Chem 121 Problem Set VII Gases ANSWERS

Transcription

1 Chem roblem Set II Gases ASWERS a. We are dealin with the same sample of as at two different altitudes, so but so 0.98atm.L0.70atm 3.L b. We are dealin with the same sample of as at two different temperatures, so Chem roblem set II Solutions but so but 3 so 93K 879K or 606 C 93K 93K n c. Since the temperature and pressure are constant, R n n L 8.0 or n n mass L but there already was 6. in the balloon, so mass needed to add is a. Let be the volume of the first flask. We are dealin with the same sample of as and the temperature is constant 3.6atm.7atm ( + 4.9L) 5L b. We are dealin with the same sample of as and the volume is constant 43K 50.0psi 40.psi 303K c. We are dealin with the same sample of as and can assume that the volume is constant so: but we do not know. We find by usin the ideal as equation for the initial conditions. 6.0atm 0.48L 6.7K nr now the temperature is increased by 400 C or 400 K so final temperature is 46.7 K, so 46.7K 6.0atm 49.7atm 50.atm 6.7K d. From the percentae composition we can calculate the empirical formula, but to et the ecular formula we need the ar mass. We can find the ar mass from the ideal as equation. atm 0.476L es of cyanoen, n 750torr.9 0 R 760torr K but.9 x 0 - of cyanoen has a mass of.00 so ar mass.00 /.9 x / o find the empirical formula, we will assume we have of cyanoen so Element Relative mass Relative number of es (atoms) Divide by the smallest number C O he empirical formula is C x y C and the empirical mass is

2 Chem roblem set II Solutions - he number of empirical formula units in the ecular formula is ecular mass empirical mass 6.0 Molecular formula is (C) C e. (i) o find the e fraction we need the total es of as 4 H es ecules 6.64 es H ecules.06 total es of ases in container H O 4.6 X X X H O (ii) total pressure from total es in ideal as equation, nr K 58.5atm 8.L (iii) artial pressure of each as X atm 7atm X atm 7.9atm X atm 4atm H H O O f. k.m.s 3R K rms(h ) 97m.s where is the ar mass in k 3 µ.06 0 k/ 3a. need to find es of carbon dioxide needed to produce.00 k of suar 000 suar CO k 34.3 suar es CO.00k suar nr K volume of CO from ideal as equation 843L.0atm 00L air 6 3b. volume of air843l CO. 0 L 0.040L CO 4. From the percentae composition we can calculate the empirical formula, but to et the ecular formula we need the ar mass. We can find the ar mass (M) from the ideal as equation. mass n dr mass mass R R where density, d and es, n dr 4.05.L K atm o find the empirical formula, we will assume we have of cyanoen: Element Relative mass () Relative number of es (atoms) Divide by the smallest number C H F he empirical formula is C x H y F z C H F and the empirical mass is he number of empirical formula units in the ecular formula is ecular mass empirical mass Molecular formula is (CH F) C H 4 F

3 Chem roblem set II Solutions We can calculate the total pressure of ases from the ideal as equation, as we are iven the volume and temperature but we need to find the total number of es of as at the completion of the reaction. nce we must evaluate the es of each species at the completion of the reaction. ote that water will be a as as the temperature is 7 C. First balance the equation: C 3 H 8 () + 5 O () 3 CO () + 4 H O() ow to find the limitin reaent. Moles of CO formed from: propane 3 CO O 3 CO propane oxyen propane O oxyen is limitin and 5.65 x 0 - CO formed. ow to find the es of the other species at the completion of the reaction. O 4 H O es water formed O O propane es propane used O propane 3 es propane remainin propane total es of ases, n 5.65 x 0 - CO x 0 - H O x 0-3 propane + 0 O 0.35 nr K.00L 5.55atm 6. We are dealin with the same sample of as so -3 atm 75atm (0.0 0 L) 738torr 760torr 0.39L 073K 69K 7. n R.4 0 atm L n O3 O3 9 O 3 O 3 R K 8a) Since and n (es of CO and H ) are constant: f f i i and we usin Dalton s Law we can treat each as as if it was the only as present. ote that the total volume or f 8.35 L + 4. L. i i 45torr 4.L CO : f 4.45torr (8.35L+ 4.L) f i i 35torr 8.35L H : f 34.0torr f (8.35L+ 4.L) CO + H 4.45 torr torr 376 torr 8b) We obtain the e fraction of H from the other form of Dalton s Law: H X H H 34.03torr XH tor 9a) S is 73 K and atm but we need to know the es of O to find the volume. We find the es of oxyen usin the ideal as equation on the data iven. Since the O is collected over water we need to correct for the vapour pressure of water. 0.3ka 3.4ka 97.9ka O atm HO

