Exam 2-solutions - Version 348 Exam 2 sutcliffe (51045) 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 348 Exam 2 sutcliffe (51045) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which one of the following pairs of acids and their conjugate bases is INCORRECTLY matched? 1. HClO : ClO 2. NH + 4 : NH 2 correct 3. H 2 O : OH 4. H 3 O + : H 2 O 5. HF : F Explanation: An acid is a proton donor and a base is a proton acceptor. The only difference between the acid and its conjugate base is that the base has one less H atom. NH 2 has two fewer H atoms than the NH + 4 ; therefore, it is incorrectly matched. 002 (part 1 of 2) 10.0 points Consider the reaction HCHO(g) H 2 (g) + CO(g). 1.0 mol of HCHO, 1.0 mol of H 2 and 1.0 mol of CO exist in equilibrium in a 2.0 L reaction vessel at 600 C. a) Determine the value of the equilibrium constant K c for this system. 1. 2.0 2. K c cannot be determined for gaseous equilibria. 3. 0.50 correct 4. 1.0 5. Some other value Explanation: [HCHO] = 1 . 0 mol 2 L = 0 . 50 M [H 2 ] = 1 . 0 mol 2 L = 0 . 50 M [CO] = 1 . 0 mol 2 L = 0 . 50 M The expression for K c here is K c = [H 2 ] [CO] [HCHO] , so we need to know the equilibrium concen- trations of each of the species. Since they are all 0 . 50 M, then K c = [0 . 50 M] [0 . 50 M] [0 . 50 M] = 0 . 50 003 (part 2 of 2) 10.0 points 2.0 moles of HCHO and 1.0 mol of CO are then added to this system. b) Which of the following statements about this reaction is now true? 1. The reaction mixture is not at equilib- rium, but will move toward equilibrium by forming more HCHO. 2. The reaction mixture remains at equilib- rium. 3. The forward rate of this reaction is the same as the reverse rate at these new concen- trations. 4. The reaction mixture is not at equilib- rium, but will move toward equilibrium by using up more HCHO. correct 5. The reaction mixture is not at equilib- rium, but no further reaction will occur. Explanation: The new concentrations for the species are [HCHO] = 3 . 0 mol 2 . 0 L = 1 . 5 M [CO] = 2 . 0 mol 2 . 0 L = 1 . 0 M [H 2 ] = 0 . 50 M Solving for Q , we have Q = [H 2 ] [CO] [HCHO] = (0 . 5 M) (1 . 0 M) 1 . 5 M = 0 . 33 Version 348 Exam 2 sutcliffe (51045) 2 Since Q is less than K c , the reaction will proceed forward, which means that HCHO will be consumed in order to form more prod- ucts. 004 10.0 points A bottle labeled 0.150 M sulfuric acid contains a solution with what concentration in sulfate ions? 1. 0.161 M 2. 0.075 M 3. 0.011 M 4. 0.150 M correct 5. 0.139 M Explanation: [Sulfuric acid] = 0 . 150 M Sulfuric acid is H 2 SO 4 . Molarity (M) is moles solute/L solution. We can write 0.150 M H 2 SO 4 as the ratio . 150 mol H 2 SO 4 1 L soln ....
View Full Document

Page1 / 7

Exam 2-solutions - Version 348 Exam 2 sutcliffe (51045) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online