<<

THERMOCHEMISTRY – 2 AND OF REACTION Dr. Sapna Gupta CAPACITY

is the amount of heat needed to raise the of the sample of substance by one degree or . q = CDt • : heat capacity of one of substance. • : Quantity of heat needed to raise the temperature of one of substance by one degree Celsius (or one Kelvin) at constant .

q = m  s  Dt (final-initial) • Measured using a – it absorbed heat evolved or absorbed.

Dr. Sapna Gupta/-2-Calorimetry 2 EXAMPLES OF SP. HEAT CAPACITY

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

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 3 CALORIMETRY: EXAMPLE - 1

Example: A piece of 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  0.388 J q  35.8 g   8.00C = 111J  gC 

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 4 CALORIMETRY: EXAMPLE - 2

Example:

Nitromethane, CH3NO2, an organic solvent burns in according to the following reaction: 3 3 1 CH3NO2(g) + /4O2(g)  CO2(g) + /2H2O(l) + /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: q   C Δt Heat evolved rxn cal  3.044 kJ  qrxn    28.81C  22.23C   20.03 kJ  C 

20.03 kJ 61.04 g CH32 NO Convert the heat evolved to per mol. qrxn  = -709 KJ 1.724 g CH3 NO 2 1mol CH 3 NO 2 Now the equation: 3 CH3NO2(l) + ¾O2(g)  CO2(g) + /2H2O(l) + ½N2(g); DH = –709 kJ Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 5 CALORIMETRY: EXAMPLE - 3

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

Energy = qwater = msDT = (150.00 g) (4.184 J/gC) (3.7C) = 2300 J (water gained energy) = -2300 J (pellet released energy) Heat capacity of pellet: q = CDT

C = q/DT = 2300 J/65.7C = 35 J/C Note: we could have calculated the sp. heat of the Specific heat of pellet: pellet in one calculation using 35 J/oC q = m  s  Dt, s = = 0.41J/goC 85.00g But since the question was also about Heat Capacity, we calculated that first and then just divide that by to get Dr. Sapna Gupta/Thermochemistry-2-Calorimetry the sp. heat of the metal. 6 CALORIMETRY: EXAMPLE - 4

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 the water temp rose by 2.70C. Calculate the energy in kJ/g for the chip. Solution:

qrxn = − Ccal DT = −(38.57 kJ/C) (2.70C) = − 104 kJ Energy content is a positive quantity. = 104 kJ/2.36 g = 44.1 kJ/g Convert J to Cal. Food : 10.5 Cal/g

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 7 HESS’S LAW

• The change in 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 x2

2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ leave as is 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

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 8 HESS’S LAW: EXAMPLE

Example: Given the following equations:

H3BO3(aq)  HBO2(aq) + H2O(l) DHrxn = 0.02 kJ x 2

H2B4O7(aq) + H2O(l)  4 HBO2(aq) DHrxn = 11.3 kJ reverse, ÷2

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

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 9 STANDARD ENTHALPY OF FORMATION

• Standard state refers to the standard thermodynamic conditions for the substances when listing or comparing thermodynamic data: the atm pressure and temperature (25 oC). • Standard conditions are indicated with a degree sign “o”: eg DH and DHo – the 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). o • The standard enthalpy of formation, DHf , is the enthalpy change for formation of one mole of substance from its elements in their reference o forms and in their standard states. The DHf for an element in its reference state is zero. 1 • Example: H2(g) + /2O2(g)  H2O(l) DHf° = –285.8 kJ • The equation below can be used to calculate the enthalpy. (n = number of

moles)    ΔHreaction  nΔHf  nΔHf products reactants Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 10 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.

kJ For liquid methanol : ΔH   238.7 f mol kJ For gaseous methanol : ΔH   200.7 f mol   kJ    kJ  ΔHvap  1 mol   200.7   1 mol   238.7  = + 38.0 KJ   mol    mol 

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 11 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 of formation, DHf°:

CH3OH(aq): −245.9 kJ/mol

HCHO(aq): −150.2 kJ/mol    ΔHreaction  nΔHf  nΔHf H2O(l): −285.8 kJ/mol products reactants

o kJ   kJ  ΔHreacton  2 mol   150.2   2 mol   285.8  mol   mol  kJ   kJ  2 mol   245.9   1 mol  0  mol   mol 

o ΔHreaction    300.4 kJ   571.6 kJ  491.8 kJ

o o ΔHreaction   872.0 kJ   491.8 kJ ΔHreaction  380.2 kJ

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 12 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: 127 C45H86O6(s) + /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.

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 13 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.

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 14 KEY CONCEPTS

• Measuring heats of reaction - calorimetry • Hess’s Law • Standard enthalpy of formation

Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 15