Ch. 6 Energy Relationships and Chemical Reactions

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Energy

The capacity to do work (exerting force over distance)

Work

Directed energy change resulting from a process. (not a state function)
w = -P∆V
P = F/d²
V = d³
- in a vacuum, p=0 and therefore w=0

Types of energy:

Kinetic energy- energy of motion
Radiant energy- solar energy
Thermal energy- energy associated with random motion of atoms and molecules; more movement=more energy
Chemical energy- stored within structural units of chemical substances; when chem. reaction occurs, energy is released, stored, or converted to another form of energy
Potential energy- energy available because of an objects position, ex: rock on a cliff edge

Law of Conservation of Energy

The total quantity of energy in the universe is assumed constant; as energy is used, it converts to different forms all the time

Heat

The transfer of thermal energy between two bodies that are at different temperatures; nearly all chemical reactions absorb or produce (release) energy in the form of heat; q (not a state function)

Thermochemistry

the study of heat change in chemical reactions

What is a system?

The specific part of the universe that is of interest to us.

The 3 types of systems are:

Open system- can exchange mass and energy, usually in the form of heat with its surroundings (ex: water in an open container)
Closed system- allows the transfer of energy(heat), but not mass (ex: closed container)
Isolated system- does not allow the transfer of either mass or energy (ex:insulated, closed container)

Explain system/surroundings relationship as it relates to chemical reactions.

In a chemical reaction, the reacting mixture would be the system and the rest of the universe the surroundings. Because energy cannot be created or destroyed, energy lost by the system must be gained by the surroundings (exothermic). Conversely, in some reactions, heat can be supplied to the system by the surroundings (endothermic).

Exothermic process

any process that gives off heat- transfers energy to the surroundings
(ex: combustion of hydrogen gas in oxygen)
2 H₂(g) + O₂(g) → 2 H₂O(l) + energy

Endothermic process

heat has to be supplied to the system by the surroundings
(ex: decomposition of mercury (II) oxide at high temp)
energy + 2 HgO(s) → 2 Hg(l) + O₂(g)

Thermodynamics

scientific study of the interconversion of heat and other kinds of energy

What determines the state of a system?

the values of all relevant macroscopic properties (composition, energy, temperature, pressure, and volume)

What are state functions?

properties that are determined by the state of the system, regardless of how that condition was achieved
(energy, temperature, pressure, volume)- when system state changes, the magnitude of change in any state function depends only on the initial and final states of the system and not on how the change is accomplished
ex: ∆V = Vf - Vi

Define the First Law of Thermodynamics:

energy can be converted from one form to another, but cannot be created or destroyed

Explain the two components of a system's internal energy:

Kinetic energy- various types of molecular motion and the movement of electrons within molecules
Potential energy- determined by attractive interactions between electrons and nuclei and by repulsive interactions between electrons and nuclei in individual molecules, as well as by interaction between molecules

To determine changes in internal energy...

If the reaction gives off heat (releases energy), then the change in internal energy change (∆U) is negative. If reaction absorbs heat, internal energy change is positive.
Because of 1st Law of Thmdcs, if energy is absorbed/released by the system, then energy is equally and oppositely released/absorbed by the surroundings. ∆Usys = -∆Usurr
In determining energy changes in the system, we want to calculate the sum of the heat exchange (q) between the sys and surr and the work done (w) on or by the system.
∆Usys = q + w
- q is positive for endothermic process, negative for exothermic
- w is positive for work done on the sys by the surr, negative for work done by sys on surr
--- q and w are NOT state functions

1 L atm

101.3 J

ex: A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands a) against a vacuum and b) against a constant pressure at 1.2 atm.

a) w = 0
b) w = - P ∆V = -1.2atm * 4.0 L = -4.8 Latm =
-4.9 x 10² J

ex: The work done when a gas is compressed in a cylinder is 387 J. During this process, there is a heat transfer of 152 J from the gas to the surroundings. Calculate the energy change for this process.

∆U = q + w
∆U = 387 J + -152 J = 235 J
q is positive bc work was done on the gas (system)
w is negative bc heat was released to the surroundings

Explain gas expansion/compression when pressure is constant:

If pressure remains constant and the reaction results in a net increase in the number of moles of a gas, then the system does work on the surroundings by expanding.
If more gas molecules are consumed than produced, while pressure remains constant, work is done on the system by the surroundings through compression.
If there is no net change in number of molecules then no work is done.
** at constant pressure: q = ∆U + P∆V

What is enthalpy?

Quantity equivalent to the total heat content of a system.
It is equal to the internal energy of the system plus the product of pressure and volume.
-Enthalpy (H) is a state function
H = U + PV
∆H = ∆U + (∆)P(∆)V

** if there is either constant volume or constant pressure, q = ∆H = U + (∆)P(∆)V

How do you find the enthalpy of a reaction?

∆H = H(products) - H(reactants)
-For endothermic processes, reaction enthalpy is positive because heat is absorbed by the system.
-For exothermic processes, reaction enthalpy is negative because heat is released to the surroundings.

Thermochemical equations

show the enthalpy changes as well as the mass relationships

Steps in writing thermochemical equations:

1. Always specify the physical states of all reactants and products (otherwise it could result in different enthalpy changes)
2. If both sides of the equation are multiplied by 'n,' then ∆H must be multiplied by 'n'
3. When reversing an equation, the roles of the reactants and products change so the magnitude of ∆H is the same but the sign changes.
- by reversing the eq, the process changes from endothermic to exothermic or exothermic to endothermic

ex: Given the thermochemical equation
2 SO₂(g) + O₂(g) → 2 SO₃(g)
∆H = -198.2 kJ/mol
calculate the heat evolved when 93.4g SO₂ (molar mass = 64.07 g/mol) is converted to SO₃.

