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AP Chemistry 12 Thermochemistry Part I. 1st Law of Thermodynamics

Thermodynamics is the study of energy flow to and from systems. Most notable energy flow is in the form of heat and work. Thermochemistry is the branch of thermodynamics that studies energy flow within chemical systems and their reactions. System: the part of the universe that is the focus of attention. Surroundings: everything else in the universe. • Open systems allow transfer of matter and energy with surroundings while. • Closed systems only allow energy transfer. • Isolated systems allow no transfer of matter or energy. Energy: the capacity to do work or to produce heat. • Work (w): force acting over distance

o PV-work is the work done by (or done to) the system that results in a change in volume • Heat (q): transfer of energy between systems due to temperature difference Pathway: the conditions in which a reaction (or more generally, an event) occurs. Certain properties of a system depends on the pathway of the reaction. Properties that do not depend on the pathway are called state functions.

E.g., Volume, enthalpy, temperature, entropy, mass, density, #moles

State functions only depend on the current state of the system, not how the system became that way. Energy is a state function. • To calculate the kinetic energy of an object, we simply need to know its current mass and velocity (KE = ½

mv2) • To calculate the gravitational potential energy of an object, we only need to know its mass and height ( PE

= mgh ) Work (w) and Heat (q) are not state functions and depends on the pathway of the reaction.

Analogy: Although there may be multiple ways to go home, the amount of “work” you do to get home depends on the path you take. OR If a ball is to be lifted onto a table, the amount of work needed depends on how it is done, but the overall change in height will remain the same. Because state functions only depend on the current state, we can calculate the change in any state function by only considering the initial state and final state.

Ex. ΔT = Tf – Ti Ex. ΔV = Vf – Vi Ex. ΔS = Sproducts – Sreactants

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In reactions, we have seen the term enthalpy (H) used often. Enthalpy is related to the total energy of the system and is also a state function. We have seen examples of these in the Kinetics unit when we calculated ΔH for reactions.

Here are two proposed mechanism for overall reaction: Na+

(aq) + Cl-(aq) NaCl(s).

Since enthalpy is a state function, ΔH is the same for both mechanisms or “pathways”.

You cannot calculate non-state functions (path-dependent functions) in this way. Ex. To calculate the heat loss due to friction when a roller coaster travels on its tracks, we need to consider the heat loss at every moment, and not only the initial and final thermal state . Use of Δw or Δq does not really make sense since these are not state functions (“Δq”≠ qf – qi).

Example: Determine if the following are state functions.

o Height o Distance and Displacement o Time o Energy absorbed by a reaction

First Law of Thermodynamics

The First Law of Thermodynamics (“First Law”) is also known as the Law of Conservation of Energy. To understand the First Law, we need to define a new term: Internal Energy (U). Internal Energy (U) of a system is the sum of all types of energy in that system.

(i.e., U = PE + KE + heat + …) Any energy type you can think of is included in internal energy; for example, chemical potential energy (bonds), kinetic energy and thermal energy, and radiant (light) energy.

In Zumdahl and some other texts, E may be used for internal energy. However, U is the correct symbol and E is only used for convenience.

The First Law of Thermodynamics is stated as follows: The energy of the universe is constant.

We may have heard the phrase “energy can neither be created nor destroyed” before, and in fact, these really mean the same thing. If the energy in the entire universe is constant, it means that energy can only transfer from one system to another (or to the surroundings). Since energy transfers(ΔU) in terms of Work(w) or Heat(q), if we are only looking at a particular system, we observe this relationship:

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ΔU = q + w

This equation tells us that any changes in the internal energy of the system must be the result of one of two things (or both):

Heat (flow), q Work, w

Where q = heat added to the system w = work done on the system (i.e., work done by the surrounding on the system)

Remember that heat flow (q) is not the same as temperature(T) or thermal energy, the latter two being state functions (you can describe a system’s temperature or amount of thermal energy). Heat is the transfer of energy between two objects and hence is a process.

