Thermochemistry The study of heat changes that occur

Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state

Energy – the ability to do work or produce heat n Exists in 2 forms: n Kinetic energy – energy of motion n Potential energy – energy at rest or energy of position n Heat (q) = the energy that transfers from one object to another because of a temperature difference between them n Heat always flows from a warmer object to a cooler object n

Energy Kinetic energy – in a chemical reaction temperature is the determining factor n The higher the temperature…the faster the particles move…the higher the average kinetic energy n Temperature is a measure of the average kinetic energy n Kelvin scale: 0 K = -273 °C n

Law of Conservation of Energy n Law of Conservation of Energy – Energy is neither created nor destroyed

Heat (q) Heat or energy can be in joules, calories, kilocalories, or kilojoules n The SI unit is the joule n 1 Cal = 1000 cal = 1 kcal n 1 cal = 4. 186 J n 1 kcal = 4186 J n 1 J = 0. 239 cal n

Heat Capacity The amount of heat it takes to change an object’s temperature by 1ºC n Depends on an object’s mass n Ex. A cup of water has a greater heat capacity than a drop of water. n

Specific Heat (C) – the amount of heat required to raise 1 gram of a substance by 1 C n Specific heat is an intensive property, and therefore does not depend on size n Every substance has its own specific heat n Ex. Water = 4. 18 J/(g x ºC) Glass = 0. 50 J/(g x ºC) n

Specific Heat Units for C = J / (g x ºC) (joules per gram degree Celsius) n Equation for Specific Heat: C = q / (m Δ T) n C = specific heat; q = heat; m = mass and ΔT = change in temperature n This equation can be rearranged to solve for heat (q) q= CmΔT n

Specific Heat n A 10. 0 g sample of iron changes temperature from 25. 0 C to 50. 4 C while releasing 114 joules of heat. Calculate the specific heat of iron.

Example C= q/ (m∆T) n C=114 J/ (10. 0 g x 25. 4°C) n C = 0. 45 J/g C n

Another example n If the temperature of 34. 4 g of ethanol increases from 25. 0 C to 78. 8 C how much heat will be absorbed if the specific heat of the ethanol is 2. 44 J/g C

Another example First, rearrange the specific heat formula to solve for heat n q = Cm T n q = (2. 44 J/g°C)(34. 4 g)(78. 8°C - 25°C) n q = 4515. 76 J n

Yet another example n 4. 50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25. 0 C & the specific heat of the gold is 0. 129 J/g C

Yet another example n n n n C= q/ (m∆T); rearrange to find ∆T = q / (C x m) ∆T = 276 J / (. 129 J/g°C x 4. 50 g) T = 475. 45 C T = Tf-Ti 475. 45 = Tf-25 Tf = 500. 45 C