thermochemistry 2 calorimetry and heats of...

THERMOCHEMISTRY – 2CALORIMETRY AND

HEATS OF REACTIONDr. Sapna Gupta

HEAT CAPACITY

• Heat capacity is the amount of heat needed to raise the temperature of the sample of substance by one degree Celsius or Kelvin.

q = CDt

• Molar heat capacity: heat capacity of one mole of substance.

• Specific Heat Capacity: Quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure.

q = m s Dt (final-initial)

• Measured using a calorimeter –

it absorbed heat evolved or absorbed.

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 2

EXAMPLES OF SP. HEAT CAPACITY

The higher the number the higher the energy required to raise the temp.


CALORIMETRY: EXAMPLE - 1

Example:

A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C).

Solution

m = 35.8 g

s = 0.388 J/(g°C)

Dt = 28.00°C – 20.00°C = 8.00°C

q = m s Dt

= 111J C8.00Cg

J0.388g35.8

q



Example:

Nitromethane, CH3NO2, an organic solvent burns in oxygen according to the following reaction:

CH3NO2(g) + 3/4O2(g) CO2(g) + 3/2H2O(l) + 1/2N2(g)

You place 1.724 g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction.

Solution:

Heat evolved

Convert the heat evolved to per mol. = -709 KJ

Now the equation:

CH3NO2(l) + ¾O2(g) CO2(g) + 3/2H2O(l) + ½N2(g); DH = –709 kJ

kJ 20.03 C22.23C28.81C

kJ3.044

Δ

rxn

calrxn

q

tCq

3 2rxn

3 2 3 2

20.03 kJ 61.04 g CH NO

1.724 g CH NO 1mol CH NOq



Example:

A metal pellet, 85.00 grams at an original temperature of 92.5C is dropped into a calorimeter with 150.00 grams of water at an original temperature of 23.1C. The final temperature of the water and the pellet is 26.8C. Calculate the heat capacity and the specific heat for the metal.

Solution:

Energy = qwater = msDT = (150.00 g) (4.184 J/gC) (3.7C)

= 2300 J (water gained energy) = -2300 J (pellet released energy)

Heat capacity of pellet: q = CDT

C = q/DT = 2300 J/65.7C = 35 J/C

Specific heat of pellet:

s =35 J/oC

85.00g= 0.41J/goC

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry

Note: we could have calculated the sp. heat of the pellet in one calculation usingq = m s Dt,But since the question was also about Heat Capacity, we calculated that first and then just divide that by mass to get the sp. heat of the metal. 6


Example:

A snack chip with a mass of 2.36 g was burned in a bomb calorimeter. The heat capacity of the calorimeter 38.57 kJ/C. During the combustion the water temp rose by 2.70C. Calculate the energy in kJ/g for the chip.

Solution:

qrxn = − Ccal DT = −(38.57 kJ/C) (2.70C)

= − 104 kJ

Energy content is a positive quantity.

= 104 kJ/2.36 g

= 44.1 kJ/g

Convert J to Cal.

Food Calories: 10.5 Cal/g


HESS’S LAW

• The change in enthalpy when reactants are converted to products is the same whether the reaction occurs in one step or a series of steps.

• Usually used when heat of reaction cannot be determined directly.

• Manipulation of equations is the same as before.

• Add all the DH values after equations are manipulated.

• Suppose you want the DH for the reaction:

2C(graphite) + O2(g) 2CO(g)

However two other reactions are known:

C(graphite) + O2(g) CO2(g); DH = -393.5 kJ

2CO2(g) 2CO(g) + O2(g); DH = – 566.0 kJ

In order for these to add to give the reaction we want, we must multiply the first reaction by 2 (Note that we also multiply DH by 2.)

Then add the two heat of reactions along with the DH. (-787+(-566.0))=1353 kJ


x2

leave as is

8

HESS’S LAW: EXAMPLE

Example:

Given the following equations:

H3BO3(aq) HBO2(aq) + H2O(l) DHrxn = 0.02 kJ

H2B4O7(aq) + H2O(l) 4 HBO2(aq) DHrxn = 11.3 kJ

H2B4O7(aq) 2 B2O3(s) + H2O(l) DHrxn = 17.5 kJ

Find the DH for this overall reaction.

