helpsheet_08_103

1 CHEMISTRY 103 – Help Sheet #8 Gases (Part II) – Module 5 Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) Nuggets: Partial Pressure; Kinetic Energy; Velocity; Relative Rates of Gases; Diffusion/effusion; Parts per million; Kinetic Theory of Gases; Gas Equations: Summary PARTIAL PRESSURES: The pressure of one gas in a mixture of gases. PARTIAL PRESSURE P total = P 1 + P 2 + ... where P 1 , P 2 , etc., are the individual gas pressures; Dalton’s Law c 1 = mol fraction gas 1 ; c is mol% in decimal form; n total = n 1 + n 2 + ... P 1 = c 1 P total this means if 30% of the moles of a gas mixture are Gas 1 (i.e., c 1 = 0.30), then 30% of the total pressure comes from Gas 1 S c i = 1.00 sum of the mole fractions totals 1.00 (i.e., 100%) P 1 V = n 1 RT applying the Ideal Gas Law to one gas in a mixture of gases P total V = n total RT applying the Ideal Gas Law to the sum of all gases in a mixture of gases Example 1: a. If a glass bulb contains 110.g CO 2 and 140.g CO with a total pressure of 6.60atm, what is the partial pressure of CO 2 ? b. What is the partial pressure of CO? Answer 1: a. P CO 2 = 2.20atm b. P CO = 4.40atm {use P 1 = c 1 P total ; first find mol CO 2 and mol CO; ; ; now find mol fraction, c , for both CO 2 and CO; ; ; ; ; now find partial pressures: P CO 2 = c CO 2 P total ; P CO 2 = (0.333)(6.60atm) = 2.20atm; (a way to think about this: 0.333 or 1 / 3 of the molecules are CO 2 and 0.333 or 1 / 3 of the pressure is due to CO 2 ); P CO = c CO P total ; P CO = (0.667)(6.60atm) = 4.40atm; (a way to think about this: 0.667 or 2 / 3 of the molecules are CO and 0.667 or 2 / 3 of the pressure is due to CO); could also have calculated P CO from P total = P 1 + P 2 = P CO + P CO 2 ; P CO = 6.60atm – 2.20atm = 4.40atm} Example 2: A 50.0-L steel tank contains He and Ne. The container is at 25.0˚C and contains 56.0g He. a. What is the partial pressure of He? b. If the total pressure is 9.75atm, how many grams of Ne are in the tank? Answer 2: a. P He = 6.85atm b. 120.g Ne {Use P 1 V = n 1 RT; ; ; T = 25.0 + 273.15 = 298.15K; ; from Dalton’s Law: P total = P He + P Ne ; P Ne = 9.75 – 6.85 = 2.90atm; use P Ne V = n Ne RT to find n Ne ; ; } Example 3: When a hydrocarbon gas is collected in a 5.0-L vessel at 35.0˚C over water the pressure recorded was 825torr. The vapor pressure of water at 35.0˚C is 42.2torr. How many moles of the hydrocarbon were present? Answer 3: n hydrocarbon = 0.203mol {P T = P H2O + P hydrocarbon ; 825torr = 42.2 + P hydrocarbon ; P hydrocarbon = 782.8torr; use P 1 V = n 1 RT and solve for n hydrocarbon ; ; } χ 1 = n 1 n total = P 1 P total 110.g CO 2 1mol CO 2 44.0g CO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.50mol CO 2 140.g CO 1mol CO 28.0g CO ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 5.00mol CO χ CO 2 = molCO 2 molCO 2 + molCO χ CO 2 = 2.50 2.50 + 5.00 = 0.333 χ CO = molCO molCO 2 + molCO χ CO = 5.00 2.50 + 5.00 = 0.667 P He = n He RT V n He = 56.0g He 1mol He 4.00g He ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 14.0mol He P He = (14.0mol)(0.0821Latm / molK)(298.15K) 50.0L = 6.85atm n Ne = ( 2 . 90 atm)( 50 . 0 L) ( 0 . 0821 Latm / molK)( 298 . 15 K) = 5 . 924 mol Ne 5.924mol Ne 20.18g Ne 1mol Ne ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 119.5g Ne n hydrocarbon = P hydrocarbon V RT n hydrocarbon = ( 782 . 8 torr)( 1 atm / 760 torr)( 5 . 0 L) ( 0 . 0821 Latm / molK)( 35 . 0 + 273 . 15 K) = 0 . 203 mol hydrocarbon

