Final 1 thermochemistry

thermochemistry

INTRODUCTION

Thermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions.

Heat is the transfer of thermal energy.

Heat is either absorbed or released during a process.

Chemical Reactions

• All chemical reactions involve bond breaking and bond forming.

• Energy is needed to break bonds and released when bonds are formed.

• Chemical reactions are accompanied by a change in energy, mainly in the form of heat.

HEAT REACTION=Exothermic and Endothermic

• A reaction in which heat is given out is exothermic.

• A reaction in which heat is taken in is endothermic.

2H2(g) + O2(g) 2H2O(l) + energy

energy + 2HgO(s) 2Hg(l) + O2(g)

Heat of:

• Formation- is the heat change in kilojoules, when one mole of a substance is formed from its elements in their standard state.

• Of reaction- heat in kilojoules released or absorbed when the number of moles of reactants indicated, in the balanced equation describing the reaction, react completely.

• Combustion- is the heat change in kilojoules when one mole of the substance is completely burned in excess oxygen

cont

• Heat of Neutralization-? The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

• Heat of vaporization:is the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure (often atmospheric pressure, as in STP).

http://en.wikipedia.org/wiki/Enthalpy

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Atmospheric_pressure

http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

The First Law of ThermodynamicsThe First Law of ThermodynamicsThe first law of thermodynamics states that energy can be converted from one form to another, but cannot be created or destroyed./The law of conservation of energy states that energy cannot be created or destroyed, but only changed from one form to another.

ΔU is the change in the internal energy.

“sys” and “surr” denote system and surroundings, respectively.

ΔU = Uf – Ui; the difference in the energies of the initial and final states.

ΔUsys + ΔUsurr = 0

ΔUsys = –ΔUsurr

Work and HeatWork and Heat

The overall change in the system’s internal energy is given by:

q is heatq is positive for an endothermic process (heat absorbed by the system)

q is negative for an exothermic process (heat released by the system)

w is workw is positive for work done on the system

w is negative for work done by the systemCalculate the overall change in internal energy, ΔU, (in joules) for a system that absorbs 188 J of heat and does 141 J of work on its surroundings.Solution The system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative.

ΔU = q + w = 188 J + (-141 J) = 47 J

ΔU = q + w

Enthalpy changes of exothermic and endothermic reactions

Heat of: • HEAT Of reaction- heat in kilojoules released or

absorbed when the number of moles of reactants indicated, in the balanced equation describing the reaction, react completely.

• For an exothermic reaction ΔH is negative(-)• For an endothermic reaction ΔH is positive(+) • HEAT OF Combustion- is the heat change in kilojoules

when one mole of the substance is completely burned in excess oxygen

• HEAT OF Formation- is the heat change in kilojoules, when one mole of a substance is formed from its elements in their standard state

• HEAT OF Neutralization-?

Heat of Combustion of Different Fuels

• Methane (natural gas) -890 kj mol-1 • Propane (LPG) -2219 kj mol-1 • Hydrogen -286 kj mol-1 • Petrol (octane) -5470 kj mol-1

Bond Energy• Bond energy is the amount of energy in

kilojoules needed to break one mole of bonds of the same type, all species being in the gaseous state.

• The average C-H bond energy in methane is 412kj mol-1 i.e. E(C-H) = 412kj mol-1 .

• The energy of a particular bond type can vary.• It is usual to quote the average bond energies.

Recall, by definition a change in energy equals heat transferred (q) plus work (w):

E = q + w

Consider a process carried out at constant pressure. At constant pressure, work involves only a change in volume. We can then substitute -PV for w.

E = qp - PV

Then if we want to solve for the heat transferred, qp, at constant pressure, we simply rearrange the equation.

qp = E + PV

At Constant Pressure

EnthalpyEnthalpy

Sodium azide detonates to give a large quantity of nitrogen gas.

Under constant volume conditions, pressure increases:

2NaN3(s) 2Na(s) + 3N2(g)

EnthalpyEnthalpy

Pressure-volume, or PV work, is done when there is a volume change under constant pressure.

w = −PΔV

P is the external opposing pressure.

ΔV is the change in the volume of the container.

