Module 3 Problem Set Quiz

Module 3 Problem Set

• Question 1

  Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7^oC is added to a 36.8 gram sample of water at 26.3^oC in a coffee cup calorimeter.

  c (water) = 4.184 J/g ^oC

  Correct Answer

  - (m_{warn H2O} x c_{warn H2O} x ∆t_{warn H2O}) = (m_{cool H2O} x c_{cool H2O} x ∆t_{cool H2O})

  - [40.5 g x 4.184 J/g ^oC x (T_mix - 85.7^oC)] = [(36.8 g x 4.184 J/g ^oC x (T_mix - 26.3^oC)]

  - [169.452 J/^oC x (T_mix - 85.7^oC)] = [(153.9712 J/^oC x (T_mix - 26.3^oC)]

  T_mix = 57.4^oC

• Question 2

  Show the calculation of the energy involved in melting 120 grams of ice at 0^oC if the Heat of Fusion for water is 0.334 kJ/g.

  Correct Answer

  q_l↔s = m x ∆H_fusion = 120 g x 0.334 kJ/g = 40.08 kJ (since heat is added) = + 40.08 kJ

• Question 3

  Thermochemical Equation Problems

  Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:

              2 S     +     3 O₂    →     2 SO₃          ΔH = - 792 kJ

  If 42.8 g of S is reacted with excess O₂, what will be the amount of heat given off?

  Correct Answer

  ΔH_rx is for 2 mole of S

  reaction uses 42.8 g S = 42.8/32.07 = 1.335 mole S

  q = -792 kJ x 1.335 mole S / 2 mole S = -528.7 kJ

• Question 4

  Thermochemical Equation Problems

  Hydrosulfuric acid (H₂S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation:

              2 H₂S    +     3 O₂    →     2 SO₂     +     2 H₂O

  What is the ΔH of the reaction if 26.2 g of H₂S reacts with excess O₂ to yield 431.8 kJ?

  Correct Answer

  ΔH_rx is for 2 mole of H₂S; q = -431.8 kJ

  reaction uses 26.2 g H₂S = 26.2/34.086 mole = 0.7686 mole H₂S

  -431.8 kJ = ΔH_rx x 0.7686 mole H₂S / 2 mole H₂S

  ΔH_rx = -1123.6 kJ

• Question 5

  Measuring Heat of Reaction Problems

  A sample of benzene (C₆H₆), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:

   2 C₆H₆ (l)     +     15 O₂ (g)     →     12 CO₂ (g)     +     6 H₂O (l)

  The heat given off was absorbed by 500 g of water  and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 ^oC. The heat capacity of water = 4.18 J/g/^oC and the heat capacity of the calorimeter = 10.5 kJ/^oC. (1) what is the ΔH of the reaction?  Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.

  Correct Answer

  1)

  q _water = s (specific heat of water) x mass x Δt = 4.18 J / g / ^oK x 500 g x  (53.13 ^oC - 25.00^oC)  = - 58,791 J

  q _calorimeter = heat capacity x Δt = (10.5 kJ/^oC) x (53.13 ^oC - 25.00^oC) = - 295.365 kJ x 1000 J/1 kJ= - 295,365 J

  q _reaction = -  58,791 J + (-295,365 J) = - 354,156 J = -  354,156 J x 1 kJ / 1000 J = - 354.16 kJ

   (new) moles C₆H₆ = 7.05 g / 78 = 0.0904 mole C₆H₆

   ΔH = q / (new moles / original moles)

  ΔH = - 354.16 / (0.09048 / 2) = - 7828 kJ / mole

   (2) Isolated system (bomb calorimeter)

  (3) Exothermic (temperature of water rises due to heat given off by combustion reaction)

• Question 6

  Measuring Heat of Reaction Problems

  A sample of 14.5 g of sodium bicarbonate (NaHCO₃) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction

   NaHCO₃ (s)    →     Na⁺ (aq)     +     HCO₃^- (aq)

   If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/^oC) decreases from 25.00^oC to 19.86^oC, (1) what is the ΔH of the reaction?  Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.

