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Thermochemistry

- Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state.

Heat (q)

- Heat is the energy that transfers from one object to another because of a temperature difference between them.
- Heat ALWAYS flows from a warmer object to a cooler one.

Heat movement

- Heat moves between the system (reaction) and the surroundings
- *** must obey the law of conservation of energy (heat (energy) is never created nor destroyed, just transferred)
- Thermochemical equations tell you the direction of heat flow by the “sign”, + or -

Endo vs. Exo-

- Endothermic reactions: absorbs heat from surroundings (+).
- If you touch an endothermic reaction it feels COLD

- Exothermic reactions: release heat to the surroundings (-)
- If you touch an exothermic reaction it feels HOT

HEAT energy

- UNIT of energy = JOULE (J)
or Calorie (cal) 1 cal = 4.184 J

- Heat capacity is how much heat (energy) is needed to increase the temperature of an object by 1 C.
- *** heat capacity of an object depends on both its mass and its chemical composition
- The greater the mass, the greater its heat capacity

Example

- On a sunny day, a 20 kg puddle of water may be cool, while a nearby 20 kg iron sewer cover may be too hot to touch.
- Both have the same mass, BUT the are made of different materials (chemical composition) BECAUSE they have different SPECIFIC HEAT CAPACITIES

SPECIFIC HEAT (C)

- SPECIFIC Heat Capacity (specific heat, C) = specific to a substance
- Amount of heat it takes to raise the temp. of 1 g of a substance by 1 C
- Units = (J/gC)
- WATER has a specific heat of 4.184 J/gC

- Units = (J/gC)

- Amount of heat it takes to raise the temp. of 1 g of a substance by 1 C

CALCULATING HEAT

- q = mCT
- Where:
- q = heat (Joules)
- m = mass (grams)
- C = specific heat (J/gC)
- T = change in temperature (C) Tf-Ti

Calculations

- How much heat is absorbed when a 95.4 g piece of copper increases from 25.0 C to 48.0 C. The specific heat of copper is 0.387 J/gC.

- m= 95.4 g C= 0.387 J/gC
- T= 48.0 C - 25.0 C= 23 C
- q= (95.4) (.387) (23)
- 849 J

Ex. 2

- How much heat is absorbed when 3.4 g of olive oil is heated from 21.0 C to 85.0 C. The specific heat of olive oil is 2.0 J/g C.

- m= 3.4 g C= 2.0 J/gC
- T= 85.0 C - 21.0 C= 64 C
- q= (3.4) (2.0) (64)
- 435 J

Ex. 3

- How much heat is required to raise the temperature of 250.0 g of mercury 52.0 C? The specific heat of mercury is 0.14 J/g C.

- m= 250 g C= 0.14 J/gC
- T= 52 C
- q= (250) (0.14) (52)
- 1820 J = 1.8 kJ

Ex. 4

- How many joules (J) of heat are absorbed when 1000.g of water is heated from 18.0 C to 85.0 C? C for water = 4.184 J/g C.

- m= 1000 g C= 4.184 J/gC
- T= 85.0 C - 18.0 C= 67 C
- q= (1000) (4.184) (67)
- 280000 J =2.80 x 105 J

Practice Problems

1. What is the specific heat of a substance that has a mass of 25.0 g and requires 2197 J of energy to raise its temperature by 15.0 C?

2. Suppose 100.0 g of ice absorbs 1255.0 J of heat. What is the corresponding temperature change? The specific heat of ice is 2.1 J/g C.

3. How many Joules of heat energy are required to raise the temperature of 100.0 g of aluminum by 120.0 C. The specific heat of aluminum is 0.90 J/g C.

ANSWERS

- 1) 5.86 J/g°C
- 2) 6.0 °C
- 3) 10800 J = 11,000 J or 1.1 x 104 J

STOP

- Work on Specific Heat Calcs WS

ENTHALPY

- Enthalpy = a type of chemical energy, sometimes referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction)
- exothermic reactions (feels hot):
- q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative values)

- endothermic reactions (feels cold):
- q = ΔH > 0 (positive values)

Thermochemical Equations

- A chemical equation that shows the enthalpy (H) is a thermochemical equation.

Rule #1

The magnitude (value) of H is directly proportional to the amount of reactant or product.

H2 + Cl22HClH = - 185 kJ

* meaning there are 185 kJ of energy RELEASED for every:

1 mol H2

1 mol Cl2

2 moles HCl

Rules of Thermochemistry

Example 1:

H2 + Cl22HClH = - 185 kJ

Calculate H when 2.00 moles of Cl2 reacts.

Rules of Thermochemistry

Example 2: Methanol burns to produce carbon dioxide and water:

2CH3OH + 3O2 2CO2 + 4H2O H = - 1454 kJ

What mass of methanol is needed to produce 1820 kJ?

Rule #2

H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction.