Worked Example 10.2

Strategy Determine change in volume (ΔV), identify the external pressure (P), and use w = −PΔV to calculate w. The result will be in L∙atm; use the equality 1 L∙atm = 101.3 J to convert to joules.

Determine the work done (in joules) when a sample of gas extends from 552 mL to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm, (b) against a constant pressure of 1.00 atm, and (c) against a vacuum (1 L∙atm = 101.3 J).

Solution ΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm, (c) P = 0 atm.

(a) w = -(1.25 atm)(339 mL)

(b) w = -(1.00 atm)(339 mL)

1 L1000 mL

101.3 J1 L∙atm

= -42.9 J

1 L1000 mL

101.3 J1 L∙atm

= -34.3 J

Worked Example 10.2 (cont.)

Solution(c) w = -(0 atm)(339 mL) 1 L

1000 mL101.3 J1 L∙atm

= 0 J

Think About It Remember that the negative sign in the answers to part (a) and (b) indicate that the system does work on the surroundings. When an expansion happens against a vacuum, no work is done. This example illustrates that work is not a state function. For an equivalent change in volume, the work varies depending on external pressure against which the expansion must occur.

EnthalpyEnthalpy

Pressure-volume, or PV work, is done when there is a volume change under constant pressure.

w = −PΔV

When a change occurs at constant volume, ΔV = 0 and no work is done.

ΔU = q + w

qV = ΔU

ΔU = q − PΔV

substitute

EnthalpyEnthalpy

Under conditions of constant pressure:

ΔU = q + w

qP = ΔU + PΔV

ΔU = q − PΔV

EnthalpyEnthalpy

The thermodynamic function of a system called enthalpy (H) is defined by the equation:

H = U + PV

A note about SI units:

Pressure: pascal; 1Pa = 1 kg/(m . s2)

Volume: cubic meters; m3

PV: 1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 J

Enthalpy: joules

U, P, V, and H are all state functions.

Enthalpy and Enthalpy ChangesEnthalpy and Enthalpy Changes

For any process, the change in enthalpy is:

ΔH = ΔU + Δ(PV) (1)

ΔH = ΔU + PΔV

If pressure is constant:

(2)

ΔU = ΔH + PΔV

Rearrange to solve for ΔU:

(3)

qp = ΔU + ΔV

Remember, qp:

(4)

qp = (ΔH − PΔV) + PΔV

Substitute equation (3) into equation (4) and solve:

(5)

qp = ΔH for a constant-pressure process

Recall our original definition of enthalpy: H = E + PV

Then for a change in enthalpy:H = E + (PV)

If we set P constant, then: H = E + P V Since

qp = E + PVThen

H = qp

The change in enthalpy, H, is then equal to the heat transferred at constant pressure, qp.

Enthalpy = Heat Transferred

Enthalpy and Enthalpy ChangesEnthalpy and Enthalpy Changes

The enthalpy of reaction (ΔH) is the difference between the enthalpies of the products and the enthalpies of the reactants:

Assumes reactions in the lab occur at constant pressure

ΔH > 0 (positive) endothermic process

ΔH < 0 (negative) exothermic process

ΔH = H(products) – H(reactants)

Thermochemical EquationsThermochemical Equations

Concepts to consider:Is this a constant pressure process?

What is the system?

What are the surroundings?

ΔH > 0 endothermic

H2O(s) H2O(l) ΔH = +6.01 kJ/mol


Concepts to consider:Is this a constant pressure process?

What is the system?

What are the surroundings?

ΔH < 0 exothermic

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol


Enthalpy is an extensive property.

Extensive properties are dependent on the amount of matter involved.

H2O(l) → H2O(g) ΔH = +44 kJ/mol

2H2O(l) → 2H2O(g) ΔH = +88 kJ/mol

Units refer to mole of reaction as written

Double the amount of matter Double the enthalpy


The following guidelines are useful when considering thermochemical equations:

1) Always specify the physical states of reactants and products because they help determine the actual enthapy changes.

CH4(g) + 2O2(g) ΔH = −802.4 kJ/molCO2(g) + 2H2O(g)

CH4(g) + 2O2(g) ΔH = +890.4 kJ/molCO2(g) + 2H2O(l)

different states different enthalpies


The following guidelines are useful when considering thermochemical equations:

2) When multiplying an equation by a factor (n), multiply the ΔH value by same factor.