∆H = 93.4 g SO₂ ⋅ ( 1 mol SO₂/64.07 g SO₂) ⋅ (-198.2 kJ/ 2 mol SO₂) = -144 kJ

∴ The heat released to the surroundings is 144kJ. (exothermic)

What is the relationship between ∆H and ∆U for a process?

In a gas, ∆H is slightly smaller than ∆U because when a chemical reaction occurs, some of the internal energy is used to do gas expansion work.
Another way to calculate ∆U in a gaseous reaction:
∆U = ∆H - RT∆n
∆n = moles product gases - moles reactant gases

ex: Calculate the change in internal energy when 2 moles CO are converted to 2 moles CO₂ at 1atm and 25°C:
2 CO(g) + O₂(g) → 2 CO₂(g)
∆H = -566.0 kJ/mol

∆U = (-566.0 kJ/mol) - (8.314 J/K)(1 kJ/1000 J)(298 K)(-1)
∆U = -563.5 kJ/mol

What is ∆U for the formation of 1 mole of CO at 1atm and 25°C?
C(graphite) + (1/2) O₂(g) → CO(g)
∆H = -110.5 kJ/mol

∆n = 1 - 1/2
(∆n only considers moles of gaseous reactants and products)

∆U = (-110.5 kJ/mol) - (8.314 J/K)(1 kJ/1000 J)(298 K)(1/2)
∆U = -111.7 kJ/mol

What is calorimetry?

The measure of heat changes.

What is specific heat? (s)

the amount of heat required to raise the temperature of one gram of the substance by 1°C.
-intensive property

What is heat capacity? (C)

the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celcius.
-extensive property

What is the relationship between specific heat and heat capacity?

C = ms

C = heat capacity (J/°C)
m = mass of substance in grams
s = specific heat (J/g ⋅ °C)

What are the equations for calculating heat change?

q = ms∆t
q = C∆t

(∆t = t(final) - t(initial))
(∆t is the change in the sample's temp)
(q is the amount of heat released (-, exothermic) or absorbed (+, endothermic) in a particular process)

ex: A 394g sample of water is heated from 10.75°C to 83.20°C. Calculate the amount of heat absorbed (in kJ) by the water.

m = 394 g
s = 4.184 J/g⋅°C
∆t = 83.20 - 10.75 = 72.45°C
q = (394)(4.184)(1/1000)(72.45)
q = 119 kJ
(+, so process is endothermic)

What is the constant-pressure calorimeter used for?

-isolated system used to determine the heat changes for non-combustion reactions.
-when pressure is constant, heat change for the process (q(rxn)) is equal to the enthalpy change (∆H).

ex: A lead (Pb) pellet having a mass of 26.47g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temp. rose from 22.50°C to 23.17°C. What is the specific heat of the lead pellet?

q(Pb) + q(H₂O) = 0
q(H₂O) = ms∆t = (100.0g)(4.184J/g⋅°C)(23.17°C-22.50°C) = 280.3 g
q(Pb) = -280.3 g = ms∆t
-280.3 g = (26.47 g)(s)(23.17°C-89.98°C)
s = 0.158 J/g⋅°C

ex: A quantity of 1.50 x 10² mL of 0.350 M HCl was mixed with 1.50 x 10² mL of 0.350 M NaOH in a constant pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 23.25°C, and the final temperature of the mixed solution was 25.60°C. Calculate the heat change for the neutralization reaction on a molar basis
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
Assume that the densities and specific heats of the solutions are the same as for the water (1.00g/mL and 4.184 J/g⋅°C respectively).

q(sys) = q(soln) + q(rxn) = 0
q(rxn) = -q(soln)
q(soln) = ms∆t = (1.50 x 10²g + 1.50 x 10²g)(4.184J/g⋅°C)(25.60°C-23.25°C) = 2.95 x 10³ J
q(soln) = 2.95 kJ
q(rxn) = -2.95 kJ

(0.350 mol/ 1L) x 0.150 L = 0.0525 mol

heat of neutralization = q(rxn)/moles of soln
= (-2.95)/(0.0525) = -56.2kJ/mol

What is standard enthalpy of formation?

∆H°ƒ - the heat change that results when 1 mole of the compound is formed from its elements at a pressure of 1 atm
-ground reference point for all enthalpy expressions.
-standard state conditions: 1atm (usually at 25°C)
-the standard enthalpy of formation of any element in its most stable form is zero
ex: O₂ is the most stable form oxygen can stake so
∆H°ƒ = 0 (while the ∆H°ƒ for O₃ = 142.2 kJ/mol and for O = 249.4 kJ/mol)

How do you calculate the standard enthalpy of reaction?

∆H°(rxn) can be calculated through this method:
if the reaction is represented as
aA + bB → cC + dD
∆H°(rxn) = [c∆H°ƒ(C) + d∆H°ƒ(D)] - [a∆H°ƒ(A) + b∆H°ƒ(B)]
or
∆H°(rxn) = ∑n∆H°ƒ(products) - ∑m∆H°ƒ(reactants)

How do you determine the value of ∆H°ƒ using the direct method?

Used when compounds in the reaction are in their most stable form,

ex: C(graphite) + O₂(g) → CO₂(g)
∆H°(rxn) = -393.5 kJ/mol
∆H°(rxn) = ∆H°ƒ(CO₂, g) - [∆H°ƒ(C, graphite) +
∆H°ƒ(O₂, g)]
>>grafite and O₂ are in their stable forms, so their ∆H°ƒ = 0

∴ ∆H°(rxn) = ∆H°ƒ(CO₂, g) = -393.5 kJ/mol

How do you determine the value of ∆H°ƒ using the indirect method?

- Using Hess's Law,

Explain Hess's Law:

When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

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