Chapter 6 Thermochemistry

Final Exam. ( May7, 2014 Wednesday) Instructional Complex :15 AM. –12:15 PM

Study of energy and its transformation ― thermodynamics Thermodynamics of chemical reactions ― thermochemistry Energy ― capacity to do work (W) and transfer heat (Q) unit: J W and Q ― two ways to transfer energy

Work = Force x Distance W = F d SI unit: W ― J, F ― N, d ― m J = N m

Two forms of energy Kinetic energy Potential energy Kinetic energy: energy from motion potential energy: energy from the interaction between objects. depends on the objects and the relative distance.

BrNO + BrNO  2NO +Br 2 exothermic reaction

CH 4 + 2O 2  2H 2 O + CO 2 Exothermic Process

systemsurroundings universe Energy of the universe is constant. Or say energy of the universe is conserved. First law of thermodynamics

∆E = Q + W E: energy of the system, including E k and E p. Or called internal energy of the system. ∆E: change of internal energy of the system. ∆E = E f − E i System absorbs heat from surroundings, Q > 0. System releases heat to surroundings, Q < 0. System does work on surroundings, W < 0. Surroundings do work on system, W > 0.

Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

State Function A property of a system whose value is determined only by specifying the system’s state, not on how the system arrived at that state. The state of a system is specified by parameters such as temperature, pressure, concentration and physical states (solid, liquid, or gas).

EiEi EfEf ∆E = E f − E i

E: State FunctionsSystem: a battery

State functions: E, P, V, T, do not depend on path. W, Q: not state functions, depend on process.

W = −P∆V PV work:

To inflate a balloon you must do pressure–volume work on the surroundings. If you inflate a balloon from a volume of L to 1.85 L against an external pressure of 1.00 atm, how much work is done (in joules)? Example 6.4, page 245

Enthalpy: H = E + PV Only PV work, constant P ∆H = Q p

∆H > 0 ↔ endothermic ∆H < 0 ↔ exothermic Recall ….

Enthalpy

Identify each process as endothermic or exothermic and indicate the sign of  H. (a) sweat evaporating from skin (b) water freezing in a freezer (c) wood burning in a fire Example 6.6, page 250

Q = c m ∆T c: specific heat (capacity) m: mass ∆T: change of temperature g °C J·°C −1 ·g −1 same sign as Q

Q = c m ∆T C = c m C: heat capacity, J ·°C −1 Q = C ∆T

Suppose you find a copper penny (minted before 1982, when pennies were almost entirely copper) in the snow. How much heat is absorbed by the penny as it warms from the temperature of the snow, which is –8.0  C, to the temperature of your body, 37.0  C? Assume the penny is pure copper and has a mass of 3.10 g. Example 6.2, page 242

A 32.5-g cube of aluminum initially at 45.8  C is submerged into g of water at 15.4  C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.) Example 6.3, page 243

A constant-pressure coffee-cup Calorimeter ΔH rxn = − Q soln = − c m ΔT Read example 6.8, page 253

∆E = Q + W PV work: W = − P∆V PV work only, constant P: ∆H = Q p PV work only, constant V: ∆E = Q v

Read example 6.5, page 248 A constant-volume Calorimeter ∆E rxn = −Q cal = −C cal ΔT

2H 2 (g) + O 2 (g)  2H 2 O(g)∆H = −483.6 kJ ∆H = H products − H reactants 1.It is important to specify the state of each species in a thermochemical reaction. ∆H: enthalpy (change) of reaction, heat of reaction 2.∆H of the reverse reaction is the negative of the original reaction. 3.∆H depends on how the reaction is written. Characteristics of enthalpy change of a reaction

An LP gas tank in a home barbeque contains 13.2 kg of propane, C 3 H 8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. Example 6.7, page 252

HAHA HBHB ∆H AB = H B − H A HCHC ∆H AB = ∆H AC + ∆H CB ∆H AC = H C − H A ∆H CB = H B − H C

 H 1 =  H 2 +  H 3 Hess’s Law: ∆H for the overall reaction is equal to the sum of the enthalpy changes of each individual step.

The enthalpy of reaction for the combustion of C to CO 2 is −393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO 2 is −283.0 kJ/mol CO: C(s) + O 2 (g)  CO 2 (g)∆H 1 = −393.5 kJ CO(g) + ½ O 2 (g)  CO 2 (g) ∆H 2 = −283.0 kJ Using these data, calculate the enthalpy for the combustion of C to CO: C(s) + ½ O 2 (g)  CO(g)∆H = ?

Calculate the ∆H for the reaction 2C(s) + H 2 (g)  C 2 H 2 (g) Given the following chemical equations and their respective enthalpy changes: C 2 H 2 (g) + 5/2 O 2 (g)  2CO 2 (g) + H 2 O(l)∆H 1 = − kJ C(s) + O 2 (g)  CO 2 (g) ∆H 2 = −393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O(l) ∆H 3 = −285.8 kJ Try example 6.9 and for practice 6.9, page

Sea level Atlanta, GA Reno, NV 1000 ft 4500 ft 3500 ft Standard 0 ft

H = H (P, T, phase), phase = s, l, g Standard state P = 1 atm T = temperature of interest, often 25 °C state = most stable form The standard enthalpy of formation of a pure substance, ∆H° f, is the change in enthalpy for the reaction that forms one mole of the substance from its elements, with all substances in their standard states.

“Sea level” : free elements at standard state

Calculate the standard enthalpy change for the following reaction 4NH 3 (g) + 5O 2 (g)  4NO (g) + 6H 2 O(g) Example 6.11, page 261

ΔH rxn ° = Σn p ΔH f °(products) − Σn r ΔH f ° (reactants)

Calculate the standard enthalpy change for the following reaction 2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s) Al(s) + ½ Fe 2 O 3 (s)  ½ Al 2 O 3 (s) + Fe(s) For practice 6.11, page 261