CH4(g) + 2O2(g) ΔH = − 802.4 kJ/molCO2(g) + 2H2O(g)

2CH4(g) + 4O2(g) ΔH = − 1604.8 kJ/mol2CO2(g) + 4H2O(g)

3) Reversing an equation changes the sign but not the magnitude of ΔH.

CH4(g) + 2O2(g) ΔH = − 802.4 kJ/molCO2(g) + 2H2O(g)

CO2(g) + 2H2O(g) ΔH = +802.4 kJ/molCH4(g) + 2O2(g)

Worked Example 10.3

Strategy The thermochemical equation shows that for every mole of C6H12O6 produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed for the production of 75.0 g of C6H12O6. We must first find out how many moles there are in 75.0 g of C6H12O6.

The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is

75.0 g C6H12O6 ×

We will multiply the thermochemical equation, including the enthalpy change, by 0.416, in order to write the equation in terms of the appropriate amount of C6H12O6.

Given the thermochemical equation for photosynthesis,

6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g) ΔH = +2803 kJ/mol

calculate the solar energy required to produce 75.0 g of C6H12O6.

1 mol C6H12O6

180.2 g C6H12O6

= 0.416 mol C6H12O6


Solution (0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]

and (0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives

2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g) ΔH = +1.17×103 kJ

Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are canceled when we multiply the thermochemical equation by the number of moles of C6H12O6.

Think About It The specified amount of C6H12O6 is less than half a mole. Therefore, we should expect the associated enthalpy change to be less than half that specified in the thermochemical equation for the production of 1 mole of C6H12O6.

Heat Capacity, C

T

q

turein tempera increase

absorbedheat C

“C” is an extensive property; so a large object has a larger heat capacity than a small object made of the same material.

Using the Equation:

Looking at the figures on the left, it can be seen that the temperature change is constant, but the heat absorbed by the larger object is greater.

This results in a larger heat capacity for the larger object because more heat is absorbed.

……..of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1°C.

Specific heat capacity: The energy (joules) required to raise the temperature of 1 gram of substance by 1C

Unit: J g-1K-1 or J g-1C-1

Molar heat capacity: The energy (joules) required to raise the temperature of 1 mol of substance by 1C

Unit: J mol-1 K-1 or J mol-1C-1

m

CCs

n

CCm

SubstanceSpecific Heat, Cs

(cal/gram°C) (J/kg °C)

Pure water 1.00 4,186*

Wet mud 0.60 2,512

Ice (0 °C) 0.50 2,093

Sandy clay 0.33 1,381

Dry air (sea level) 0.24 1,005

Quartz sand 0.19 295

Granite 0.19 294

1 calorie = 4.186 joules

*The high heat capacity of water makes it ideal for storing heat in solar heating systems.

Neutralization

HClaq + NaOHaq NaClaq + H2O

The reaction between an acid and a basewhich results in a salt plus water.

Another example, cyanic acid and a hydroxide ion.

For example, hydrochloric acid and sodium hydroxide:

acid + base salt + water

Heat of Neutralization

Energy released by reaction = Energy absorbed by solution

Cs = q / [(mass) (Tfinal-Tinitial)]

Net ionic equation for neutralization:

H+(aq) + OH-(aq) H2O(l)

Specific heat capacity, Cs, is defined as the quantity of heat transferred, q, divided by the mass of the substance times the change in temperature. A value of Cs is specific to the given substance.

q = Cs (mass) (Tfinal-Tinitial)

This can then be rearranged to solve for the heat transferred.

Enthalpy of Fusion (Melting)Enthalpy of Fusion is defined as the heat that is absorbed when the melting occurs at constant pressure. If the substance freezes, the reaction is reversed, and an equal amount of heat is given off to the surroundings; i.e.,

ΔHfreez = - ΔHfus

Melting (fusion) is an endothermic process

solid liquid

)()( solidHliquidHH mmfus

Heat Capacity of Calorimeter

• To determine the heat capacity of coffee cup calorimeter:– A 25.0-g sample of warm water at 40.0oC was

added to a 25.0-g sample of water in a Styrofoam coffee cup calorimeter initially at 20.0oC. The final temperature of the mixed water and calorimeter was 29.5oC and the specific heat capacity of water is 4.184 J/g.oC. Calculate the heat capacity, Ch, of the calorimeter.