Heat (q) is normally associated with temperature in every day experiences. However, when heat flows into the system, it does not necessarily mean the temperature of the system will increase. The system can “use” this heat flow to increase its energy in other ways (for example, melting an ice cube at 0oC)

q is positive when heat flows into the system q is negative when heat flows out of the system

How do we measure q? As we soon will see, q is related to the change in enthalpy, ΔH, of a reaction.

Work (w) can come in many forms (electrical work, gravitational work, magnetic work, as a few examples). The type of work that is most relevant to chemical systems is pressure-volume work (mechanical work).

Imagine a rigid steel cylinder containing gas molecules. On the top side of the cylinder is a movable piston that allows changes in gas volume. The piston is held down by external pressure, either by atmospheric gas pressure

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or by some mass.

When the gas expands and pushes against the piston, increasing its volume, we say that the system is doing work on its surroundings. If the gas compresses and the piston lowers, decreasing the gas volume, we say that work is being done on the system

w is positive when work is being done on the system (i.e., compression) When work is done onto the system, the system gains energy and the internal energy increases

w is negative when work is done by the system (i.e., expansion) When work is done by the system, the system loses energy and the internal energy decreases

In mechanical engineering (and other disciplines in physics), it is common to swap the convention (this results in the equation becoming

ΔU = q - w’

Using ΔU = q + w,

Work is defined as force(F) acting over a distance(d), but how does this relate to expanding gases?

W= force x distance = Fd = -Pexternal ΔV = -Pext(Vf - Vi)

In expanding (or compressing) gases, work is equal to the volume change against an external pressure (not the pressure of the gas itself).

Derivation p.216 Zumdahl Notice that the negative sign is necessary to make it agree with how we have defined work: When a gas is expanding in the system • ΔV is positive for the system • But work is negative for the system. • There must be negative sign for Pexternal

Note on Units for Work

ΔU = q + w w= - Pext ΔV

U and q are measured in joules (or kJ) P and V are commonly measured in units of atm and L Need conversion factor for work.

Conversion factor: 101.3J / L atm (not included in the formula sheet, but will likely be provided.)

Summary of First Law:

ΔU = q + w w= - Pext ΔV

q is heat flow (positive when heat flows into the system) w is work, usually pressure-volume work (negative when gas expands)

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Example 1: A system is undergoing an endothermic process where 30.0 kJ of heat flows into the system and 5.0 kJ of work is done on the system. Calculate ΔU.

Example 2: When 0.10 moles of benzene (C6H6) combusts, 327.0 kJ of heat is released. If ΔU is 227.0 kJ, did the container expand or compress?

Example 3: Calculate the work (both in L*atm and J) associated when 13.0 L of gas expands to 63.0 L against an external pressure of 15.0 atm. Draw initial and final states of the scenario. Example 4: When a gas at 5.0 atm is compressed by an external pressure of 10.0 atm from 30.0 L to 15.0 L, 12.0 kJ of heat is also released. What is ΔU for this process? Draw initial and final states of the scenario, labelling the direction of heat flow and work.

àQuiz 1

Enthalpy, Specific Heat, Calorimetry, and Phase Changes

The definition of enthalpy is the sum of the internal energy of the system and PV-work required to create the system .

H = U + PV How are we supposed to measure enthalpy? At constant pressure (denoted by subscript p):

ΔU = qp + w and w= - Pext ΔV Combining the two to get

qp = ΔU + Pext ΔV

Since H = U + PV, ΔH = ΔU + Δ(PV)

At constant P,

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ΔH = ΔU + PΔV Therefore,

qp = ΔU + PΔV = ΔH and

ΔH = qp The change in enthalpy (ΔH ) is defined as equal to the heat flow (qp ) at constant pressure. Most reactions are done in open atmosphere, meaning that the pressure remains constant!

If we could measure the heat flow (qp) of a reaction, we could determine the enthalpy change (ΔH) for that particular reaction. Recall enthalpy is a state function.