2H3BO3(aq) B2O3(s) + 3H2O(l)

Solution:

2H3BO3(aq) 2HBO2(aq) + 2H2O(l) DHrxn = 2(−0.02 kJ) = −0.04 kJ

2HBO2(aq) 1/2H2B4O7(aq) + 1/2H2O(l) DHrxn = +11.3 kJ/2 = 5.65 kJ

1/2H2B4O7(aq) B2O3(s)+ 1/2H2O(l) DHrxn = 17.5 kJ/2 = 8.75 kJ

2H3BO3(aq) B2O3(s) + 3H2O(l) DHrxn = 14.36 kJ


x 2

reverse, ÷2

÷ 2

9

STANDARD ENTHALPY OF FORMATION

• Standard state refers to the standard thermodynamic conditions for the substances when listing or comparing thermodynamic data: the atmpressure and temperature (25 oC).

• Standard conditions are indicated with a degree sign “o”: eg DH and DHo –the second one is the standard enthalpy at 1 atm and 25 oC.

• Elements can exist in more than one physical state e.g. carbon as graphite or diamond (allotropes).

• The standard enthalpy of formation, DHfo, is the enthalpy change for

formation of one mole of substance from its elements in their reference forms and in their standard states. The DHf

o for an element in its reference state is zero.

• Example: H2(g) + 1/2O2(g) H2O(l) DHf° = –285.8 kJ

• The equation below can be used to calculate the enthalpy. (n = number of moles)

reactants

f

products

freaction ΔΔΔ HnHnH


STANDARD ENTHALPY: EXAMPLE

Calculate the DH° for the following reaction:

CH3OH(l) CH3OH(g)

Solution:

Find the DHf° values for all the compounds and subtract product from reactant.

= + 38.0 KJ

mol

kJ200.7Δ:methanolgaseousFor

mol

kJ238.7Δ:methanolliquidFor

f

f

H

H

mol

kJ238.7 mol 1

mol

kJ200.7 mol 1Δ vapH


STANDARD ENTHALPY: EXAMPLE

Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate DH° for the following reaction: 2CH3OH(aq) + O2(g) 2HCHO(aq) + 2H2O(l)

Standard enthalpies of formation, DHf°:

CH3OH(aq): −245.9 kJ/mol

HCHO(aq): −150.2 kJ/mol

H2O(l): −285.8 kJ/mol

reactants

f

products

freaction ΔΔΔ HnHnH

oreacton

kJ kJΔ 2 mol 150.2 2 mol 285.8

mol mol

kJ kJ2 mol 245.9 1 mol 0

mol mol

H

kJ 491.8kJ 571.6kJ 300.4 Δ oreaction H

kJ 491.8kJ 872.0Δ oreaction H kJ 380.2Δ o

reaction H


APPLICATIONS: FOOD

Foods are fuel needed for the body. They undergo a combustion process in the body.

For glucose, a carbohydrate:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l); DHf° = –2803 kJ

For glycerol trimyristate, a fat:

C45H86O6(s) + 127/2O2(g) 45CO2(g) + 43H2O(l); DHf°= –27,820 kJ

The average value for carbohydrates is 4.0 Cal/g and for fats is 9.0 Cal/g.


APPLICATIONS: FUEL

• Fuels originate from organic substances: plants and animals, that have decayed over millions of years. This provides the coal, petroleum products and natural gas.

• Coal is about 30% of US energy consumption. Variety of coal determines how much carbon is in it: anthracite – 80% carbon; bituminous coal – 45-65% carbon.

• Natural gas is about 23% of energy. It is a fluid and easy to transport. Pure natural gas is methane, CH4 with small amounts of ethane, propane and butane.

• Petroleum is a mixture of hydrocarbon compounds.

• Natural gas and petroleum products are in short supply while coal will last longer.

• All these are used for energy sources.


KEY CONCEPTS

• Measuring heats of reaction - calorimetry

• Hess’s Law

• Standard enthalpy of formation


thermochemistry 2 calorimetry and heats of...

Documents