General Chemistry and Qualitative Analysis
General Chemistry and Qualitative Analysis CHEM 104
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This 8 page Class Notes was uploaded by Francisca Sipes DVM on Tuesday October 27, 2015. The Class Notes belongs to CHEM 104 at University of Wisconsin - Milwaukee taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/230299/chem-104-university-of-wisconsin-milwaukee in Chemistry at University of Wisconsin - Milwaukee.
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Date Created: 10/27/15
A or N205 Concentration molL FirstOrder Reaction Plot of N 0 Concentration as a Function of Time A12 100 molL39 r action 50 complete after 1200 sec 20 min 2 N205 gt 4 N02 93 029 mAmAD kz A A10 A0 initial reaction is quot50 025 reaction is completequot 0125 A0 is quot875 0 1200 2400 3600 4800 6000 Timet sec A or NOCI Concentration molL SecondOrder Reaction Plot of NOCI Concentration as a Function of Time A10 446 molL reaction 50 complete after 582 hours NOCI g gt NO 9 Clzg A0 is quot50 025 reaction is completequot 6 12 18 Time t hours NOCI 50 complete 75 complete RATE LAWS Summary ORDER ZeroOrder FirstOrder SecondOrder A A Rate Law rate L k A0 k rate k A1 rate k A2 At At At Rate Constant k unit mo39 7 L Ls 3 mol s AO 1n2 H Ifo t 7 t 7 a Ie 12 t1 2k 12 k 12 Ab k It thtL A A kt 1A1A kt 1 1 he rae ae aw n n 77 9 0 0 A1 A10 A10 k t A A1 A10 e 1 1AOkt slope k slope k 1A0 1 1 1 2 The concentration of N204 at equilibrium is 00750 molL Hence the number of moles of this compound is 00750 molL 50 L 0375 mol and the change in moles of this compound 3 As there is twice the number of N02 molecules compared with N204 the number of N02 molecules must be increasing by 0250 mol 2 0500 mol Accordingly the number of N02 1 PROBLEM The brown gas N02 and the colorless gas N204 exist in equilibrium 2 N02 3 N204 In an experiment 0625 mol of N204 was introduced into a 50 L vessel and was allowed to decompose until equilibrium was reached The concentration of N204 at equilibrium was 00750 M Calculate Kc for the reaction 1 SOLUTION Start solving this problem using the lCE Table 2 N02 1 N204 No of moles initially 0000 mol 0625 mol Changes in moles No of moles at equilibrium Equilibrium Conc molL 00750 molL from initial to equilibrium condition is 0375 mol 0625 mol 0250 mol 2 N02 1 N204 No of moles initially 0000 mol 0625 mol Changes in moles 0250 mol No of moles at equilibrium 0375 mol Equilibrium Conc molL 00750 molL molecules at equilibrium must be 0500 mol 2 N02 1 N204 No of moles initially 0000 mol 0625 mol Changes in moles 0500 mol 0250 mol No of moles at equilibrium 0500 mol 0375 mol Equilibrium Conc molL 01000 moIL 00750 molL K N022 012 001 N204 0075 0075 75 Calculate equilibrium constant by using equilibrium concentrations as follows PROBLEM Phosgene COCI2 a poisonous gas decomposes according to the equation COCI2g S COg Cl2g Calculate Kp for this reaction if Kc 0083 at 900 C Solution 1 The equilibrium constant Kp is defined to be KP Kc R TW 7 where An is the difference of moles between products and reactants Hence An 1 2 Using R 008206 and T 900 273 K 1173 K the equilibrium constant Kp is Kp 0083 008206 11731280 PROBLEM If one starts with pure N02g at a pressure of 0500 atm the total pressure inside the reaction vessel when 2 NOgg S2NOgOgg reaches equilibrium is 0674 atm Calculate the equilibrium partial pressure of N02 1 0 1 U1 SOLUTION Start solving this problem using the lCE Table 2 N02 2 NO 1 Oz TOTAL Partial pressures initially atm 0500 0000 0000 0500 Changes in partial pressure Equ partial pressures atm 0674 As some part of N02 decomposes its concentration and therefore its partial pressure must be decreasing by X Accordingly its equilibrium partial pressure must be 0500 X 2 N02 2 NO 1 Oz TOTAL Partial pressures initially atm 0500 0000 0000 0500 Changes in partial pressure x Equ partial pressures atm 0500 X 0674 As the ratio of moles N02 NO 02 is 2 2 1 the changes in partial pressure must be X for NO and 05 X for 02 remember NO and 02 concentrations are increasing with respect to initial conditions Therefore the overall change in pressure from initial to equilibrium conditions must be X X 05 X 05 X and the equilibrium partial pressures of NO and 02 are X and 05 X 2 N02 2 NO 1 Oz TOTAL Partial pressures initially atm 0500 0000 0000 0500 Changes in partial pressure X X 05x 05x Equ partial pressures atm 0500 X X 05 X 0674 The total pressure of a gas miXture is the sum of all partial pressures of gases in this miXture Ptotal PNO PM Po2 0500 96 x 05x 0674 Pm 0500 05x 0674 05x 0674 0500 0174 x 017405 0348 The equilibrium partial pressure of N02 therefore is PNO2 0500 0348 0152 atm PROBLEM Equilibrium is established for the reaction 2Xs Yg 3 229 at 500 K Kc 100 Determine the concentration of Z in equilibrium with 02 mol X and 050 M Y at 500 K 1 N Solution Please note that the concentration of Xs has no impact on the equilibrium constant of this gas reaction because X is a solid Hence equilibrium constant reads 22 KC Y Rearranging this equation for Z reads 22 Kc in Z ch 1V 100050 71 molL PROBLEM Sodium carbonate Na2C03 s can be prepared by heating sodium bicarbonate NaHC03s 2 NaHCO3s S Na2003s C02g H20g Kp 023 at 100 C If a sample of NaHC03 is placed in an evacuated flask and allowed to achieve equilibrium at 100 C what will the total gas pressure be SOLUTION 1 Please note that the concentrations of sodium carbonate and sodium bicarbonate have no impact on the equilibrium constant of this gas reaction because both compounds are solids The equilibrium constant therefore reads K p PCOZ PH20 023 2 As equal amounts moles of C02 and H20 are present PCOZ PH0 x 3 Hence x2 023 x M 048 4 The total pressure of a gas mixture is the sum of all partial pressures of all gases involved PM PCO2 PHZO 048 atm 048 atm 096 atm