2 Example 4: (A more challenging question!) 10.0g N 2 (g) reacts with 15.0g O 2 in the reaction below. If the reaction goes to completion, the temperature is 55.0˚C, and container volume is 2.0L, what are the P N 2 , P O 2 , P N 2 O 5 , and P total at the end of the reaction ? 2N 2 (g) + 5O 2 (g) ® 2N 2 O 5 (g) Answer 4: ; ; ; P total = 4.81atm {Limiting reagent (LR) problem since 2 reactant quantities given; need to find how many mol of each chemical are left over at the end of the reaction; do g A ® mol B two times to determine mol N 2 O 5 produced and which reactant is the LR; ; ; The smaller quantity of N 2 O 5 can be produced: 0.1875mol N 2 O 5 is produced; O 2 is the LR since it produced the smaller amount of N 2 O 5 so there are 0.0 mol of O 2 left over at the end of the reaction; find out how many moles of excess reactant, N 2 , is left over at the end of the reaction; left over = starting amount – amount used; find starting amount N 2 in mol: ; find amount N 2 used in mol: LR ® EX: ; left over = 0.3571 – 0.1875 = 0.1696mol N 2 ; use P 1 V = n 1 RT and solve for P 1 for each gas; ; ; ; P O 2 = 0.0atm because O 2 is the limiting reagent; P total = P N 2 + P O 2 + P N 2 O 5 = 2.2846 + 0 + 2.5257 = 4.8103atm} VELOCITY KE = 1 / 2 mv 2 ® for a single molecule (KE = kinetic energy); for one mole of molecules this becomes: (this equation not emphasized) (J) is average KE; M is molar mass of the gas (kg/mol); is the mean of the squared velocity; Another equation for : ; KE µ T ; the KE calculated using 3 / 2 RT is for 1 mol of gas Set the two equations equal to one another: ; solve for root mean square velocity, : where M = molar mass in kg/mol not g/mol (M is not molarity!) ; R = 8.314J/(molK) (same as R = 0.0821 Latm/(molK) but has different units!) ; T in K; v rms in m/s Graphical Gas Distribution Velocities : The velocity of the gas calculated in different ways. The most probable gas velocity , , occurs at the peak, the average speed , , appears at a different location, and the root mean square velocity (V rms ) appears at a third location on the distribution of molecular velocities versus probability. P N 2 = 2.28atm P N 2 O 5 = 2.53atm P O 2 = 0atm 10.0gN 2 1molN 2 28.0gN 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2molN 2 O 5 2molN 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.3572molN 2 O 5 15.0gO 2 1molO 2 32.0gO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2molN 2 O 5 5molO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.1875molN 2 O 5 10.0g N 2 1mol N 2 28.0g N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.3571mol N 2 15.0gO 2 1molO 2 32.0gO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2molN 2 5molO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.1875molN 2 used P N 2 = n N 2 RT V P N 2 = ( 0 . 1696 mol)( 0 . 0821 Latm / molK)( 55 . 0 + 273 . 15 K) ( 2 . 0 L) = 2 . 2846 atm N 2 P N 2 O 5 = ( 0 . 1875 mol)( 0 . 0821 Latm / molK)( 55 . 0 + 273 . 15 K) ( 2 . 0 L) = 2 . 5257 atm N 2 O 5 KE average = KE = 1 2 Mv 2 KE v 2 KE KE = 3 2 RT KE 1 2 Mv 2 = 3 2 RT v 2 v 2 = v root mean square = v rms = 3RT M V most probable = V peak V average = V