ΔH = qp = Hf - Hi = Hproduct - Hreactant

The unit of enthalpy (H) is the same as the unit of heat flow, kJ (or J). However, changes in enthalpies (ΔH) are often shown in tables as kJ/mol. This is because ΔH depends on the number of moles reactant (extensive property).

For example: When 1.0 mole of hydrogen gas combusts, 290 kJ of heat is released When 2.0 moles of hydrogen gas combust, 580 kJ of heat is released

In both cases, the change in enthalpy is different, but the change in enthalpy per mole is the same.

Example 1: When 12.0 grams of ethane (C2H6) is combusted at constant pressure, 625 kJ of heat is released. Determine the change in enthalpy (ΔH) for this process in kJ. Calculate the enthalpy of combustion (ΔHcomb) in kJ/mol, including the appropriate sign.

Many ΔH values for specific reactions are tabulated. Specific types of reactions have their own name, so these enthalpies are named as such. Ex. Enthalpy of combustion, enthalpy of neutralization, enthalpy of condensation, or simply enthalpy of reaction… Since enthalpy and heat are equal at constant pressure, sometimes these two terms are interchanged.

Reaction Change in Enthalpy

H+ + OH- à H2O(l) ΔH neutralization = -57.62kJ/mol

H2(g) + ½ O2(g) à H2O(l) ΔH formation = -285.8kJ/mol

H2(g) + ½ O2(g) à H2O(l) ΔH combustion= -285.8kJ/mol

H2O(s) à H2O(l) ΔH fusion= +6.005kJ/mol

H2O(l) à H2O(g) ΔH vapourization= +40.66kJ/mol

NaCl(s) à Na+ + Cl- ΔH dissolution= +3.87kJ/mol

H2(g) + F2(g) à 2HF(g) ΔH reaction= -271kJ/mol

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Formation Reactions Formation reactions are particularly important because most of the thermodynamic data that we have is on this type of reaction. Formation reactions are reactions in which a particular compound is formed from its elements at standard state (symbol of o is used, for example, ΔHf

o). Recall that standard state means: • Elements are in its reference form; state is determined at 25oC • All pressures of gases are at 105 Pa = 1 bar (This is relatively newer… most old texts use 1.01325 x 105 Pa = 1 atm) • All concentrations of solutions are at 1 M • All gases and solutions are ideal Although no temperature is specified at standard state, 25oC = 298.15 K is often used.

To write a formation reaction, we have to determine:

• What are the elements in the species we are forming. • What are the reference states of these elements.

o Most metals (s) except for Hg (l) o Most non-metals (g) except for Br2 (s) and S, P, B, I2 (s) and C (s, graphite)

• Balance the reaction so that the species formed has a coefficient of 1. The formation reaction of CO2(g): The elements are carbon and oxygen. Carbon is a solid (graphite) in its reference form. Oxygen is a diatomic gas in its reference form. It is already balanced, and coefficient of CO2 is 1. Looking up data, the enthalpy of formation of CO2:

ΔHfo

(CO2(g)) = -393.5kJ/mol The formation reaction of CH3COOH(l) The elements are carbon, hydrogen, and oxygen. Carbon is a solid (graphite), oxygen is a diatomic gas, hydrogen is a diatomic gas.

Balance as normal:

Coefficient of CH3COOH is 1. Looking up the enthalpy change, ΔHf

o(CH3COOH(l)) = -484.3kJ/mol

The formation reaction of HF(g)

H2(g) + F2(g) à 2HF(g) ½ H2(g) + ½ F2(g) à HF(g) ΔHf

o(HF(g)) = -273.3kJ/mol

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The formation reaction of C(s, diamond) C(s, graphite) à C(s, diamond)

ΔHfo

(C(s, diamond)) = +1.9 kJ/mol

The enthalpy of formation of an element already in its reference state must be zero. Example: What is formation reaction and enthalpy of formation of H2(g)?

H2(g) à H2(g) Since the reactants and products are the same, their energy levels (and thus enthalpy) must be the same.