4 97.9ka 0.6L atm 0.3ka 3 es O, no R K 3 n O R K volume O at S 0.L atm Chem roblem set II Solutions - 4 9b) he KClO 3 is impure so we cannot calculate the es of KCl from the e of KClO 3. We need to calculate the mass of KCl formed from the es of oxyen formed. But first we must balance the equation KClO 3 (s) KCl(s) + 3 O () KCl mass KCl O O KCl 9c) o find the percent purity of the KClO 3 we must find the mass of pure KClO 3 from the es of O formed. 3 KClO3.5 mass KClO O O KClO ercent purity % a) he averae speed (µ rms ) is inversly proportional to the mass of the ecules (and proportional to the temperature). 3R µ rms ar masses: F is ; O is ; is so the heaviest (F ) is the slowest and the lihtest ( ) is the fastest and the order is F, O, 0b) We have equal masses of each as in a.00 L tank. Let us say we have 00. of each as. F O es F es O es total es.63 F O es F.63 e fraction F, XF 0.8 total es c) o find the partial pressure of we need to find the e fraction of X X atm 0.383atm For this problem we use Graham's Law of Diffusion r r r r 75mL.hr.0 0 ml.hr he rate of effusion of is.00 ml in 75 s.00 ml/75 s 5.74 x 0-3 ml.s - he rate of effusion of the CO-CO mixture.00 ml/00 s 5.00 x 0-3 ml.s - 3 r ml s r r CO CO CO CO CO-CO ml r CO CO s CO CO

5 Chem roblem set II Solutions - 5 but M CO and M CO let x be the fraction of CO in the mixture and so the fraction of CO is ( x) 8.0 x ( x) x x x 7.43 x or ercent CO in the mixture is 46% 3. Since the total es of as before and after mixin are unchaned we have O f f O n + ( + ) or + R R R R f 900mL 500mL 700torr + 400mL + 950torr f 8torr n e fraction of X R n f f f f R 700torr 500mL X f f 389torr f f f 900mL 8torr 389torr 4torr O f O O f f O O 4. From Dalton s Law the pressure from before and after removal of O is the same, so oriinal pressure of is 50 torr. 35torr 385torr 50torr 35torr X 0.35 X X O O total O O O total 385torr 5. KE AG for H atom is 8.00 x 0-6 J.atom - and first we want the averae kinetic enery for a e of deuterium 6 J 3 atoms 8 J KEAG/ 8.00x0 6.0x x0 atom KE AG/ 3/ R KEAG/ 8 J x0 3.86x0 K J 3R Rate Rate H where is the number of atoms and H is the number of H ecules H H 3.06 and 4.7 and 4.7 H H H H H H () H XH (+ 4.7) H H H H H 7. Mo (s) + x/ F () MoF x () 7.770atm.00L Initial es of as initial es of F 0.00 atml K K 3.885atm.00L Final es of as es excess F + es MoFx 0.00 atml K K but es of MoF x formed initial es of Mo so es excess F thus es F used

6 Chem roblem set II Solutions - 6 F so es of F in MoF x 0.50 F F and of Mo in MoF x Atom Moles Divide by smallest Closest whole number ratio Mo F and the compound is MoF 6 and consequently x 6. 8i) (a) Assumin ideal behavoir nr K 489atm.00L (b) Usin the an Der Waals equation na + ( nb) nr nr n a K (0.0) atm (-nb) (.00L L. (.00L) 8ii) (a) Assumin ideal behavoir nr K.atm.00L (b) Usin the an Der Waals equation na + ( nb) nr nr n a K (0.500) atm (-nb) (.00L L. (.00L) At low pressures (0.5 /L) ), the deviation is neative as the effect of attractive forces overwhelms the effect of ecular size and the attractive forces pull the ecules toether away from the container walls, lowerin the pressure. At hih pressures (0 /L), the deviation is positive as the effect of ecular size overwhelms the effect of attractive forces and the volume occupied by the ecules means there is less volume between the ecules, increasin the pressure. Carbon dioxide switches from a neative to a positive deviation at 600 atm (actual pressure not that calculated from the ideal as equation).