Specific Heat Capacity

• To determine the specific heat capacity of a metal using coffee cup calorimeter:– A 55.0-g sample of hot metal initially at 99.5oC

was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water and calorimeter were initially at 21.0oC. If the final temperature of mixture was 30.5oC, calculate the total heat lost by metal and the specific heat capacity of the metal. The specific heat of water is 4.184 J/(g.oC) and heat capacity of calorimeter is 10.0 J/oC.

Heat of Neutralization

• To determine the molar enthalpy of acid-base reaction using coffee cup calorimeter.– 50.0 mL of 2.0 M HCl was reacted with 50.0 mL of 2.0 M

NaOH in a coffee cup calorimeter. The reaction was exothermic, which caused the temperature of the solution to increase from 22.0oC to 35.6oC. Assume the density of solution as 1.0 g/mL, its specific heat capacity as 4.18 J/g.oC, and the heat capacity of calorimeter as 10.J/oC. Calculate the total amount of heat produced by the reaction. Calculate the enthalpy change (H, in kJ/mol) for the following reaction:

• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Calculation the enthalpy of reaction using Styrofoam cup calorimeter

• To determine the enthalpy of reaction using coffee cup calorimeter.– Suppose 100. mL of 1.0 M HCl solution is placed in a Styrofoam

coffee cup calorimeter. The initial temperature of HCl solution is 22.5oC. A 0.255-g sample of magnesium ribbon is cut to short pieces and added to the acid solution in which the following exothermic reaction occurred.

• Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

– The heat produced by the above reaction is completely absorbed by the solution and calorimeter, which attained the highest temperature of 34.2oC. Assume the acid solution has a density of 1.0 g/mL and its specific heat capacity as 4.0 J/g.oC, and the calorimeter has a heat capacity of 10. J/oC. Calculate the molar enthalpy change (H, in kJ/mol) for the above reaction.

Calculating the enthalpy of reaction using Styrofoam cup calorimeter

Reaction: Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g);

Calculations:

Heat Capacity of Calorimeter

• To determine the heat capacity of bomb calorimeter:– When a 1.200-g sample of glucose, C6H12O6, was

completely combusted in a bomb calorimeter, the temperature of the calorimeter assembly increased by 4.48oC. If the combustion of glucose produces 14.0 kJ/g of energy, how much heat energy is absorbed by the calorimeter. Calculate the heat capacity, Ch, of the calorimeter. (Assume that all of the heat produced by the combustion of glucose is absorbed by the calorimeter.)

Heat of Combustion

• To calculate the enthalpy of combustion using bomb calorimeter

– When a 1.010-g sample of sucrose (cane sugar) is completely combusted in a bomb calorimeter, the temperature of the calorimeter was increased by 4.50oC. If the heat capacity of calorimeter is 3.75 kJ/oC, how much heat was absorbed by the calorimeter? Calculate the molar enthalpy of combustion of sucrose according to the following equation:

• C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l)

Calorimetry

• Heat capacity- the quantity of heat needed to change the temperature of the system 1K– Cp = q/ T (units are J/k or J/oc)

• Specific Heat capacity – the quantity of heat need to raise the temperature of 1 gram of a substance 1 oC– q = C m T

Specific Heat

• The quantity of heat required to raise 1 gram of water 1oC.

• The higher the specific heat the harder it is to change the temperature of the substance. C (J/goC)– Al .092– Cu. .385– Ethanol 2.46– Water 4.182

Calorimetry

• Measures the amount of heat generated from a chemical reaction by letting the heat generated flow into a mass of cooler water.

Calorimetry is the measurement of heat changes.

Heat changes are measured in a device called a calorimeter

q= m T Cq = heat T = Tf – Ti m = mass C = specific heat

(J/g oC)

Specific Heat and Heat CapacitySpecific Heat and Heat Capacity

The heat capacity (C) is the amount of heat required to raise the temperature of an object by 1°C.

