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

3 Probability versus Molecular Speed at Different T or with Different Masses Comparing the molecular speed distribution of one gas at two temperatures. Note how the curve peak shifts to a faster speed (moves right on the x-axis) as the temperature increases. Comparing molecular speed distributions of 4 different gases. Note how the curves’ peaks shifts to a higher speed (moves right on the x- axis) as the molecular size decreases. The smallest gas has the fastest overall speed distribution. take home messages from above graphs: 1. Smaller gas molecules travel faster than heavier ones at same T; 2. Same gas molecules travel faster at higher T Example 5: If CO 2 (g) is at 58.5˚C, what is the root mean square velocity of the CO 2 molecules? Answer 5: 4.34 x 10 2 m/s ; ; ; Example 6: If Ar(g) atoms are moving at 509m/s what is the average T in K of these atoms? Answer 6: 415K ; ; ; ; 259,081 = 624.33T; T = 414.97K Example 7: Place the following gases all at the same T in order from slowest root mean square velocity to fastest root mean square velocity. Ar, CO 2 , Ne, O 2 Answer 7: CO 2 < Ar < O 2 < Ne since larger molecules move slower than smaller molecules at the same T, place them in order by molar mass: CO 2 (44g/mol) < Ar (40g/mol) < O 2 (32g/mol) < Ne (20.2g/mol)} Diffusion : the mixing of two gases. (other used equations: and ) rate can be velocity or rate of diffusion; T in K; M in kg/mol (M is not molarity!) PARTS PER MILLION: The number of molecules per 1 million molecules Example 8: a. If methane, CH 4 , is 0.0000722% in the atmosphere, what is the amount of methane in ppm? Answer 8: 0.722ppm Step 1: Convert % to decimal: 0.0000722%/100 = 0.000000722 Step 2: Multiply by : (note that 1 x 10 6 = 1 million so the bottom 1 x 10 6 is replaced with “one million”) v rms = 3RT M v rms = 3(8.314)(58.5 + 273.15) (44.01g / mol)(1kg /1000g) v rms = 8272.0 0.04401 v rms = 8272.0 0.04401 = 4.335 x 10 2 m / s v rms = 3RT M 509.0 = 3(8.314)(T) (39.95g / mol)(1kg /1000g) (509) 2 = 24.942(T) 0.03995 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 259,081 = 24.942(T) 0.03995 rate 1 rate 2 = M 2 M 1 rate 1 rate 2 = T 1 T 2 M 1 M 2 = T 1 T 2 1x 10 6 1x 10 6 0.000000722 1x 10 6 1x 10 6 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0.722 parts 1x 10 6 = 0.722 parts one million = 0.722parts per million = 0.722ppm