Thermochemistry Thermochemistry Thermochemistry is the study of heat

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Thermochemistry

Thermochemistry

Thermochemistry § Thermochemistry is the study of heat changes that accompany chemical reactions and

Thermochemistry § Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. § In thermochemistry, the system is the specific part of the universe that contains the reaction or process you wish to study.

Thermochemistry § Everything in the universe other than the system is considered the surroundings.

Thermochemistry § Everything in the universe other than the system is considered the surroundings. § Therefore, the universe is defined as the system plus the surroundings. universe = system + surroundings

1 st Law of Thermodynamics § 1 st Law of Thermodynamics – energy is

1 st Law of Thermodynamics § 1 st Law of Thermodynamics – energy is conserved § The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

Energy § Energy – the ability to do work or produce heat § Exists

Energy § Energy – the ability to do work or produce heat § Exists in 2 forms: § Kinetic energy – energy of motion § Potential energy – energy at rest or energy of position

Energy § A roller coaster at the top of a hill has a great

Energy § A roller coaster at the top of a hill has a great amount of potential energy. § As the rollercoaster begins to speed down the hill, the potential energy is turned into kinetic energy

Energy § The SI unit for energy is the joule (J) § 1 J

Energy § The SI unit for energy is the joule (J) § 1 J = 1 Kgm 2 / s 2 § Another unit of energy that you may be more familiar with is the calorie § calorie – amount of energy required to raise 1 g of water 1°C § 1 cal = 4. 18 J § 1000 cal = 1 Cal or kcal

Energy Conversions § Convert 15, 500 joules into Calories § 15500 J x 1

Energy Conversions § Convert 15, 500 joules into Calories § 15500 J x 1 cal x 1 Cal = § 4. 18 J 1000 cal § 3. 71 Cal

Calorimetry § A calorimeter is an insulated device used for measuring the amount of

Calorimetry § A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

Heat Calculations § q = mc T § q = heat (j) § m

Heat Calculations § q = mc T § q = heat (j) § m = mass (g) § c = specific heat (j/g C) § T = change in temperature = Tf-Ti ( C)

Heat Calculations § Specific Heat (c) – the amount of heat required to raise

Heat Calculations § Specific Heat (c) – the amount of heat required to raise 1 gram of a substance by 1 C § Specific heat is an intensive property § Every substance has its own specific heat

Heat Calculations § Specific Heat (c) – the amount of heat required to raise

Heat Calculations § Specific Heat (c) – the amount of heat required to raise the temperature of 1 g of water 1 °C § The units for specific heat are J/g°C § The specific heat of water (in a liquid form is 4. 18 J/g°C) § All substances have a particular specific heat

Specific Heat § A 10. 0 g sample of iron changes temperature from 25.

Specific Heat § 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 § q = mc T § 114 = (10. 0)c(50. 4 -25) §

Example § q = mc T § 114 = (10. 0)c(50. 4 -25) § c = 0. 449 j/g C

Another example § If the temperature of 34. 4 g of ethanol increases from

Another example § 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 § q = mc T § q = (34. 4)(2. 44)(78. 8

Another example § q = mc T § q = (34. 4)(2. 44)(78. 8 -25) § q = 4. 53 x 10 3 j

Yet another example § 4. 50 g of a gold nugget absorbs 276 J

Yet another example § 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 § q = mc T § 276 = (4. 50)(0. 129)

Yet another example § q = mc T § 276 = (4. 50)(0. 129) T § T = 475 C § T = Tf-Ti § 475 = Tf-25 § Tf = 500 C

Endothermic & Exothermic Reactions § Endothermic reactions – chemical reaction that requires energy to

Endothermic & Exothermic Reactions § Endothermic reactions – chemical reaction that requires energy to break existing bonds § Heat goes into the reaction from the surroundings § Exothermic reactions – chemical reaction in which energy is released § Heat goes out of the reaction into the surroundings

Endothermic & Exothermic Reactions § Endothermic reactions – since heat goes from the surroundings

Endothermic & Exothermic Reactions § Endothermic reactions – since heat goes from the surroundings into your system, it will feel colder § Temperature of endothermic reactions goes down § The sign for the heat change will be + § Exothermic reactions – since heat goes from the system to the surroundings, it will feel hotter § Temperature of exothermic reactions goes up § The sign for the heat change will be -

Endothermic & Exothermic Reactions

Endothermic & Exothermic Reactions

Endothermic & Exothermic Reactions

Endothermic & Exothermic Reactions

Endothermic & Exothermic Reactions § § § § Are the following reactions endothermic or

Endothermic & Exothermic Reactions § § § § Are the following reactions endothermic or exothermic? CO + 3 H 2 CH 4 + H 2 O H= -206 k. J Exothermic ( H is negative) I add magnesium metal to some hydrochloric acid. The temperature goes from 23 C to 27 C Exothermic – temperature goes up I mix together some vinegar & baking soda. The temperature goes from 28 C to 23 C Endothermic – temperature goes down

Calorimetry § When using calorimetry, you are usually trying to determine the identity of

Calorimetry § When using calorimetry, you are usually trying to determine the identity of an unknown metal by finding its specific heat § The heat lost from the metal will be gained by the water § q metal = - q water

Calorimetry § q metal = - q water § (m metal)(c metal)( T metal)

Calorimetry § q metal = - q water § (m metal)(c metal)( T metal) = - (m water)(c water)( T water)

Calorimetry Examples § A 58. 0 g sample of a metal at 100. 0

Calorimetry Examples § A 58. 0 g sample of a metal at 100. 0 °C is placed in a calorimeter containing 60. 0 g of water at 18. 0 °C. The temperature of that water increases to 22. 0 °C. Calculate the specific heat of the metal.

Calorimetry Examples § q metal = - q water § § (m metal)(c metal)(

Calorimetry Examples § q metal = - q water § § (m metal)(c metal)( T metal) = - (m water)(c water)( T water) (58. 0)(c metal)(22. 0 -100. 0) = - (60. 0)(4. 18)(22. 0 -18. 0) -4524 (c metal) = -1004 (c metal) = 0. 222 J/g°C

Calorimetry Examples § A piece of metal with a mass of 4. 68 g

Calorimetry Examples § A piece of metal with a mass of 4. 68 g at 135°C is placed in a calorimeter with 25. 0 g of water at 20. 0 °C. The temperature rises to 35. 0 °C. What is the specific heat of the metal?

Calorimetry Examples § q metal = - q water § § (m metal)(c metal)(

Calorimetry Examples § q metal = - q water § § (m metal)(c metal)( T metal) = - (m water)(c water)( T water) (4. 68)(c metal)(35. 0 -135. 0) = - (25. 0)(4. 18)(35. 0 -20. 0) -468 (c metal) = -1567. 5 (c metal) = 3. 35 J/g°C

Specific Heat & Phase Changes

Specific Heat & Phase Changes

Changes in State § Increasing or decreasing the amount of kinetic energy will cause

Changes in State § Increasing or decreasing the amount of kinetic energy will cause changes in the state of matter § Changes of State

§ Phase changes § Transitions between solid, liquid, and gaseous phases § § typically

§ Phase changes § Transitions between solid, liquid, and gaseous phases § § typically involve large amounts of energy compared to the specific heat. If heat were added at a constant rate to a mass of ice to take it through its phase changes to liquid water and then to steam, the energies required to accomplish the phase changes (called the latent heat of fusion and latent heat of vaporization ) would lead to plateaus in the temperature vs time graph. The graph below presumes that the pressure is one standard atmosphere.