May 2, 2011
1. Upon the addition of potassium nitrate to liquid water, the beaker feels cold. Explain, using appropriate terms: system, surroundings, heat, temperature, endothermic or exothermic.
The endothermic process whereby the system of KNO3 crystals dissolving in the water absorb energy from the surroundings and the measured temperature of the surroundings (beaker, air) drops.
2. The phase change from a gas to a liquid is called condensation. Is the enthalpy change, g → l, exothermic or endothermic? Explain.
The enthalpy change is exothermic. Recall that energy is always released when new bonds form. Here, the phase change involves the formation of intermolecular attractions releasing energy as the particles adopt lower-energy conformations. The strength of the intermolecular attractions between molecules, and the amount of energy released when attractions form (or the amount of energy required to overcome these attractive forces) depends on the molecular properties of the substance. Generally, the more polar a molecule is, the stronger the attractive forces between molecules are.
3. Identify the type of molar enthalpy for each reaction below: a) MgCl2 (s) → Mg2+ (aq) + 2 Cl- (aq)ΔHsol
b) CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l) ΔHneut c) H2O (l) → H2O (s) ΔHfr
4. There are three modes of molecular motion associated with energy. Identify the mode(s) of molecular motion available to helium gas (He) and compare to those of nitrogen gas (N2). Which gas has the higher molar heat capacity? Explain.
Three kinds of motion are associated with energy: translational, rotational, and vibrational. Helium, He, is a monatomic gas which only has translational motion as there is no bond about which to rotate and no bond on which to vibrate.
Nitogen, N2 , is a diatomic gas which can divert heat energy into rotational and vibrational motion. Therefore, for the same temperature increase, Nitrogen gas requires more heat energy than Helium. The molar heat capacity of N2 ( 29.1 J/mol·°C) is higher than that of He (20.8 J/mol·°C).
5. Temperature does not reflect the total thermal energy of a system. Explain using a comparative example.
A cup of hot chocolate at 85°C has a higher temperature than a bathtub at 75°C. However, the bathtub of water has many more water molecules that can store thermal energy compared to the small cup of hot chocolate.
6. a) Consider the process: H2O (s) → H2O (l) → H2O (g). When the temperature holds constant at 100 ˚C, what is the heat energy used for?
Heat energy supplied is used to overcome the strong dipole-dipole intermolecular attractive forces (Hydrogen Bonding) holding liquid molecules together.
7. Calculate the enthalpy change for the reaction:
HCl (g) + NaNO2 (s) → HNO2 (g) + NaCl (s) ∆H˚ = ?
Use the following thermochemical equations:
2 NaCl + H2O → 2 HCl + Na2O ∆H˚ = +507 kJ
NO + NO2 + Na2O → 2 NaNO2 ∆H˚ = - 427 kJ
NO + NO2 → N2O + O2∆H˚ = - 43 kJ
2 HNO2 → N2O + O2 + H2O ∆H˚ = + 34 kJ
Eq’n #1 : Rev, x ½ ΔH = -253.5 kJ
Eq’n #2 : Rev, x ½ΔH = + 213.5 kJ
Eq’n #3 : x ½ ΔH = - 21.5 kJ
Eq’n #4 : Rev, x ½ΔH = - 17 kJ
ΔH = - 78.5 kJ
8. Using standard heat of formation values (∆Hf˚)(
, calculate the molar enthalpy of combustion of propane.
(Assume the production of liquid water.)
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)
ΔH = [3( -393.5) + 4 ( -285.8)] – [- 104.7 + 5(0)] = -2219 kJ
n = m/M 1mol = 1134 mol
= 50 000g / 44.11 g/mol-2219 kJ x kJ
= 1134 molx = - 2.515...