Thermochemistry the heat energy of chemical reactions
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Thermochemistry: The heat energy of chemical reactions. heat. • Enthalpy is the amount of ________ transferred during a reaction. The symbol for the change in enthalpy is ∆H .

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Thermochemistry the heat energy of chemical reactions

Thermochemistry:The heat energy of chemical reactions

heat

•Enthalpyis the amount of ________ transferred during a reaction. The symbol for the change in enthalpy is ∆H.

  • An endothermic reaction is one that ___________ heat from the surroundings. (___∆ H) An endothermic reaction feels ______.

    Example--an “instant” ice pack

    •An exothermic process is one that _____________ heat to the surroundings. (___∆ H) An exothermic reaction feels _____.

    Example--burning paper

gains

+

cold

loses

hot


Thermochemistry how to measure heat energy changes

Thermochemistry:How to measure heat (Energy) changes

calorie (cal)

  • A ____________ is the amount of energy (heat) required to raise the temperature of one gram of water by one degree Celsius.

  • The “calorie” written on food is actually not one calorie in chemistry. It is actually 1 __________ (or ____calories) and is written with a capital C (Calorie) to keep the two separate.

  • A ____________ is the SI unit for measuring the amount of energy or heat transferred in chemistry.

  • Write down this conversion factor:

kilocalorie

1000

Joule (J)

1 cal = 4.184 J


Mcdonalds mcchicken

McDonalds McChicken


Mcdonalds mcdouble

McDonalds McDouble


Crunchy cheetos

Crunchy Cheetos


So what do we do with all these calories

So what do we do with all these Calories?

  • Your body will use these Calories as energy to do everyday activities but what if you don’t use all the calories you consume?

  • Your body will either use the energy or it will store it as fat!

  • So people who eat more food than their body can use exercise as a way of releasing the extra energy. But how much exercise do you really need to do

    in order to burn off those extra Calories?


Exercise

Exercise!

  • Running and walking are not the same!! Even though you can run OR walk a mile, there is a difference. According to David Swain, a Ph.D. in exercise physiology, “When you perform a continuous exercise, you burn five Calories for every liter of oxygen you consume and running in general consumes a lot more oxygen than walking.”

  • Running burns approximately 100 Calories per mile.

  • SO WHAT DOES THIS MEAN?


I love food but i hate running

I LOVE FOOD! But I hate running! 

  • To run off the food previously mentioned, this is approximately how long you would have to RUN!

  • 1 McChicken: 1.38 MILES

  • 1 McDouble: 3.10 MILES

  • 1 bag of Cheetos:3.30 MILES


Thermochemical reactions

Thermochemical Reactions

  • A thermochemical reaction is written as follows:

    2S + 3O2 2SO3 + 791.4 kJ

  • This equation represents an ___________ reaction since the heat is a ________.

    H2 + Br2 + 72.80 kJ  2HBr

  • This equation represents an ___________ reaction since the heat is a __________.

exothermic

product

endothermic

reactant


Calculating enthalpy h

Calculating Enthalpy (∆H)

  • There are two ways to calculate enthalpy.

  • Enthalpy can be calculated by the heat of formation of the products minus the reactants using the equation:

  • Enthalpy can also be calculated by the energy required to break bonds verses the energy give off when bonds form.

∆H = H (products) – H (reactants)

∆H = Energy of bonds broken – Energy of bonds formed


Calculating enthalpy h1

Calculating Enthalpy (∆H)

  • Calculate the heat given off by the reaction shown below:

    2 B5H9(g) + 12O2(g) 5B2O3(g) + 9H2O(g)

  • Heat of Formation

    B5H9(g) = 73.2 kJ/mol

    B2O3(g) = -1272.77 kJ/mol

    O2(g) = 0 kJ/mol

    H2O(g) = -241.82kJ/mol

∆H = H (products) – H (reactants)

[(5 mol B2O3 x -1272.77kJ/mol) +

(9 mol H2O x -241.82kJ/mol)] –

∆H =

(12 mol O2 x 0kJ/mol)]

[(2 molB5H9 x 73.2kJ/mol) +

-8686.63kJ

∆H =

[-8540.23kJ] -

[146.4kJ] =


Calculating enthalpy h2

Calculating Enthalpy (∆H)

  • Find the ∆ H for the following reaction given the following bond energies:

    2H2 + O2 2H2O

    Bond Energy

    H-H 436 kJ/mol

    O=O 499 kJ/mol

    O-H 463 kJ/mol

We have to figure out which bonds are broken and which bonds are formed.