ΔHfo

(H2(g)) = 0kJ/mol

Examples: write the formation reaction for the following species: N2H4(g) H2SO4(s) (sulfur’s reference state is S8(s)) Cl(g) Br2(l) NH4NO3(s) O3(g) CaCO3(s)

Other Named Reactions Reactions involving phase changes or other extremely common reactions also have their enthalpies tabulated. Here are some examples you may see in the future.

à Quiz 2

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Heat Flow, Heat Capacity, Calorimetry, Phase Changes Now that we have a lot of the theory down… how is heat flow measured and determined experimentally? Recall we derived ΔH = qp at a constant pressure When heat flows into a system, generally, one of two things happen: 1. The system’s temperature increases 2. The temperature does not increase, and the energy is used to change other properties (eg. phase change)

Case 1. The system’s temperature increases due to heat flow. The temperature change will depend on three factors: • Amount of heat flow, q • Amount or mass of substance heated/cooled, m • Heat capacity of substance, Cp The heat capacity (Cp) is the amount of heat (in J) required to raise the temperature of a substance by 1 K (or 1 oC). Since having more substances means more heat is required, it is more common to use specific heat capacity (in J/g Co or J/g K) or molar heat capacity (in J/mol Co or J/mol K) Water has the highest heat capacity of any liquid (it takes the most energy to heat up water than other liquids). It takes 4.184 J of energy to raise the temperature of 1 gram of water by 1oC.

Specific heat capacity of H2O(l) = 4.184 J/gCo or J/gK Molar heat capacity of H2O(l) = 75.31J/molCo or J/molK

Here is the equation that relates heat and temperature change: qp = mCpΔT (= ΔH )

qp = heat flow (at constant pressure) m = the mass of the substance Cp = the heat capacity (specific heat capacity or molar heat capacity) ΔT = the change in temperature

Ensure to match the unit for mass and temperature.

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Example 1. Determine the amount of heat required to increase 100. mL of water from 25.0oC to 100.0oC (without boiling). The density of water is 1.00g/mL. Look up the appropriate heat capacity.

31.4 kJ

Example 2. A 3.00 kg piece of copper metal is heated from 300.K to 600.K. How much heat, in kJ, is absorbed by the copper? (Cp,Cu = 0.384 J/goC) Example 3. 5.0 moles of ice is cooled from 273 K to 200 K. How much heat is released during the cooling? (Cp,ice= 2.06 J/goC)

Calorimetry (Constant Pressure) Calorimetry is the experimental process used to determine the quantity of heat flow as a reaction progresses. A calorimeter is a device used to create a closed system (or an isolated), allowing us to measure the quantity of heat flow (q). In a calorimeter, the heat produced or absorbed by the reaction affects the temperature of the water holding the reaction. We measure the temperature of the water to determine the heat flow (using q = mCΔT) Recall that at constant pressure,

qp = ΔH The signs between qrxn and qwater:

any amount of heat the reaction produces goes into the water, and any amount of heat the reaction absorbs comes from the water. Hence

ΔHrxn = qrxn = - qwater = - mwCwΔTw Double check the sign of ΔHrxn by visualizing the process. (Exothermic = temperature increases = ΔHrxn is negative) Example 1. When 100.0 mL of 1.0 M HI is mixed with 100.0 mL of 1.0 M NaOH in a calorimeter, the temperature of the mixture solution increases from 25.0oC to 35.5oC. Calculate the heat of reaction of salt, ΔHrxn (in kJ and kJ/mol of salt). Assume the total volume of solution is 200.0 mL, and the density and specific heat capacity of the resulting solution is the same as pure water, 1.00g/mL and 4.184 J/goC.