The “object” may be a given quantity of a particular substance.

Specific heat capacity has units of J/(g • °C)

Heat capacity has units of J/°C

4.184 Jheat capacity of 1 kg of water = 1000 g = 4184 J/ C

1 g C

Specific heat capacity of water heat capacity of1 kg water

Coffee Cup Calorimeter

• A coffee cup calorimeter is essentially a polystyrene (Styrofoam) cup with a lid.

• The cup is partially filled with a known volume of water and a thermometer is inserted through the lid of the cup so that its bulb is below the water surface.

• When a chemical reaction occurs in the coffee cup calorimeter, the heat of the reaction if absorbed by the water.

• The change in the water temperature is used to calculate the amount of heat that has been absorbed or evolved in the reaction.

For example,

a chemical reaction which occurs in 200 grams of water with an initial temperature of 25.0°C. The reaction is allowed to proceed in the coffee cup calorimeter. As a result of the reaction, the temperature of the water changes to 31.0°C. The heat flow is calculated:qwater = 4.18 J/(g·°C) x 200 g x (31.0°C - 25.0°C)

qwater = +5.0 x 103 J

ΔHreaction = -(qwater)

CalorimetryCalorimetry

Calculate the amount of heat required to heat 1.01 kg of water from 0.05°C to 35.81°C.

Solution:

Step 1: Use the equation q = msΔT to calculate q.

1000 g 4.184 J1.01 kg [35.81 C 0.05 C] = 151000 J = 151 kJ

1 kg g Cq


A coffee-cup calorimeter may be used to measure the heat exchange for a variety of reactions at constant pressure:

Heat of neutralization:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

Heat of ionization:

H2O(l) → H+(aq) + OH‒(aq)

Heat of fusion:

H2O(s) → H2O(l)

Heat of vaporization:

H2O(l) → H2O(g)


Concepts to consider for coffee-cup calorimetry:

qP = ΔH

System: reactants and products (the reaction)

Surroundings: water in the calorimeter

For an exothermic reaction:

the system loses heat

the surroundings gain (absorb) heat

qsys = −msΔT

The minus sign is used to keep sign conventions consistent.

qsurr = msΔT qsys = −qsurr

Worked Example 10.5

Strategy Water constitutes the surroundings; the pellet is the system. Use qsurr = msΔT to determine the heat absorbed by the water; then use q = CΔT to determine the heat capacity of the metal pellet.

mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C. The heat absorbed by the water must be released by the pellet: qwater = -qpellet, mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.

A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125 g of water originally at 25.1°C. The final temperature of both pellet and the water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.


Solution qwater =

Thus, qpellet = -3242.6 J

From q = CΔT we have

-3242.6 J = Cpellet × (-57.1°C)

Thus,Cpellet = 57 J/°C

4.184 Jg∙°C × 125 g × 6.2°C = 3242.6 J

Think About It The units cancel properly to give appropriate units for heat capacity. Moreover, ΔTpellet is a negative number because the temperature of the pellet decreases.

Constant-Volume CalorimetryConstant-Volume Calorimetry

Constant volume calorimetry is carried out in a device known as a constant-volume bomb.

qcal = −qrxn

A constant-volume calorimeter is an isolated system.

Bomb calorimeters are typically used to determine heats of combustion.

Constant-Volume CalorimetryConstant-Volume Calorimetry

qrxn = −CcalΔT

qcal = CcalΔT

qrxn = −qcal

To calculate qcal, the heat capacity of the calorimeter must be known.

Worked Example 10.6

Strategy Use qrxn = -CcalΔT to calculate the heat released by the combustion of the cookie. Divide the heat released by the mass of the cookie to determine its energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C.

A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of the cookie.

Solution qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ

Because energy content is a positive quantity, we write

energy content per gram = 1.559×102 kJ

7.25 g = 21.5 kJ/g

Think About It According to the label on the cookie package, a service size is four cookies, or 29 g, and each serving contains 150 Cal. Convert the energy per gram to Calories per serving to verify the result.

21.5 kJg ×

1 Cal4.184 kJ ×

29 gserving = 1.5×102 Cal/serving

Hess’s LawHess’s Law

Hess’s law states that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps.

