Exam 2-solutions

Exam 2-solutions - Version 182 Exam 2 McCord (53575) This...

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Version 182 – Exam 2 – McCord – (53575) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. McCord 10am For 0 = ax 2 + bx + c x = - b ± b 2 - 4 ac 2 a 001 10.0 points What happens to the concentration oF HI(g) when the total pressure on the equi- librium reaction 2 HCl(g) + I 2 (s) 2 HI(g) + Cl 2 (g) is increased (by compression)? 1. remains the same 2. Unable to determine 3. decreases correct 4. increases Explanation: Increasing the total pressure on the sys- tem by decreasing its volume will shiFt the equilibrium toward the side oF the reaction with Fewer numbers oF moles oF gaseous com- ponents. IF the total number oF moles oF gas is the same on the product and reactant sides oF the balanced chemical equation, then changing the pressure will have little or no eF- Fect on the equilibrium distribution oF species present. 002 10.0 points IF 75.0 mL oF a 0.026 M HCl solution is mixed with 25.0 mL oF a 0.15 M NaOH solution, what is the pH oF the fnal mixture? 1. 13.09 2. 4.15 3. 2.74 4. 10.37 5. 12.26 correct 6. 7.38 7. 3.13 8. 1.74 Explanation: [HCl] = 0.026 M V HCl = 75.0 mL [NaOH] = 0.15 M V NaOH = 25.0 mL To determine the pH oF the fnal mixture, we need to determine how much H or OH is leFt over aFter the reaction. Remember that For complete neutralization we need H and OH in equal molar amounts: H + + OH H 2 O ±irst calculate how many moles oF H + and OH we have: ? mol H + = 0 . 075 L × 0 . 026 mol HCl 1 L × 1 mol H + 1 mol HCl = 1 . 95 × 10 3 mol H + ? mol OH = 0 . 025 L × 0 . 15 mol NaOH 1 L × 1 mol OH 1 mol NaOH = 3 . 75 × 10 3 mol OH Now we can see that we have more OH than H + so H + will be consumed and OH will be leFt over. Let’s fnd out by how much: ? mol OH = 3 . 75 × 10 3 - 1 . 95 × 10 3 = 1 . 8 × 10 3 mol OH The next step is to calculate the [OH ]: ? M OH = 1 . 8 × 10 3 mol OH 0 . 075 L + 0 . 025 L = 0 . 018 M OH

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Version 182 – Exam 2 – McCord – (53575) 2 pOH = - log[OH ] = 1 . 74 pH = 14 - pOH = 14 - 1 . 74 = 12 . 26 003 10.0 points Pure liquids or solids do NOT appear in the equilibrium constant expression. 1. True correct 2. False Explanation: 004 10.0 points The equilibrium constant K p is 5 . 00 × 10 17 at 25 C for the reaction C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) . From this information, calculate Δ G 0 at 25 C. 1. - 43 . 9 kJ/mol 2. - 517 kJ/mol 3. 101 kJ/mol 4. - 101 kJ/mol correct 5. insu±cient information 6. +43.9 kJ/mol 7. - 996 J/mol Explanation: K p = 5 . 00 × 10 17 T = 25 C + 273 = 298 K Δ G 0 = - RT ln K = ( - 8 . 314 J / mol · K)(298 K) × ln ( 5 × 10 17 ) = - 1 . 01 × 10 5 J / mol = - 101 kJ / mol 005 10.0 points Suppose the reaction A + 3 B 2 C has a value of K = 10 . 0 at a certain tempera- ture. If 0.5 moles of A, 0.5 moles of B and 0.5 moles of C are placed in a 5 L solution, the reaction 1. shifts to the left.
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Exam 2-solutions - Version 182 Exam 2 McCord (53575) This...

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