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# Exam 2 - Version 222 – Exam 2 – mccord –(51620 1 This...

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Unformatted text preview: Version 222 – Exam 2 – mccord – (51620) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord - CH302 1pm class only - 51620 001 10.0 points Given the hypothetical reaction X(g) ⇀ ↽ Y(g) predict what will happen when 1.0 mol Y is placed into an evacuated container. 1. Δ G ◦ will decrease until Δ G ◦ = 0. 2. Q will decrease until Q = K . correct 3. Q will increase until Q = K . 4. Nothing; the products are already formed, so no reaction occurs. Explanation: The container begins with products only. The reverse reaction will predominate until equilibrium is reached. The reaction quotient Q starts out larger than K and decreases until Q = K . 002 10.0 points A 20 mL sample of 0.20 M nitric acid solution is required to neutralize 40 mL of barium hydroxide solution. What is the molarity of the barium hydroxide solution? 1. 0.0025 M 2. 0.025 M 3. 0.100 M 4. 0.200 M 5. 0.050 M correct Explanation: V HNO 3 = 20 mL [HNO 3 ] = 0.20 M V Ba(OH) 2 = 40 mL The balanced equation for this neutraliza- tion reaction is 2 HNO 3 + Ba(OH) 2 → Ba(NO 3 ) 2 + 2 H 2 O We determine the moles of HNO 3 used: ? mol HNO 3 = 0 . 020 L soln × . 20 mol HNO 3 1 L soln = 0 . 0040 mol HNO 3 Using the mole ratio from the chemical equa- tion we calculate the moles Ba(OH) 2 needed to react with 0.0040 mol of HNO 3 : ? mol Ba(OH) 2 = 0 . 0040 mol HNO 3 × 1 mol Ba(OH) 2 2 mol HNO 3 = 0 . 0020 mol Ba(OH) 2 There are 0.0020 moles Ba(OH) 2 in the 40 mL sample. Molarity is moles solute per liter of solution: ? M Ba(OH) 2 = . 0020 moles Ba(OH) 2 . 040 L solution = 0 . 050 MBa(OH) 2 003 10.0 points Consider the reaction: NO(aq) + CO 2 (aq) ↔ NO 2 (aq) + CO(aq) If at some temperature K c = 4, and the initial concentrations of NO, CO 2 , NO 2 and CO are 5, 10, 2 and 0 M respectively, what is the equilibrium concentration of CO? 1. 4 correct 2. 14 3. 1 4. 16 Version 222 – Exam 2 – mccord – (51620) 2 5. 6 6. 9 Explanation: K = parenleftbigg [NO 2 ] · [CO] [NO] · [CO 2 ] parenrightbigg = parenleftbigg (2 + x) · x (5- x) · (10- x) parenrightbigg = 4 x = 4 004 10.0 points A certain reaction has Δ H equal to 11.57 kJ/mol. This reaction is normally run at room temperature (25 ◦ C). At what new tem- perature should the reaction be run so that K is twice its value at 25 ◦ C? 1. 77 ◦ C correct 2. 31 ◦ C 3. 135 ◦ C 4. 65 ◦ C 5. 50 ◦ C 6. 94 ◦ C Explanation: T 1 = 25 ◦ C +273 = 298 K K 2 = 2 K 1 Δ H = 11570 J/mol ln K 2 K 1 = Δ H R parenleftbigg 1 T 1- 1 T 2 parenrightbigg R Δ H ln K 2 K 1 = 1 T 1- 1 T 2 T 2 = parenleftbigg 1 T 1- R Δ H ln K 2 K 1 parenrightbigg − 1 = parenleftbigg 1 298- 8 . 314 11570 ln2 parenrightbigg − 1 = 349 . 94 K = 77 ◦ C 005 10.0 points Which of the statements concerning equilib- rium is NOT true?...
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Exam 2 - Version 222 – Exam 2 – mccord –(51620 1 This...

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