CH4(g) + 2O2(g) ΔH = −890.4 kJ/molCO2(g) + 2H2O(l)


2H2O(l) ΔH = +88.0 kJ/mol2H2O(g)

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

ΔH = −890.4 kJ

ΔH = +88.0 kJ

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ

10.5

Bomb calorimetry is used to determine the enthalpy of combustion, H, for hydrocarbons:

Bomb Calorimeter

• In a coffee cup calorimeter, the reaction takes place in the water.

• In a bomb calorimeter, the reaction takes place in a sealed metal container, which is placed in the water in an insulated container.

• Heat flow from the reaction crosses the walls of the sealed container to the water.

• The temperature difference of the water is measured, just as it was for a coffee cup calorimeter.

Bomb Calorimeter

• Analysis of the heat flow is more complex than for the coffee cup calorimeter because the heat flow into the metal parts of the calorimeter must be accounted for (heat capacity, Cp):

• qreaction = - (qwater + qbomb)• where qwater = 4.18 J/(g·°C) x mwater x Δt• qbomb = Cp x Δt

Calorimetry

• A 1.5886 g sample of glucose (C6H12O6) was ignited in a bomb calorimeter. The temperature increased by 3.682oc. The heat capacity of the calorimeter was 3.56 kJ/oc, and the calorimeter contained 1.00 kg of water. Find the molar heat in kJ/molrxn

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Calorimetry

qbomb = 3.562 kJ/oc (3.682oc) = 13.12 kJqwater = 1000g (4.184 J/goc) (3.682oc) = 15400 JQtotal = qbomb + qwater

qtotal = - 28.52 kJ (exothermic)per molrxn

1.5886 g / 180.16g/mol = .0088177 mol-28.52 kJ/ .0088177 mol = -3234 kJ/molrxn

Enthalpy

• Enthalpy is a measure of the total energy of a thermodynamic system.

• Most chemistry reactions take place at constant pressure, open to the atmosphere.

So, Enthalpy (H) is used to describes these types of reactions.

H = U + PV

Enthalpy (H) – The Heat of the Reaction

• The change in enthalpy (H) is equal to the difference in the heat of the reaction.

H = Hfinal – Hinitial

= Hproducts – Hreactants

• Heat of formation (Hof) - the change in

enthalpy when a compound forms from its pure elements. (table in back of text book)

Hof of a pure element = 0

Enthalpy and Internal Energy

H = U + VP

• VP = ngasRT @ const. T and P

H = U + ngasRTU and H are very close to the same value

and are the same when no gas is generated by the reaction.

Hess’s Law

• .Example:1.What is the H for C + ½ O2 CO?

CO + ½ O2 CO2 H = -283.0 kJ

C + O2 CO2 H = -393.5 kJ

Hess’s law states that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps.

What is the H for C + ½ O2 CO?

CO2 CO + ½ O2 H = +283.0 kJ

C + O2 CO2 H = -393.5 kJ

C + ½ O2 CO H = -110.5 kJ

Hess’s LawHess’s Law

CH4(g) + 2O2(g) ΔH = −890.4 kJ/molCO2(g) + 2H2O(l)


2H2O(l) ΔH = +88.0 kJ/mol2H2O(g)

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

ΔH = −890.4 kJ

ΔH = +88.0 kJ

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ

10.5

2

Worked Example 10.7

Strategy Arrange the given thermochemical equations so that they sum to the desired equation. Make the corresponding changes to the enthalpy changes, and add them to get the desired enthalpy change.

Given the following thermochemical equations,

determine the enthalpy change for the reaction

NO(g) + O(g) → NO2(g)

NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol

O3(g) → O2(g) ΔH = –142.3 kJ/mol

O2(g) → 2O(g) ΔH = +495 kJ/mol

32


Solution The first equation has NO as a reactant with the correct coefficients, so we will use it as is.

The second equation must be reversed so that the O3 introduced by the first

equation will cancel (O3 is not part of the overall chemical equation). We also

must change the sign on the corresponding ΔH value.

