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THERMOCHEMISTRY

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THERMOCHEMISTRY

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

- 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 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 -

- 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

- 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

- 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 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

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

- 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

- 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

- 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

- 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

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.

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

- Work on Specific Heat Calcs WS

- 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)

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

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

Example 1:

H2 + Cl22HClH = - 185 kJ

Calculate H when 2.00 moles of Cl2 reacts.

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?

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)

Example 1:

Given:

H2 + ½O2 H2OH = -285.8 kJ

Reverse:

H2O H2 + ½O2H = +285.8 kJ

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

What is the H for the REVERSE RXN?

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

Given:

H2 + ½O2 H2OH = -285.8 kJ

Calculate H for the equation:

2H2O 2H2 + O2

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

- 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

Put in the alternate form

Ex:

2 NaHCO3 + 129 kJ Na2CO3 + H2O + CO2

The alternate form is this:

2 NaHCO3 Na2CO3 + H2O + CO2H =+ 129 kJ

- 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

- 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

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

- 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?

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

- 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?

- = 45,000 J

- 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?

- 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

- 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…

- 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

- 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

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

- 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 102 J

- STOP

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

*melting point

*heat of fusion

*heat of vaporization

- 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

- 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

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

temperature

added energy

Heating curves and DH

liquid

solid

gas

Heating curves and DH

temperature

added energy

liquid

solid

gas

Heating curves and DH

melting/

freezing pt

temperature

added energy

liquid

solid

gas

Heating curves and DH

boiling/

cond. pt

melting/

freezing pt

temperature

added energy

liquid

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 (DH) 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

temperature

Dt of solid

absorbing energy

added energy

temperature

DH = m x CsolidxDt

added energy

the energy absorbed

as a solid melts becomes

potential energy,

so no Dt

temperature

DH = m x CsolidxDt

added energy

DH = DHfusx m

temperature

DH = m x CsolidxDt

added energy

DH = DHfusxm

Dt of liquid

absorbing energy

temperature

DH = m x CsolidxDt

added energy

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

the energy absorbed

as a liquid boils becomes

potential energy,

so no Dt

DH = DHfusx m

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

DH = DHvapx m

DH = DHfusx m

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

Dt of gas

absorbing energy

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

DH = m x CgasxDt

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

DH = m x CgasxDt

DH = DHvapxm

DH = DHfusxm

DH = m x CliquidxDt

temperature

DH = m x CsolidxDt

added energy

The DH 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 DH 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 DH 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 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

temperature

0 oC

-20 oC

added energy

EX: 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 = 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 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 = 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

DH2=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 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

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 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 = 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 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 = 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 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 = 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 DH 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 DH 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 DH 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

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

- 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

- 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

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?

- 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

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

- Complete the problems