4 Kinetic-Molecular Theory 1. A gas is composed of molecules that are small compared to the distances between them; most of a gas volume is empty space. 2. Gases consist of tiny molecules that are in constant random motion moving at varying speeds. 3. The attractive/repulsive forces between gas molecules are minimal. 4. Molecular collisions and collisions with the container walls occur without loss of E (elastic collisions), and the collisions with the container create pressure. Gas Equations: Summary Gas Laws P 1 V 1 = P 2 V 2 ( ) (V µ T) PV = nRT STP: P = 1atm; T = 0˚C; 1mol gas = 22.4L at STP (V µ n) Stoichiometry Partial Pressure P total = P 1 + P 2 + ... c 1 = = P 1 = c 1 P total SC i = 1.00 P 1 V = n 1 RT P total V = n total RT Kinetic Molecular Theory (Velocity) (if covered) (if covered) P ∝ 1 V P V V 1 T 1 = V 2 T 2 V T P 1 V 1 T 1 = P 2 V 2 T 2 molar mass = gRT PV D = (P)(molar mass) RT V 1 n 1 = V 2 n 2 atoms or molecules B atoms or molecules A grams A grams B moles A moles B 1 mol B = 6.022 x 10 B 23 1 mol A = 6.022 x 10 A 23 molar mass A molar mass B molarity A molarity B gas A: P, V, T gas B: P, V, T PV = nRT PV = nRT M = mol /L A A A M = mol /L B B B chemical formula or chemical reaction n 1 n total P 1 P total KE average = KE = 1 2 Mv 2 KE = 3 2 RT v rms = 3RT M rate 1 rate 2 = M 2 M 1 rate 1 rate 2 = T 1 T 2 M 1 M 2 = T 1 T 2

5 Partial Pressure 1. A container has 6.0mol Ne, 4.0mol N 2 , and 5.0mol NH 3 with a total pressure of 1.5atm. What is the partial pressure of NH 3 ? 2. A sample of the atmosphere contained 120.12 g of nitrogen gas (N 2 ) and 38.72 g of oxygen gas (O 2 ). What is the partial pressure of N 2 and O 2 when the atmospheric pressure is 730.0 torr? 3. A 2.0-L container at 25˚C has both N 2 and O 2 in it. The pressure on the system is determined to be 550torr with the partial pressure due to N 2 being 450torr. How many grams of O 2 are in the flask? 4. A mixture of 2.0mol SO 2 (g), 1.0mol H 2 O(g), and 3.0mol O 2 (g) have a total pressure of 100torr. Which gas has the greatest partial pressure? 5. A 1.85g sample of ammonium nitrate (molar mass = 80.1g/mol) in an evacuated 2.12L flask is heated to 250.˚C. What are the partial pressures of H 2 O(g) and N 2 O(g), and the total pressure in the flask at 250.˚C when the ammonium nitrate has completely decomposed according to the following reaction? NH 4 NO 3 (s) ® N 2 O(g) + 2H 2 O(g) 6. The valve between a 5-L tank containing a gas at 9atm and a 10-L tank containing a gas at 6atm is opened. Calculate the final pressure in the tanks after the valve is opened. a. 10 atm b. 9 atm c. 8 atm d. 7 atm e. 6 atm 7. 25.0g N 2 (g) reacted with 45.0g O 2 (g) to completion and produced N 2 O 3 (g) in a 5.0L reaction vessel. The temperature was 70.0˚C. What is the partial pressure of N 2 (g), O 2 (g), and N 2 O 3 (g), and what is the total pressure at the end of the reaction? (A more challenging question; see Example 4 for guidance.) 2N 2 (g) + 3O 2 (g) ® 2N 2 O 3 (g) Velocity and Kinetic Energy 8. If four 1-L flasks contain He, Ne, O 2 , and CO 2 at the same T and P, which has the greatest average KE? (Hint: Minimal calculations required.) a. He b. Ne c. O 2 d. CO 2 e. All have same KE 9. What is the total kinetic energy in Joules for 2.0L of He at 175˚C and 2.0atm? 10. Order the following gases for decreasing rate of velocity: CO 2 , N 2 , O 2 , Ar, H 2 O (all the gases are at the same T). (Hint: Minimal calculations required.) highest velocity lowest velocity a. CO 2 > N 2 > O 2 > Ar > H 2 O b. H 2 O > N 2 > O 2 > Ar > CO 2 c. CO 2 > Ar > O 2 > N 2 > H 2 O d. Ar > N 2 > O 2 > H 2 O > CO 2 e. CO 2 > H 2 O > N 2 > O 2 > Ar 11. What is the root mean square velocity of methane, CH 4 (g), at 25.0˚C. a. 682m/s b. 1.46 x 10 -3 m/s c. 55.6m/s d. 4.64 x 10 5 m/s e. none of the above 12. If the root mean square velocity of NH 3 molecules is found to be 510. m/s, what is the temperature in K?