2 H-H bonds are broken

1 O=O bond is broken

4 O-H bonds are formed

(2 O-H bonds are formed per water molecule,

and there are 2 water molecules formed)

Now we can substitute the values given into the equation:

∆H = Energy of bonds broken – Energy of bonds formed

[(2 x 436) + (1 x 499)] –

[4 x 463]

= -481 kJ

∆H =


Thermochemistry problems using stoichiometry

Thermochemistry Problems using Stoichiometry

  • How much heat will be released when 6.5 moles of sulfur reacts with excess oxygen according to the following equation? Also, tell whether it will be exothermic or endothermic!

    2S + 3O2 2SO3∆H = -791.4 kJ

-791.4 kJ

6.5 mol S

_______________

X

=

-2572 kJ

2 mol S

Exothermic ~

  • ∆H means heat

  • is a product!


Thermochemistry the heat energy of chemical reactions

Phase Changes & Energy

Endothermic: melting, evaporating/boiling & sublimation

Exothermic: freezing, condensation, & deposition


Thermochemistry the heat energy of chemical reactions

“Reaction Profiles”

Endothermic

Exothermic


Calculations in thermodynamics

Calculations in Thermodynamics

  • In order to calculate how much heat is transferred by a thermochemical reaction the equation we use is

    q = mc∆T

  • q = the ______ lost or gained in the process

  • m = the _____ of the substance

  • c = the ________ _____ ________

    The Specific heat of water is 4.186 Joules/gram °C

  • ∆T = ________ Temp. – ________Temp.

heat

mass

Specific

heat

capacity

Final

Initial


Specific heat

Specific Heat

  • Specific heat or “c” has units of Joule/gram °C. This refers to the energy needed to change one gram of the substance one degree Celsius.

    Check your understanding:

  • Which has a higher specific heat? Aluminum or water?

  • Why does a pizza roll that you can pickup with your fingers still have the possibility of burning your tongue?

Aluminum. Less heat is required to change aluminum’s temperature.

The pizza sauce inside contains water with a high specific heat.


Calculations practice

Calculations Practice

  • Example 1: How many Joules would it take to raise the temperature of 250 g of ice from -20 °C to -5 °C? (The specific heat of ice is 2.108 Joule/gram °C)

q = mc∆T

q = 250g

(2.108 J/g °C)

( -5 °C – (-20 °C))

q = 250g (2.108 J/g °C)( 15 °C)

q = 7905 Joules

= 7.9 kJ


More practice

More Practice

  • Example 2: What is the specific heat of Ethyl Alcohol if 100 grams of Ethyl Alcohol was heated from 30 °C to 50 °C when 1160 calories of heat was applied? The specific heat of Ethyl alcohol is 0.58 cal/g °C)

q = mc∆T

1160 cal = 100g

c

( 50 °C - 30 °C)

1160 cal = c x (2000 g °C)

c = 0.58 cal/g °C


Thermochemistry the heat energy of chemical reactions

Calorimeters

  • Calorimeters measure heat flow. It measures changes in water temperature after a reaction is performed.

(Constant Pressure)

Bomb Calorimeter

Usually studies combustion (Constant Volume)


Thermochemistry the heat energy of chemical reactions

Heat of Fusion & Heat of Vaporization

  • Molar heat of fusion(∆Hfus) : the amount of energy required to take 1 mole of a solid to the liquid state.

  • Example: H2O(s) H2O(l) ∆Hfus = 6.01 kJ

    • -Heat of fusion is usually greater for ionic solids than molecular solids since ionic solids are more strongly held together.

    • For ice to water … ∆Hfus= 6.01 kJ/mol

  • Molar heat of vaporization(∆H vap): the amount of energy required to take 1 mole of a liquid to the gaseous state.

  • Example: H2O(l) H2O(g)∆Hvap = 40.67 kJ

  • For water to steam … ∆Hvap= 40.67 kJ/mol

These values are mainly used as conversion factors!


Heating curve of water

Heating Curve of Water


Calculations practice1

Calculations Practice

1) How much energy is required to raise 75 grams of water from --20 °C to 20 °C? (∆Hfus= 6.01 kJ/mol and ∆Hvap= 40.67 kJ/mol The specific heat of ice is 2.108 Joule/gram °C, and the specific heat of water is 4.186 joule/gram °C )

Big picture: Solid water heats up, melts, and then liquid water heats up.

q = mc∆T

q = 75g

x (2.108 J/g °C)

x (0°C – -20°C)

= 3162 Joules

= 3.2 kJ

1 mol H2O

6.01 kJ

75 g H2O

_______________

_______________

X

X

=

25.0 kJ

18.02 g H2O

1 mol H2O

q = mc∆T

q = 75g

x (4.186 J/g °C)

x (20°C –0°C)

= 6279 Joules

= 6.3 kJ

Energy required = 3.2 kJ + 25.0 kJ + 6.3 kJ = 34.5 kJ


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