November+8th+Answers

1st Link

1- a) Exothermic b) Endothermic c) Exothermic

2- This reaction is endothermic, as it is absorbing the heat it needs from the surrounding water. This cools the water down.

2nd Link

1- a) C + O2 -> CO2 ∆ H = -393.5 kJ

b) Cu + Cl2 -> CuCl2 ∆ H = -220.1 kJ

c) Cu + ½ Cl2 -> CuCl ∆ H = -137.2 kJ

d) N2 + 2H2 -> N2H4 ∆ H = 50.6 kJ

e) N2 + 2H2 + ½ Cl2 -> N2H4Cl ∆ H = -314.4 kJ

2- Molar Mass of SO2 is molar mass of S + 2O. Molar Mass is 64.1 g/mol.

25.0 g * 1 mol/64.1 g = 0.39 mol

Hf = -296.8 kJ/mol E = -296.8 * 0.39 mol E = -115.8 kJ

3- 11.5 g * 1 mol/46.1 g = 0.25 mol Hr = -950 kJ/mol E = 950 kJ/mol* 0.25 mol E = 237.5 kJ

4- 33.0 g * 1 mol/44.1 g = 0.75 mol Hr = -2220 kJ/mol E = -2220kJ/mol* 0.75 mol E = -1661 kJ

Assignment:

1.) STP – Standard Temperature and Pressure - 0oC (273.15 K, 101.325 kPa) It’s commonly used to define standard conditions for temperature and pressure which is important for the measurements of chemical or physical processes.

SATP – Standard Ambient Temperature and Pressure - 25oC (298.15 K,101 kPa) It is used as a reference in chemistry. When scientists are experimenting with gases, they try to work in the same conditions (temperature and pressure) or their results would be different. They, therefore, develop //standard conditions// when working with gases.

2.) a. 2 Hg(l) + I2 (s) ® Hg2I2(s) DH° = -28.9 kcal Exothermic: 2 Hg(l) + I2 (s) ® Hg2I2(s) + 28.9 kcal

b. N2 (g) + 3 F2 (g) ® 2 NF3 (g) + 27.2 kcal Exothermic: N2 (g) + 3 F2 (g) ® 2 NF3 (g) DH° = -27.2 kcal

c. NH4NO3(s) + 6.1 kcal ® NH4+ (aq) + NO3- (aq) Endothermic: NH4NO3(s) ® NH4+ (aq) + NO3- (aq) DH° = +6.1 kcal

d. Na(s) ® Na (g) DH° = 25.98 kcal Endothermic: Na(s) + 25.98 kcal ® Na (g)

3.) Graph 1 = Exothermic as the potential energy in its system is now released into the environment Graph 2 = Endothermic as it has absorbed the energy into its system. b.) Graph 1 would have a negative value for DH as it is exothermic.

4.) a. SO2(g) S + 02 à SO2 + 296.8kJ

b. C3H8 (g) 3C + 4H2 à C3H8 + 103.8 kJ

c. N2O (g) N2 + 1/2O2 + 82.1kJ à N2O (g)

d. Na2CO3 (s) 2Na + C + 3/2O3 à Na2CO3 (s) + 1130.7kJ

5.) A D and F would be spontaneous as they are all exothermic.

Hess' Law:

1.) #1 - 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l) ΔH° = -1170 kJ
 * 1) 2 - 4 NH3 (g) + 3 O2 (g) → 2 N2 (g) + 6 H2O (l) ΔH° = -1530 kJ


 * 1) 3. - 1/2N2 + 1/202 --> NO

Reverse Equation #2

4NH3 + 5O2 --> 4NO + 6H2O ∆ H = -1170kJ 2N2 + 6H2O --> 4NH3 + 3O2 ∆ H = +1530kJ

Add the two equations and get the result: 2N2 + 2O2 --> 4NO divide the coefficients by 4

1/2N2 +1/2 O2 --> NO DH = +90kJ