S10_Student_Lecture_10

S10_Student_Lecture_10 - Mon. 2/15/2010 Announcements HW...

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Mon. 2/15/2010 Announcements HW schedule, as usual! Register your iclicker if you are not receiving points on Blackboard Lab this week: Equilibrium Concentrations Ch. 6
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Determining Equilibrium Concentrations from K–I Problem: One laboratory method of making methane is from carbon disulfide reacting with hydrogen gas, and K this reaction at 900°C is 27.8. CS 2 (g) + 4 H 2 (g) CH 4 (g) + 2 H 2 S (g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS 2 , 1.10 mol of H 2 , and 0.45 mol of H 2 S, how much methane was formed? Plan: Write the reaction quotient, and calculate the equilibrium concentrations from the moles given and the volume of the container. Use the reaction quotient and solve for the concentration of methane. Solution: [CH 4 ] [H 2 S] 2 CS 2 (g) + 4 H 2 (g) CH 4 (g) + 2 H 2 S (g) K = = 27.8 [CS 2 ] [H 2 ] 4 [CS 2 ] = 0.250 mol 4.70 L [CS 2 ] = _________ mol/L
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Determining Equilibrium Concentrations from K–II Solution cont. [H 2 ] = = _______ mol/L 1.10 mol 4.70 L [H 2 S] = = _______ mol/L 0.450 mol 4.70 L [CH 4 ] = = K c [CS 2 ] [H 2 ] 4 [H 2 S] 2 (27.8)(0.05319)(0.23404) 4 (0.095745) 2 [CH 4 ] = = _________ mol/L = ____ M 0.009167 0.004436 Check: Substitute the concentrations back into the equation for K and make sure that you get the correct value of K K = = = 27.8 [CH 4 ] [H 2 S] 2 [CS 2 ] [H 2 ] 4 (0.485547 M )(0.095745 M ) 2 (0.05319 M )(0.23404 M ) 4 OK!
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Typical Calculations PROBLEM: Place 1.00 mol each of H PROBLEM: Place 1.00 mol each of H 2 and I and I 2 in a in a 1.00 L flask. Calc. equilibrium concentrations. 1.00 L flask. Calc. equilibrium concentrations. H H 2 (g) + I (g) + I 2 (g) (g) 2 HI(g) 2 HI(g) K K c = 55.3 = 55.3
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H (g) + I (g) 2 HI(g) K = 55.3 Step 1. Set up Step 1. Set up ICE ICE table to define table to define EQUILIBRIUM concentrations. EQUILIBRIUM concentrations. [H [H 2 ] ] [I [I 2 ] ] [HI] [HI] Initial Initial 1.00 1.00 1.00 1.00 0 0 Change Change Equilib Equilib
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H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 2. Put equilibrium concentrations Step 2. Put equilibrium concentrations into K into K c expression. expression. K c = [2x] 2 [1.00 - x][1.00 - x] = 55.3
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Step 3. Solve K Step 3. Solve K c expression - take expression - take square root of both sides. square root of both sides. x = _____ x = _____ Therefore, at equilibrium Therefore, at equilibrium H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 K c = [2x] 2 [1.00 - x][1.00 - x] = 55.3 00 . 1 2 = 44 . 7 x x - [H [H 2 ] = [I ] = [I 2 ] = 1.00 - x = 0.21 M ] = 1.00 - x = 0.21 M [HI] = [HI] =
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I 2 + I o I 3 o A solution with concentrations of [I
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S10_Student_Lecture_10 - Mon. 2/15/2010 Announcements HW...

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