(If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

Rules of Thermochemistry

Example 1:

Given:

H2 + ½O2 H2OH = -285.8 kJ

Reverse:

H2O H2 + ½O2H = +285.8 kJ

Example 2

CaCO3 (s) CaO (s) + CO2 (g) H = 178 kJ

What is the H for the REVERSE RXN?

CaO (s) + CO2 (g) CaCO3 (s) H = ?

Practice Problem: (rule 1 + rule 2)

Given:

H2 + ½O2 H2OH = -285.8 kJ

Calculate H for the equation:

2H2O 2H2 + O2

Alternate form of thermochem. eq.

- Putting the heat content of a reaction INTO the actual thermochemical eq.
- EX:
- H2 + ½O2 H2OH = -285.8 kJ
- EXOTHERMIC:
- Heat is ___________ as a ____________.

ALTERNATE FORM

- EX:
- H2 + ½O2 H2OH = -285.8 kJ
- The alternate form is this:H2 + ½O2 H2O +285.8 kJ

2 NaHCO3 + 129 kJ Na2CO3 + H2O + CO2

The alternate form is this:

2 NaHCO3 Na2CO3 + H2O + CO2H =+ 129 kJ

Put the following in alternate form

- 1. H2 + Cl2 2 HCl H = -185 kJ
- 2. 2 Mg + O2 2 MgO + 72.3 kJ
- 3. 2 HgO 2 Hg + O2H = 181.66 kJ

CALORIMETRY

- The enthalpy change associated with a chemical reaction or process can be determined experimentally.
- Measure the heat gained or lost during a reaction at CONSTANT pressure

Calorimeter

- Device used to measure the heat absorbed or released during a chemical or physical process

Example

- If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?
- Why is there a difference in temperature between the two objects?

Because…

- Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).

Need to know how to calc. heat (review)

- Heat (q) = mCT
- If the specific heat of Al is 0.90 J/gC, how much heat is required to raise the temperature of 10,000 g of Al from 25.0 C to 30.0 C?

Calculator says……..

- = 45,000 J

For example (review):

- If 418 J is required to increase the temperature of 50.0 g of water by 2.0 C, what is the specific heat of water?

What happens in a calorimeter

- One object will LOSE heat, and the other will ABSORB the heat
- System loses heat to surroundings = EXO = -q
- System absorbs heat from surroundings = ENDO = +q

EXAMPLE

- A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble?
The specific heat of water is 4.184 J/g C.

- Which was warmer? The pebble or the water?

- The pebble because the water heated up from 25.0 C to 26.4 C.
- Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic)
- Do the calc…

All values are for WATER

- A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble?
- Calc the heat absorbed by the water (+q).
- The heat (J) released by the warm pebble = - of the heat absorbed by the water.
q water = - q pebble

PRACTICE 1 (LAB type of CALC)

- Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.
The specific heat of water is 4.184 J/g C.

- Heat gained by water = Heat lost by the Al
- q of water = - q Al
- Make a chart

We don’t know specific heat of Al, but we know all the values for water

- So, calc q for WATER
- The q for water is the same for q of Al
- (the value of q is the same)

0.879 J/g °C values for water

Practice 2 values for water

- A lead mass is heated and placed in a foam cup calorimeter containing 40.0 g of water at 17.0C. The water reaches a temperature of 20.0 C.
How many joules of heat were released by the lead?

The specific heat of water is 4.184 J/g C.

- 502 J or 5.0 x 10 values for water2 J

Work on WS: Calorimetry values for water

- STOP

Review Heat in Changes of STATE values for water

- Heating and Cooling Curve
- LABEL:*physical states
*boiling point

*melting point

*heat of fusion

*heat of vaporization

Need to calculate heat for the WHOLE process of changing physical state

- Review
- Specific heat is the amount of heat that must be added to a stated mass of a substance to raise its temperature by 1°C, with NO change in state.
- CHANGING STATE requires “heat of vaporization” and “heat of fusion”
- Heat of vaporization = amount of heat that must be added to 1 g of a liquid at its boiling point to convert it to vapor with NO temp. change
- Heat of fusion = amount of heat needed to melt 1 g of a solid at its melting point

MAKE YOUR OWN HEATING/COOLING CURVE physical state

- Specific heat of water = 4.184 J/g°C
- Specific heat of ice =2.09 J/g°C
- Specific heat of steam = 2.03 J/g°C
- Heat of fusion = 334 J/g
- Heat of vaporization = 2260 J/g

Now let’s take it a step further…. physical state

- How can you use a heating curve to explain specific heat?

liquid physical state

solid

gas

Heating curves and DH

boiling/

cond. pt

melting/

freezing pt

temperature

added energy

liquid physical state

solid

gas

Heating curves and DH

boiling/condensing

occurring here

melting/freezing

occurring here

boiling/

cond. pt

melting/

freezing pt

temperature

added energy

How is the total enthalpy change ( physical stateDH) calculated for a substance whose temperature change includes a change in state?

