 0, w = 0....
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Exam 4 - Version 155 – Exam 4 – Mccord –(52385 1 This...

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