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Example 2. When 4.50 grams of NaOH is dissolved in 200.0 mL of water at 30.0oC in a calorimeter, the temperature of the mixture raises to 36.2oC. (Cp = 4.184 J/goC, d =1.00g/mL). Determine the heat of reaction of NaOH , ΔHrxn in J/g and kJ/mol of NaOH. Example 3. The burning of a 50.0 g candle (C30H62) releases heat onto a 100.0 g copper jar, containing 300.0 mL of water, both initially at 20.0oC. When the mass of the candle is 49.2 g, the burning is stopped and the temperature of the water (and copper) is measured to be 48.0oC. Calculate the heat of combustion (ΔHcomb) in kJ/mol. (dwater = 1.00g/mL, Cp, water = 4.184 J/goC, Cp,copper= 0.385 J/goC) (Hint: ΔHcomb = qrxn = -qp = -(qCu + qH2O)

Summary Case 1. The system’s temperature increases due to heat flow. The temperature change will depend on three factors: • Amount of heat flow, q • Amount or mass of substance heated/cooled, m • Heat capacity of substance, Cp

ΔHrxn can be calculated by using qp = mCpΔT

àQuiz 3

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Recall that when heat flows into a system, generally, one of two things happen: 1. The system’s temperature increases 2. Its temperature does not increase, and the heat is used to change other properties During phase changes, heat that enters or exits the system does not cause a change in temperature. When ice melts, heat that enters the system is used to rearrange molecules, and break intermolecular forces, allowing molecules to move more freely. When all the ice have melted completely, additional heat will cause temperatures to rise.

Review a heating curve of water (temperature vs heat added) below.

“fus” = fusion = older term for melting “vap”= vapourization = boiling

In general, there are five phases for a normal heating (or cooling) curve • Solid phase • Solid-Liquid equilibrium phase • Liquid phase • Liquid-Gas equilibrium phase • Gas phase

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During the equilibrium phase (s ⇄l, l ⇄g), temperature does not change. Hence, we cannot use ΔH= qp = mCpΔT .We calculate the heat required to completely change the phase.

ΔH = qp = nΔHfus or vap

Notice that because ΔHfus and ΔHvap are both in kJ per mol, to determine the overall heat required, we need to multiply by the number of moles of substance, n. Be mindful of the sign for ΔH!

Melting and boiling = exothermic and positive ΔH Freezing and condensation = endothermic and negative ΔH

Example 1. How much heat does 100.0 g of ice require to change completely to water at 0oC? (ΔHfus = 6.005 kJ/mol)

33.33kJ Check the sign!

Example 2. How much heat is released when 30.0 g of NH3(g) completely condenses into NH3(l)? (ΔHvap = 23.35 kJ/mol)

-41.1 kJ Check the sign!

Let us consider double phase changes! Consider heating 50.0 g of H2O from ice at -40oC to water vapour at 130oC. To calculate the total amount of heat required, we need to make separate calculations

• Solid phase (q1 = mCiceΔT) • Solid-Liquid equilibrium phase (q2 = nΔHo

fus) • Liquid phase (q3 = mCwaterΔT) • Liquid-Gas equilibrium phase (q4 = nΔHo

vap) • Gas phase (q5 = mCw.vapΔT)

We will need to use the phase change enthalpies and heat capacities for each phase:

Boiling Point: 100oC Melting Point: 0oC ΔHvap: 40.66 kJ/mol ΔHfus: 6.005 kJ/mol Cp,ice = 2.06 J/goC Cp,water = 4.184 J/goC Cp,w.vap = 2.03 J/goC

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qtotal = q1+q2+q3+q4+q5 = • Solid phase (q1 = mCiceΔT = • Solid-Liquid equilibrium phase (q2 = nΔHo

fus = • Liquid phase (q3 = mCwaterΔT = • Liquid-Gas equilibrium phase (q4 = nΔHo

vap = • Gas phase (q5 = mCw.vapΔT =

Common mistakes:

Trying to calculate qtotal = 50.0g(2.06 J/goC)(130oC-(-40oC)) Forgetting that each phase has a different heat capacity

Example 1. Calculating the total heat released when 10.0 grams of gaseous ethanol (C2H5OH) at 500 K is cooled to solid ethanol at 100 K. Draw a cooling curve for this process.