7 ANSWERS 1. 0.50atm {use P NH 3 = c NH 3 P T and solve for P NH 3 ; ; P NH 3 = (0.33)(1.5atm) = 0.50atm} 2. 569.3 torr N 2 and 160.7 torr O 2 {use P i = c i P T ; P T = 730.0torr; convert g ® mol; ; n T = 1.210 + 4.288 = 5.498mol; ; ; P N 2 = (0.7799)(730.0torr) = 569.3torr; P O 2 = (0.2200)(730.0torr) = 160.7torr} 3. 0.34g O 2 {use P i V = n i RT; ; find P O 2 from P T = P O 2 + P N 2 ; P O 2 = 550-450 = 100torr; ; } 4. O 2 (g) {since there are more moles of O 2 (g), it will have a greater c O 2 and from that a greater partial pressure} 5. P H 2 O = 0.936atm; P N 2 O = 0.468atm; P T = 1.40atm; {use P H 2 O V = n H 2 O RT; ; find mol H 2 O: ; find mol N 2 O: 0.04619mol H 2 O x (1mol N 2 O/2mol H 2 O) = 0.02310mol N 2 O; P total = P H 2 O + P N 2 O ; P total = 0.9358 + 0.4680 = 1.4038atm} 6. d {This problem can be done in various ways: an easy way is to run through P 1 V 1 = P 2 V 2 two times; I will use “1” as initial conditions and “2” as final; first gas (call it X): P 1 V 1 = P 2 V 2 ; (9)(5) = (P 2 )(10+5); P 2 for X = 3.0atm; second gas (call it Y): P 1 V 1 = P 2 V 2 ; (6)(10) = (P 2 )(10+5); P 2 for Y = 4.0atm; P T = P X + P Y = 3.0 + 4.0 = 7.0atm} χ NH 3 = mol NH 3 mol Total = 5.0 6.0 + 4.0 + 5.0 = 0.33 120.12g N 2 1mol N 2 28.02g N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4.288mol N 2 38.72g O 2 1mol O 2 32.00g O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1.210mol O 2 χ N 2 = mol N 2 mol T = 4.288 5.498 = 0.7799 χ O 2 = mol O 2 mol T = 1.210 5.498 = 0.2200 n O 2 = P O 2 V RT n O 2 = PV RT = ( 100 torr)( 1 atm / 760 torr)( 2 . 0 L) ( 0 . 0821 Latm / molK)( 25 + 273 . 15 K) = 0 . 01075 mol O 2 0.01075mol O 2 32.00g O 2 1mol O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.3440g O 2 P H 2 O = n H 2 O V RT n H 2 O = 1.85gNH 4 NO 3 ( ) 1molNH 4 NO 3 80.1gNH 4 NO 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2molH 2 O 1molNH 4 NO 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.04619molH 2 O P H 2 O = nRT V = ( 0 . 04619 mol)( 0 . 0821 Latm / molK)( 250 . + 273 . 15 K) 2 . 12 L = 0 . 9358 atm H 2 O P N 2 O = nRT V = ( 0 . 02310 mol)( 0 . 0821 Latm / molK)( 250 . + 273 . 15 K) 2 . 12 L = 0 . 4680 atm N 2 O