We can use DH in our specific heat equation in place of “q”

DH = m x C xDT

the energy absorbed physical state

as a solid melts becomes

potential energy,

so no Dt

temperature

DH = m x CsolidxDt

added energy

D physical stateH = DHfusxm

Dt of liquid

absorbing energy

temperature

DH = m x CsolidxDt

added energy

the energy absorbed physical state

as a liquid boils becomes

potential energy,

so no Dt

DH = DHfusx m

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

D physical stateH = DHvapx m

DH = DHfusx m

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

D physical statet of gas

absorbing energy

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

DH physical state = m x CgasxDt

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

DH physical state = m x CgasxDt

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

The physical stateDH of any substance

being heated will be the sum of the

DH of any Dt occurring plus

DH of any phase change occurring

DH = m x CgasxDt

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

EX: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

DH = m x CgasxDt

DH = DHvapx m

DH = DHfusx m

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

EX: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

50 oC

temperature

0 oC

-20 oC

added energy

EX: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 334J/g, CH2O liq = 4.186 J/goC

50 oC

temperature

0 oC

-20 oC

added energy

EX: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC

50 oC

DH3 = m x CliquidxDt

DH2 = DHfusx m

temperature

0 oC

DH1 = m x CsolidxDt

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC

50 oC

DH3 = m x CliquidxDt

DH2 = DHfusx m

temperature

0 oC

DH1 = 10g x 2.1 J/goC x20oC

-20 oC

added energy

D physical stateH2=10 g x 1mol/18g x 6.01kJ/mol

EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 334J/g, CH2O liq = 4.186 J/goC

50 oC

DH2=334J/g x 10g

DH3 = m x CliquidxDt

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 334J/g CH2O liq = 4.186 J/goC

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC

total DH = DH1 + DH2 + DH3

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC

total DH = DH1 + DH2 + DH3

= 420 J + 3340 J + 2093 J

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

use the following values:

Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = 4.186 J/goC

total DH = DH1 + DH2 + DH3

5853 J = 420 J + 3340 J + 2093 J

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

It takes 5853 joules to heat up 10 grams

of water from -20 oC to +50 oC.

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20 oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

It takes 5853 joules to heat up 10 grams

of water from -20 oC to +50 oC.

DH3 = 10g x 4.186 J/goC x 50 oC

50 oC

DH2=334J/g x 10g

temperature

0 oC

DH1 = 10g x 2.1 J/goC x 20 oC

-20 oC

added energy

EXAMPLE: What is physical stateDH for 10 g water with a total Dt from -20 oC to +50 oC?

It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC.

50 oC

temperature

0 oC

-20 oC

5853 J

added energy

USE YOUR HEATING/COOLING CURVE to help guide your calculations

1. How much heat is released by 250.0 g of water as it cools from 85.0 °C to 40.0 °C?

ANSWER calculations

- 1. How much heat is released by 250.0 g of water as it cools from 85.0 °C to 40.0 °C?
- Calculation is within the liquid phase of water (no phase changes)
q = mcΔT = (250.0 g)(4.184 J/g°C)(40.0°C - 85.0°C)

=-47070 J

Example 2 calculations

- How much heat energy is required to bring 135.5 g of water at 55.0 °C to its boiling point and then vaporize it?
- THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE?Use you heating and cooling curve to help you understand the steps!

Step 1 = raise temp of water from 55.0 °C to 100.0 °C calculations

Step 2 = vaporize water using heat of vaporization

q = mcΔT

q = (135.5g)(4.184 J/g°C)(100.0°C-55.0°C) = 25511.94 to (3 sig figs) = 25500 J

q = mass x heat of vaporization

q = (135.5g)(2260 J/g) =

= 306230 to (3 sig figs) = 306000 J

***ADD UP THE 2 NUMBERS TO GET THE OVERALL HEAT REQUIRED FOR THIS PROCESS

How much heat energy is required to bring 135.5 g of water at 55.0 °C to its boiling point and then vaporize it?

ANSWER (2 steps)Example 3 calculations

- How much heat energy is required to convert 15.0 g of ice at -12.5 °C to steam at 123.0 °C?
- THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE??

331500 J calculations

How much heat energy is required to convert 15.0 g of ice at -12.5 °C to steam at 123.0 °C? = 5 steps

q = (15.0 g)(2.09 J/g°C)(0.0--12.5°C) = 392 J

q = (15.0 g)(334 J/g) = 5010 J

q = (15.0 g)(4.184 J/g°C)(100.0-0.0°C) = 6280 J

q = (15.0 g)(2260 J/g) = 33900 J

q = (15.0 g)(2.03 J/g°C)(123.0-100.0°C) = 700. J

46282 J

WS: PROBLEM SET (heat transfer) -12.5 °C to steam at 123.0 °C? =

- Complete the problems