1 THERMODYNAMICS Energy changes in reactions Text chapter 3, 4, 5, 6 & 7
TERMINOLOGY: Thermodynamics: study of heat changes that occur during chemical reactions. Energy (J): Cannot be seen, touched, smelled, or weighed. Cannot be created or destroyed (1 st law of thermodynamics) It is recognized by its effects; heat, light, electricity etc Useful forms of energy in chemistry: Radiant (sun) Thermal (associated with the temperature Q=mc T) Chemical (stored in molecular bonds) Kinetic (movement E K = ½ mv 2 ) & Potential (E P =mgh) 2
3 HEAT is a transfer of thermal energy. Eg. A change in temperature between a system & its surroundings. System: The part we are observing. ENERGY Surrounding. Everything else! Heat will flow from a warmer body to a cooler body, until both bodies have the same temperature. (2 nd law of thermodynamics)
4 TYPES OF SYSTEMS: Open system: Energy & matter can easily enter or leave. Eg. Beaker Closed system: Energy can enter or leave, but not matter. Eg. Closed bottle Isolated system: Neither energy nor matter can enter or leave. Eg. Insulated container or calorimeter.
5 BOMB CALORIMETRY Is a device used to measure heat transfer. An inner chamber (aka bomb ) where the rxn happens. A thermometer. An outer chamber that holds water (or another substance of known specific heat capacity) Insulation to prevent loss of heat to the surroundings. When the water changes temperature, thermal energy has been absorbed or released. Q system vs Q surroundings
Exothermic reactions: Release heat into the surroundings Q system = neg Eg campfire Endothermic reaction: Absorbs heat from the surroundings. Q system = pos Boiling water (liquid to gas) 6
7 ENDOTHERMIC EXOTHERMIC E + NH 4 NO 3(s) NH 4 (aq) + NO 3 CaCl 2 (s) Ca + + 2 Cl - (aq) + E Sodium Acetate. Supersaturated solution.
8 Q=mc T When an object absorbs heat, its temperature increases. Q = thermal energy (joules) m = mass (g) c = specific heat capacity (J/gºC) (how well it absorbs heat) T = change in temperature (in ºC) = T final T initial Q gained by the surrounding = Q lost by the system (rxn)
SPECIFIC HEAT CAPACITY= 9 Amount of energy required to raise the temperature of 1g of a substance by 1 C. Substance Liquid Water Water vapour Ice (solid water) Ethylene glycol Aluminum Copper Glass Air (dry) Specific Heat Capacity 4.19 J/(g C) 1.41 J/(g C) 2.05 J/(g C) 2.20 J/(g C) 0.90 J/(g C) 0.39 J/(g C) 0.84J/(g C) 1.02 J/(g C)
EX.1 10 If 125g of copper forms a layer on the bottom of a frying pan, how much heat is need to raise the temp of the copper from 25 o C to 300 o C. The specific heat capacity for Cu is 0.39J/g o C. m = 125g Ti = 25 o C Tf = 300 o C c = 0.39J/g o C Q = m c ΔT Q Cu = (125g)(0.39J/g o C)( 300 o C - 25 o C) Q Cu =13406.25J or 1.3 x 10 4 J Heat is absorbed by the frying pan.
11 EX.2 HCl was added to of NaOH in a calorimeter, producing 50mL of water. At the start, the solutions were at 25 o C. During the reaction, the highest temp observed was 32 o C. What is the heat of the reaction? m = 50g Ti = 25 o C Tf = 32 o C c = 4.19 J/g o C Recall H 2 O density of 1g/mL So, 50mL x 1g/mL = 50g
12 Q = m c ΔT Q H2 O = (50g)(4.19J/g o C)( 32 o C - 25 o C) Q H2 O = 1466.5 J or 1.5 x 10 3 J Heat is absorbed by the water.
13 EX.3 Calculate the heat involved when a 5.5g iron nail is cooled from 37 o C to 25 o C. The heat capacity of iron is 0.45J/g o C. Q = m c ΔT Q Fe = (5.5g)(0.45J/g o C)( 25 o C - 37 o C) m = 5.5g Ti = 37 o C Tf = 25 o C c = 0.45 J/g o C Q Fe = -29.7 J Heat is given off by the iron nail
LAWS OF THERMODYNAMICS 1 st law: Energy cannot be created or destroyed. 2 nd law: Heat will flow from a warmer body to a cooler body. Entropy always increases. 3 rd law: Entropy of a system approaches a constant value as the temperature approaches absolute zero. Entropy is the measure of a system s thermal energy per unit of temperature that is unavailable for doing useful work. Measure of disorder. 14
HEAT TRANSFER CALCULATIONS 15 The heat lost by a system is equal numerically but opposite in sign to the heat gained by the surroundings. Q 1 = Q 2 Consider adding milk to coffee. 1=milk 2=coffee Q 1 = Q 2 m 1 c 1 T 1 = m 2 c 2 T 2 m 1 c 1 (T 1f T 1i ) = m 2 c 2 (T 2f T 2i ) The coffee cools down the milk heats up. But what is the final temperature?
EX.1 What volume of milk at 2.0 o C, must be added to 160.0mL of coffee in order to obtain the perfect cup of cappuccino at 55.5 o C? The initial coffee temperature is 80.5 o C. Assume milk & coffee have the same density as water 1.00g/mL m 1 c/ 1 (T 1f T 1i ) = m 2 c/ 2 (T 2f T 2i ) 160 ml(55.5 80.5 ) = m 2 (55.5 2.0 ) 16 160 ml( 25.0 ) = m 2 (53.5 ) 4000 / ml / 53.5 / = m / 2 m 2 = 74.7664 ml 75 ml
EX2 17 500ml of 10 o C water is mixed with 250ml of 60 o C water. What is the final temperature of the water mixture?
18 500ml of 10 o C water + 250ml of 60 o C water. What is the final temperature of the water mixture? Q = -Q mc T = -mc T (500)(4.19)(x-10) = -(250)(4.19)(x-60) 500x - 5000 = -250x + 15000 750x = 20000 T x =26.67 f =27 o C
19 P137 2, 3 &6 P141 4, 6 & 8 Due next class.