  Correct Answer

  q _water = s (specific heat of water) x mass x Δt = 4.18 J / g / ^oK x 100 g x 5.14^oK = 2148.52 J

  q _calorimeter = heat capacity x Δt = 150 x 5.14^oK = 771 J

  q _reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ

  (1)       ΔH _reaction = 2.92 kJ / 14.5 g NaHCO₃ / 84 g NaHCO₃ / mol NaHCO₃ = + 16.9 kJ / mol

  (2)       The calorimeter + contents (no lid) = open system, (3) Endothermic process

• Question 7

  Hess' Law Problems

  The combustion of ammonia by the following reaction yields nitric oxide and water

              4 NH₃ (g)    +     5 O₂ (g)   →     4 NO (g)     +     6 H₂O (g)

   Determine the heat of reaction (ΔH_rxn) for this reaction by using the following thermochemical data:

  N₂ (g)    +     O₂ (g)   →     2 NO (g)          ΔH = 180.6 kJ

  N₂ (g)    +     3 H₂ (g)   →     2 NH₃ (g)      ΔH = - 91.8 kJ

  2 H₂ (g)    +     O₂ (g)   →     2 H₂O (g)      ΔH = - 483.7 kJ

  Correct Answer

  2 (2 NH₃ (g)       →     3 H₂ (g)   +     2 N₂ (g)      ΔH = + 91.8 kJ)

  3 (2 H₂ (g)    +     O₂ (g)   →     2 H₂O (g)             ΔH = - 483.7 kJ)

  2 (N₂ (g)    +     O₂ (g)   →     2 NO (g)                 ΔH = 180.6 kJ)

   4 NH₃ (g)     +    5 O₂ (g)     →     4 NO (g)     +     6 H₂O (g)                ΔH_rxn = - 906.3 kJ

   

  ΔH_rxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6)  =  - 906.3 kJ

• Question 8

  Heat of Formation Problems

  Determine the heat of reaction (ΔH_rxn) for the combustion of ethanol (C₂H₅OH) by using heat of formation data:

              C₂H₅OH (l)    +     3 O₂ (g)     →    2 CO₂     +    3 H₂O (g)

  Correct Answer

  Determine the heat of reaction (ΔH_rxn) for the combustion of ethanol (C₂H₅OH) by using heat of formation data:

  C₂H₅OH (l)    +     3 O₂ (g)     →    2 CO₂ (g)     +    3 H₂O (g)

  C₂H₅OH (l)   →   2 C (graphite)   +   3 H₂ (g)   +   1/2 O₂ (g)   ΔH_f⁰ (C₂H₅OH) = - (- 277.6 kJ/mole)

  2 (C (graphite)     +     O₂     →     CO₂ (g)                                  2 ΔH_f⁰ (CO₂) = 2 (-393.5 kJ/mole)

  3 (H₂ (g)    +    1/2 O₂ (g)    →     H₂O (g)                                    3 ΔH_f⁰ (H₂O) = 3 (- 241.8 J/mole)

   C₂H₅OH (l)   +   3 O₂ (g)   →   2 CO₂ (g)   +   3 H₂O (g)           

   ΔH_rxn =  - (- 277.6) + 2 (- 393.5) + 3 (- 241.8)

  ΔH_rxn = - 1234.8 kJ/mole

   OR more simply and preferably:            

  ΔH_rxn = Σ n ΔH_f⁰ (products) - Σ m ΔH_f⁰ (reactants)

  ΔH_rxn = 2 ΔH_f⁰ (CO₂) + 3 ΔH_f⁰ (H₂O) - ΔH_f⁰ (C₂H₅OH) – 3 ΔH_f⁰ (O₂) 

  ΔH_rxn = 2 (- 393.5) + 3 (- 241.8) - (- 277.6) - 3 (0) = - 1234.8 kJ/mole

• Question 9

  A gas sample has an original volume of 680 ml when collected at 720 mm and 28^oC.  What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 55^oC?

  Correct Answer

  P_i  x  V_i    =    P_f  x  V_f

     T_i                     T_f

  680 ml/1000 = 0.680 liters = V_i

  720 mm/760 = 0.947 atm = P_i                  820 mm/760 = 1.08 atm = P_f

  28^oC + 273 = 301^oK = T_i                            55^oC + 273 = 328^oK = T_f

  Put in the data:       (0.947) x (0.680)    =    (1.08) x V_f:

                                             (301)                        (328)

   Solve for V_f:

  0.002139402    =    0.003292683 x  V_f:                 V_f: = 0.002139402 = 0.650 liter

                                                                                          0.003292683

• Question 10

  A gas sample containing 0.546 mole collected at 700 mm and 25^oC. would occupy what volume?