Boiling Point: 352 K Melting Point: 159 K ΔHvap = 42.3 kJ/mol ΔHfus = 4.90 kJ/mol Cp,s = 2.42 J/goC Cp,l = 2.44 J/goC Cp,g = 1.90 J/goC

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Another way to represent phase change is by using a phase diagram. It is commonly known that water boils at 100oC and freezes at 0oC. However, boiling points and melting points depend on the pressure! Following phase diagrams show what state a substance is in at a particular pressure and temperature.

At different pressures, these transitions happen at different temperatures The normal boiling point and normal melting point are the temperatures where the substance changes phases at 1.00 atm. At different pressures, these transitions happen at different temperatures. If we move across the phase diagram at a constant pressure and draw a horizontal line, that is called an isobar. Likewise, if we move up or down the phase diagram at a constant temperature and draw a vertical line, that is called an isotherm. Moving along an isobar or isotherm tells us which phases we expect to see that those pressures and temperatures.

The triple point is the point where all three phases coexist (0.06 atm and 0.01oC for H2O). The critical point is the point where the liquid and gaseous phase are indistinguishable (supercritical fluid). Gas cannot be liquefied with increasing pressure in a supercritical fluid.

For CO2, at atmospheric pressure, solid carbon dioxide sublimates (solid à gas) at -78.5oC. CO2 does not first melt, and then vapourize at 1 atm Example 1. At what temperature and pressure do all three phases of CO2 coexist? Example 2. Between which two pressures can an equilibrium between a liquid and gas form for CO2?

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à Quiz 4

Bond Dissociation Energy and Hess’s Law So far, we have used calorimetry to determine the enthalpy of reaction.

1. ΔH= qp = mCpΔT There are additional methods to determine the enthalpy of reaction 2. Using Bond Dissociation Energy 3. Hess’s Law 4. ΔHf

o (Enthalpy of formation) 2. Bond Dissociation Energy

When reactant molecules collide to react and form products, the reactant bonds must be broken (requires energy, ΔH>0) and product bonds must be formed (releases energy, ΔH<0).

ΔHrxn = Hproduct - Hreactant = BDEreactant + BDEproduct = (Positive) + (Negative)

For example, consider the reaction between H2 and Cl2 to form HCl (all gases)

H2(g) + Cl2(g) à 2HCl(g) Which and how many bonds must be broken/formed?

First, consider Lewis Diagram version of the reaction. Separating each steps,

Overall: H2(g) + Cl2(g) à 2HCl(g), ΔH= ?? Lewis Overall: H-H + Cl-Cl à H-Cl+ H-Cl

Energy required to break bonds:

H-H: 436 kJ/mol Cl-Cl: 243 kJ/mol Total =

Energy released when bonds form: H-Cl: 2 * 432 kJ/mol Total =

ΔHrxn = BDEreactant + BDEproduct = 679 + (-864)kJ = _________. Hence this reaction is exothermic and releases energy.

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Example: Calculate ΔHrxn.

2CO2 à O2 + 2CO Lewis Overall is:

Energy required to break bonds:

C=O: 4 * 732 kJ/mol Total =

Energy released when bonds form: O=O: 1 * 498 kJ/mol C ≡O: 2* 1079kJ / mol Total =

ΔHrxn = BDEreactant + BDEproduct = 2928 + (-2656)kJ = _________. Hence, this reaction is endothermic and requires energy input. Example: Calculate ΔHrxn.

C4H9Br + H2 à C4H10 + HBr If we draw the structure and calculate all the bonds in the reactant and products, that would take a long time. What strategy should you take?

Energy required to break bonds:

Energy released when bonds form:

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Examples: Calculate ΔHrxn for the following using bond dissociation energy:

1. HI decomposes into its molecular form.

2. Synthesis of water from its molecular form.

3. CH4 + 2Br2 à CBr4 + 2H2.

4. Reacting 1-decene and chlorine gas to make 1,2-dichlorodecane.

5. Given that the following reaction is exothermic and releases 137 kJ of energy, what is the BDE of C≡N? (BDE(H-H)= 436kJ/mol, BDE(C-H)= 410kJ/mol, BDE(N-H)=390 kJ/mol, BDE(C-N)=300kJ/mol).