8 7. P N 2 = 0atm; P O 2 = 0.378atm; P N 2 O 3 = 5.03atm; P T = 5.41atm {limiting reagent problem since 2 reactant quantities are given; determine how many mol N 2 O 3 can be made, which reactant is the LR and how many mol of the excess reactant remains. ; ; N 2 is the LR since it yielded the smaller amount; O 2 is the EX; 0.8929mol N 2 O 3 can be produced; find left over O 2 : left over = starting amount – amount used; find amount O 2 used in mol: LR ® EX: used ; starting amount O 2 = ; left over O 2 = 1.4063 – 1.3393 = 0.06701mol O 2 left over; P N 2 (after reaction) = 0.0atm (LR = no N 2 left); ; P T = P N 2 + P O 2 + P N 2 O 3 = 0 + 0.3776 + 5.0311 = 5.4087atm} 8. e {since average KE µ T ® same T means same average KE} 9. 608J {KE = 3 / 2 RT per mol of gas ; start by finding mol: PV = nRT; ; find KE: KE = ( 3 / 2 )(8.314J/molK)(175 + 273.15K) = 5589J/mol; KE for gas present = (5589J/mol)(0.1087mol) = 607.5J} 10. b {heavier molecules move slower; find the molar masses: CO 2 (44g/mol), N 2 (28g/mol), O 2 (32g/mol), Ar (40g/mol), H 2 O (18g/mol) – arrange so lightest is on the left side (highest velocity) and heaviest is on the right side (slowest velocity) of the series; H 2 O > N 2 > O 2 > Ar > CO 2 } 11. a {use ; = 681.6m/s} 12. 177K {use ; ; square both sides: ; solve for T; T = 177.3} 13. 601K {use ; (7.50kJ)(1000J/1kJ) = (1.50mol)(8.314J/molK)(T); solve for T; T = 601.4K} 14. e {use: ; ; ; square both sides: ; 2.746M 1 = 0.044; M 1 = 0.0160kg/mol(1000g/1kg) = 16.0g/mol; O 2 : 32g/mol; N 2 : 28g/mol; SO 2 : 64g/mol; Ne: 20g/mol; CH 4 : 16g/mol; CH 4 is the only molecule with this molar mass} 25.0g N 2 1mol N 2 28.0g N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2mol N 2 O 3 2mol N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.8929mol N 2 O 3 45.0g O 2 1mol O 2 32.0g O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2mol N 2 O 3 3mol O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.9375mol N 2 O 3 25.0g N 2 1mol N 2 28.0g N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3mol O 2 2mol N 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1.3393mol O 2 45.0g O 2 1mol O 2 32.0g O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1.4063mol O 2 P O 2 = nRT V = ( 0 . 06701 mol)( 0 . 0821 Latm / molK)( 70 . 0 + 273 . 15 K) 5 . 0 L = 0 . 3776 atm O 2 P N 2 O 3 = nRT V = ( 0 . 8929 mol)( 0 . 0821 Latm / molK)( 70 . 0 + 273 . 15 K) 5 . 0 L = 5 . 0311 atm N 2 O 3 n = PV RT = ( 2 . 0 atm)( 2 . 0 L) ( 0 . 0821 Latm / molK)( 175 + 273 . 15 K) = 0 . 1087 mol v rms = 3RT M v rms = 3(8.314J/molK)(25.0+273.15K) (16.0g/mol)(1kg/1000g) v rms = 3RT M 510 = 3(8.314J/molK)T (17.0g/mol)(1kg/1000g) 260,100 = (3)(8.314)T 0.0170 KE = 3 2 RT rate 1 rate 2 = M 2 M 1 406 245 = 0.044 M 1 1.657 = 0.044 M 1 1.657 ( ) 2 = 0.044 M 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 2.746 = 0.044 M 1