CHM 1045
Dr. Michael Blaber

Name_________________________

SS #__________________________

(Due in Recitation Tues Oct 5)

1. A single hydrogen atom is moving at 99% the speed of light and another is moving at 90% the speed of light. How much kinetic energy does each H atom have (in Joules) and how much more kinetic energy does the faster H atom have compared to the slower one? (1 point)

Kinetic energy = ½mv²

½*(1.008amu*1.66 x 10^-24g/amu)*(1 kg/1000g)*(0.9*3x10⁸ m/s)² = 6.10x10^-11kg m² s^-2

6.10x10^-11kg m² s^-2 * (1J/1 kg m² s^-2) = 6.10x10^-11 J for H atom at 90% speed of light

½*(1.008amu*1.66 x 10^-24g/amu)*(1 kg/1000g)*(0.99*3x10⁸ m/s)² = 7.38x10^-11kg m² s^-2

7.38x10^-11kg m² s^-2 * (1J/1 kg m² s^-2) = 7.38x10^-11 J for H atom at 90% speed of light

(7.38x10^-11 J/6.10x10^-11 J) = 1.21

The faster H atom has 1.21 x the kinetic energy of the slow H atom

2. A certain sample of aluminum requires 42 Joules to raise its temperature from 0°C to 37°C. How many moles of aluminum are in the sample? (The specific heat of aluminum is 0.90 J g^-1 K^-1). (1 point)

 



The temperature in Kelvin is (T_final – T_initial) = (273+37) – (273+0) = 37 K

Thus:

grams aluminum = 42 J / (0.90 J g^-1 K^-1 * 37 K) = 1.26 g

1.26 g (1 mole/27.0g) = 0.0467 moles

3. 52 mls of an aqueous solution of hydrochloric acid (HCl) and 18 mls of sodium hydroxide (NaOH) are allowed to react in a constant pressure calorimeter. After the reaction, it is noted that the solution has increased its temperature from 278K to 358K. Calculate the enthalpy change for the reaction. (assume density and specific heat of the solution is the same as for water) (1 point)

DT solution = (T_final – T_initial) = 358 K – 278 K = 80 K

specific heat of liquid H₂O = 4.18 J g^-1 K^-1

mass = (52 + 18) mls * 1 g/ml = 70 g

DH_solution = 4.18 J g^-1 K^-1 * 70 g * 80 K = 23,408 J

DH_rxn = -DH_solution = -23,408 J or –23.4 kJ

4. Given the following data:

C₂H₅OH(l) + 3O₂(g) -> 2CO₂(g) + 3H₂O(l) DH = -1367 kJ

C(s) + O₂(g) -> CO₂(g) DH = -394 kJ

2H₂(g) + O₂(g) -> 2H₂O(l) DH = -572 kJ

Calculate DH for the reaction:

2C(s) + 3H₂(g) + ½O₂(g) -> C₂H₅OH(l) (2 points)

2C(s) + 2O₂(g) -> 2CO₂(g) DH = -788 kJ

3H₂(g) + 1½O₂(g) -> 3H₂O(l) DH = -858 kJ

3H₂O(l) + 2CO₂(g) -> C₂H₅OH(l) + 3O₂(g) DH = +1367

2C(s) + 3H₂(g) + 3½O₂(g) -> C₂H₅OH(l) + 3O₂(g) DH = -279 kJ

2C(s) + 3H₂(g) + ½O₂(g)-> C₂H₅OH(l) DH = -279 kJ

5. Draw an enthalpy diagram for the reaction in problem #4 where initially appropriate amounts of carbon, diatomic oxygen and diatomic hydrogen are converted to two molecules of carbon dioxide and three molecules of water. Then, the carbon dioxide and water are converted to liquid ethanol (C₂H₅OH) and three molecules of diatomic oxygen. Show the reaction enthalpy for each reaction, be sure to show the reaction enthalpy for the overall reaction of carbon, oxygen and hydrogen to produce ethanol (3 points)



6. Calculate the enthalpy change for the combustion of 0.8 moles of sucrose. Use the Standard Enthalpies of Formation from the chart on page 166 of your book. (2 points)