CH302 Holcombe Exam #1 '12

CH302 Holcombe Exam #1 '12 - Version 056 – Exam #1 –...

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Unformatted text preview: Version 056 – Exam #1 – holcombe – (51160) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The colligative effects of 1 molal sugar solu- tion would be ? 1 molal sodium chloride solution. 1. the same as 2. greater than 3. less than correct Explanation: Sugar is a non-electrolyte so the concen- tration of particles is the concentration of dissolved sugar. Soduim chloride is an electrolyte which dis- sociates in solution, giving twice as many par- ticles as dissolved: NaCl-→ Na + + Cl − Since colligative properties depend on the number of particles in solution, the colligative effect doubles for the NaCl(aq), compared to any identical concentration of sugar. 002 10.0 points Consider the following system at equilibrium. H 2 (g) + I 2 (g) ⇀ ↽ 2 HI(g) + heat Which response includes all the following that will shift the equilibrium to the left, and no others? I) increasing the temperature II) decreasing the temperature III) increasing the pressure IV) decreasing the pressure V) removing some HI VI) adding some HI VII) removing some I 2 VIII) adding some I 2 1. II, IV, VII, and VIII only 2. II, V, and VIII only 3. I, III, V, and VII only 4. I, VI, and VII only correct 5. II only Explanation: Increasing temperature adds heat and will shift equilibrium to the left. Changing pres- sure will not affect equilibrium because the number of gas molecules is the same on both sides of the equation. Removing a reactant or adding a product will shift equilibrium to the left. 003 10.0 points Calculate the standard reaction enthalpy for the reaction NO 2 (g) → NO(g) + O(g) Given: O 2 (g) → 2 O(g) Δ H ◦ = +498 . 4 kJ / mol NO(g) + O 3 (g) → NO 2 (g) + O 2 (g) Δ H ◦ =- 200 kJ / mol 3O 2 (g) → 2 O 3 (g) Δ H ◦ = +285 . 4 kJ / mol . 1. +93 . 5 kJ / mol 2. +306 kJ / mol correct 3. +555 kJ / mol 4. +413 kJ / mol 5. +355 kJ / mol 6. +592 kJ / mol 7. +192 kJ / mol Explanation: O 2 (g) → 2 O(g) Δ H ◦ = +498 . 4 kJ / mol NO(g) + O 3 (g) → NO 2 (g) + O 2 (g) Δ H ◦ =- 200 kJ / mol Version 056 – Exam #1 – holcombe – (51160) 2 The standard formation of ozone is 3 2 O 2 (g) → O 3 (g) ΔH = +142 . 7 kJ / mol We calculate the ΔH rxn using Hess’ Law: To combine the reactions and get the de- sired reaction, reverse the second and third equations and add half of the first one: NO 2 (g) + O 2 (g) → NO(g) + O 3 (g) ΔH = +200 kJ / mol O 3 (g) → 3 2 O 2 (g) ΔH =- 142 . 7 kJ / mol 1 2 O 2 (g) → O(g) ΔH = 1 2 (498 . 4 kJ / mol) NO 2 (g) → NO(g) + O(g) ΔH rxn = 306 . 5 kJ / mol 004 10.0 points 005 10.0 points Consider the reaction 2 Fe 2 O 3 (s) + 3 C(s) → 4 Fe(s) + 3 CO 2 (g) , Δ H ◦ = 462 kJ , Δ S ◦ = 558 J · K − 1 . Calcu- late the equilibrium constant for this reaction at 525 ◦ C....
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This note was uploaded on 02/12/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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CH302 Holcombe Exam #1 '12 - Version 056 – Exam #1 –...

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