  Correct Answer

  P  x  V   =   n  x  R  x  T

  0.546 mole  = n                                           R  =  0.0821

  700 mm/760 = 0.921 atm = P                   25^oC + 273 = 298^oK = T

               (0.921) x V   =   (0.546) x (0.0821) x (298)

   (0.921) x V   =   13.358                   V   =   13.358   =   14.5 liters

                                                                     0.921

• Question 11

  GAS VOLUME STOICHIOMETRY PROBLEMS

  The combustion of ethanol (C₂H₅OH) takes place by the following reaction equation.

              C₂H₅OH (l)    +     3 O₂ (g)     →    2 CO₂ (g)    +    3 H₂O (g)

  What is the volume of CO₂ gas produced by the combustion of excess ethanol by 23.3 grams   of O₂ gas at 25^oC and 1.25 atm?

  Correct Answer

   (MW = 46)                   (MW = 32)                  (MW = 44)                 (MW = 18) 

              C₂H₅OH (l)     +          3 O₂ (g)          →        2 CO₂ (g)       +          3 H₂O (g)

   

                                                  23.3 grams               9.5 liters

                                                              ↓                         ↑ by V = nRT / P = (0.4854)(0.0821)(298) /1.25

                                                  0.7281 mol    →        2/3 x 0.7281 mol

• Question 12

  PARTIAL PRESSURE - MOLE FRACTION PROBLEMS

  A mixture of gases consists of 4.00 moles of He, 2.00 moles of H₂, 3.00 moles of CO₂ and 5.00 moles of Ar.  The total pressure of the mixture is 2900 mm.  Determine the mole fraction of each gas in the mixture. Determine the mole percent of each gas in the mixture.  Determine the partial pressure of each gas in the mixture.

  Correct Answer

  X_He = 4.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2857          Mole%_He = 100(X_He) = (100) 0.286 = 28.57%

  X_H2 = 2.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.1429          Mole%_H2 = 100(X_H2) = (100) 0.143 = 14.29%

  X_CO2 = 3.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2143        Mole%_CO2 = 100(X_CO2) = (100) 0.214 = 21.43%

  X_Ar = 5.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.3571          Mole%_Ar = 100(X_Ar) = (100) 0.357 = 35.71%

   P_He =   X_He (2900 mm) = 0.2857 (2900 mm) = 828.53 mm

  P_H2 =   X_H2 (2900 mm) = 0.1429 (2900 mm) = 414.41 mm

  P_CO2 = X_CO2 (2900 mm) = 0.2143 (2900 mm) = 621.47 mm

  P_Ar = X_Ar (2900 mm) = 0.3571 (2900 mm) = 1035.59 mm

• Question 13

  COLLECTION OF GASES OVER WATER PROBLEMS

  A sample of hydrogen (H₂) gas is collected over water at 35^oC and 725 mm.  The volume of the gas collected is 72.0 ml.  How many moles of H₂ gas has been collected?  How many grams of H₂ gas has been collected?

  Correct Answer

  P_H2 = 725 - P _H2O (from table) = 725 - 42.2 = 682.8 x 1 atm / 760 mm = 0.898 atm H₂

   from Ideal Gas Law:  n _H2 = PV / RT = (0.898 atm) (72.0 ml x 1 liter / 1000 ml) / (0.0821) (308^oK)

   n _H2 = 0.002557 moles

   grams _H2 = moles x MW = 0.002557 moles x 2.016 grams / 1 mole = 0.00515 grams

• Question 14

  EFFUSION AND DIFFUSION PROBLEMS

  The rate of effusion of nitrogen gas (N₂) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the unknown gas?

  Correct Answer

  (r_N2 /   r_unknown)²  =  MW_unknown / MW_N2

  (1.253/1)² = MW_unknown / 28.02                              

  MW_unknown = (1.253)² x 28.02 = 43.99