HCN + 2H2 à CH3NH2

Methods covered so far to determine enthalpy of reaction (ΔHrxn): 1. ΔH= qp = mCpΔT 2. ΔHrxn = BDEreactant + BDEproduct 3. Hess’s Law 4. ΔHf

o (Enthalpy of formation)

3. Hess’s Law In the Kinetics Chapter, we added reaction “steps” (mechanism) to obtain an overall reaction. In the Equilibrium Chapter, we were also able to combine individual, sequential equilibrium reactions and calculate the equilibrium constant for the overall reaction. Hess’s Law is similar: The total enthalpy change during a chemical reaction is the same whether the reaction is one step or multistep. After all, enthalpy change is a state function and the “middle” pathway does not affect the overall enthalpy change.

ΔHrxn = ΔH1+ΔH2+ΔH3+…

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Why not just measure the heat of reaction for the overall (ΔH), rather than finding individual ΔH? Hess’s Law is often used to determine the heat of reaction for extremely slow reactions. Individual reactions might all be relatively faster, or already studied well.

A few reminders before we continue:

When we reverse a reaction, the sign of the ΔH value is flipped. If A à B (ΔH = -30 kJ) Then B à A (ΔH = +30 kJ)

If we multiply a reaction by any number (to change coefficients), the value of the ΔH is also multiplied

If C à 2 D (ΔH = +20 kJ) Then 2 C à 4D (ΔH = +40 kJ) And also ½ C à D (ΔH = +10 kJ)

If we add two reactions together, we add their ΔH values

If A à B (ΔH = -30 kJ) , B à C (ΔH = +80 kJ) Then A à C (ΔH = +50 kJ) And also 2 A à B+ C (ΔH = ??? kJ)

Example 1. Calculate ΔHrxn

o for the following reaction with the given information:

Overall: C(s, graphite) + 2H2(g) à CH4(g)

C(s, graphite) + O2(g) à CO2(g) , ΔHo= - 393.5kJ H2(g) + ½ O2(g) à H2O(l), ΔHo= - 285.8kJ

CH4(g) + 2O2(g) à CO2(g) + 2H2O(l), ΔHo= - 890.3kJ Example 2. Calculate ΔHrxn

o for the following reaction with the given information:

Overall: N2(g) + 2O2(g) à N2O4(g)

N2(g) + O2(g) à 2NO(g), ΔHo= + 180kJ 2NO(g) + O2(g) à 2NO2(g), ΔHo= - 112kJ

N2O4(g) à 2NO2(g) , ΔHo= + 58kJ

Methods covered so far to determine enthalpy of reaction (ΔHrxn): 1. ΔH= qp = mCpΔT 2. ΔHrxn = BDEreactant + BDEproduct 3. ΔHrxn = ΔH1+ΔH2+ΔH3+… 4. ΔHf

o (Enthalpy of formation)

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4. Enthalpy of Formation

We have already seen formation reactions and enthalpy of formation. Recall that many formation reactions and their ΔH values are tabulated. Using enthalpy of formation to determine overall enthalpy of reaction is a special case of Hess’s Law.

Calculate the ΔHrxn:

C2H4(g) + HCl(g) à C2H5Cl(g)

Knowing this is a special case of Hess’s Law, it is not necessary to check that sequential reactions lead to overall.

Consider the unbalanced reaction and Calculate the ΔHrxn

C2H4(g) + O2(g) à CO2(g) + H2O(l)

Summary of Change in Enthalpy

4 Methods to calculate ΔHrxn: 1. ΔH= qp = mCpΔT, used for a reaction with known temperature change or phase change. 2. ΔHrxn = BDEreactant + BDEproduct, used for a simple reaction with known BDE. 3. ΔHrxn = ΔH1+ΔH2+ΔH3+…, used for a sequential reactions. 4. ΔHf

o (Enthalpy of formation), used for a reaction with known formation enthalpies.

à Quiz 5

ΔHrxno = ΔHproduct

o∑ − ΔHreact...o∑