These two steps sum to give:


O2(g) → O3(g) ΔH = +142.3 kJ/mol32

NO(g) + O3(g) → NO2(g) + O2(g)

O2(g) → O3(g)32

ΔH = –198.9 kJ/mol

ΔH = +142.3 kJ/mol

NO(g) + O2(g) → NO2(g) ΔH = –56.6 kJ/mol12

+ O2(g) 12


Solution We then replace O2 on the left with O by incorporating the last

equation. To do so, we divide the third equation by 2 and reverse its direction. As a result, we must also divide ΔH value by 2 and change its sign.

Finally, we sum all the steps and add their enthalpy changes.


O2(g) → O3(g) ΔH = +142.3 kJ/mol32

+

O(g) → O2(g) ΔH = –247.5 kJ/mol12

O(g) → O2(g) ΔH = –247.5 kJ/mol12

NO(g) + O(g) → NO2(g) ΔH = –304 kJ/mol

Think About It Double-check the cancellation of identical items–especially where fractions are involved.

12

Standard Enthalpies of FormationStandard Enthalpies of Formation

The standard enthalpy of formation (ΔH f°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.

Elements in standard states

C(graphite) + O2(g) CO2(g) ΔH f° = −393.5 kJ/mol

1 mole of product

10.6


The standard enthalpy of formation (ΔH f°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.

The superscripted degree sign denotes standard conditions.1 atm pressure for gases1 M concentration for solutions

“f” stands for formation.

ΔH f° for an element in its most stable form is zero.

ΔH f° for many substances are tabulated in Appendix 2 of the textbook.


The standard enthalpy of reaction (ΔH °rxn) is defined as the enthalpy of a reaction carried out under standard conditions.

aA + bB → cC + dD

n and m are the stoichiometric coefficients for the reactants and products.ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants)

ΔH °rxn = [cΔH f°(C) + dΔH f°(D) ] – [aΔH f°(A) + bΔH f°(B)]

Worked Example 10.8

Strategy Use ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants) and ΔH f° values from Appendix 2 to calculate ΔH °rxn. The ΔH f° values for Ag+(aq), Cl-(aq), and AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.

Using data from Appendix 2, calculate ΔH °rxn for Ag+(aq) + Cl-(aq) → AgCl(s).

Solution ΔH °rxn = ΔH f°(AgCl) – [ΔH f°(Ag+) + ΔH f°(Cl-)]

= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)]

= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/molThink About It Watch out for misplaced or missing minus signs. This is an easy place to lose track of them.

Worked Example 10.9

Strategy Arrange the equations that are provided so that they will sum to the desired equation. This may require reversing or multiplying one or more of the equations. For any such change, the corresponding change must also be made to the ΔH °rxn value. The desired equation, corresponding to the standard enthalpy of formation of acetylene, is

2C(graphite) + H2(g) → C2H2(g)

Given the following information, calculate the standard enthalpy of formation of

acetylene (C2H2) from its constituent elements:

C(graphite) + O2(g) → CO2(g) ΔH °rxn = –393.5 kJ/mol

H2(g) + O2(g) → H2O(l) ΔH °rxn = –285.5 kJ/mol

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ΔH °rxn = –2598.8 kJ/mol

12

(1)

(2)

(3)


Solution We multiply Equation (1) and its ΔH °rxn value by 2:

We include Equation (2) and its ΔH °rxn value as is:

We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):

Summing the resulting equations and the corresponding ΔH °rxn values:

2C(graphite) + 2O2(g) → 2CO2(g) ΔH °rxn = –787.0 kJ/mol


2CO2(g) + H2O(l) → C2H2(g) + O2(g) ΔH °rxn = +1299.4 kJ/mol

12

52

2C(graphite) + 2O2(g) → 2CO2(g) ΔH °rxn = –787.0 kJ/mol


2CO2(g) + H2O(l) → C2H2(g) + O2(g) ΔH °rxn = +1299.4 kJ/mol

12

52

2C(graphite) + H2(g) → C2H2(g) ΔH °f = +226.6 kJ/mol

Think About It Remember that a ΔH °rxn is only a ΔH °f when there is just one product, just one mole produced, and all the reactants are elements in their standard states.

Final 1 thermochemistry

Education

Transcript of Final 1 thermochemistry