Brainscape's Knowledge GenomeTM


Top Flashcards (Certified)
All Flashcards

chemistry 1A > thermodynamics introduction > Flashcards

thermodynamics introduction Flashcards

1
Q

open system

A

an unsealed flask
there will be transfer of matter (molecules) and transfer of energy between the system and surroundings.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

closed system

A

a sealed flask
there will be transfer of energy between the system and surrounding but no transfer of matter.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

isolated system

A

a sealed flask in a vacuum flask
there will be no transfer of energy or matter between the reaction system and surroundings.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

reaction equations note

A

when writing down a reaction equation at a specific temperature, the products must be in their specific state at that temperature, even if it produces the product in a different state, since the product will then change state to its normal state at that temperature.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

1st law of thermodynamics

A

the internal energy of an isolated system is always constant.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

2nd law of thermodynamics

A

the entropy of a reaction system and its surroundings is always increasing for a spontaneous process.

This basically means that a reaction will proceed spontaneously if the total entropy of the universe will be increased by the reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

3rd law of thermodynamics

A

the entropy of a perfect crystalline material at 0K = 0

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

1st law helpful development

A

the change in internal energy of a closed system is equal to the heat and work done by the system.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

internal energy equation

A

ΔU = q + w
change in internal energy = heat energy + work

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

what is the potential energy of a reaction system

A

the stored energy of chemical bonds in the reaction system.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

how do we calculate the potential energy of the reaction system

A

we use the mean molar bond enthalpies,

bonds broken are endothermic and bonds formed are exothermic, so do reactants - products.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

mean molar bond enthalpy definition

A

the enthalpy change when a specific bond is broken within a molecule.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

what is the kinetic energy of a molecule determined by

A

the number of degrees of freedom that the molecule has.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

degrees of freedom definition

A

distinct ways in which molecules can move.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

3 types of degrees of freedom

A

rotational
vibrational
translational

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

translational degrees of freedom

A

translational degrees of freedom are movements of a molecule through space in the x, y, and z axis.
there are always 3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

rotational degrees of freedom

A

the degrees of freedom produced by bond rotations, there will be 3 in non linear molecules and 2 in linear molecules.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

vibrational degrees of freedom

A

the degrees of freedom produced by the molecules bonds absorbing energy to bend and stretch. (normal modes)
3N - 5 for linear molecules
3N - 6 for non linear molecules.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

kinetic energy per degree of freedom per molecule formula

A

E = 1/2 kb x T
energy = 1/2 Boltzmann’s constant x Temperature

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Boltzmann’s constant

A

1.381 x 10^-23 J K-1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

energy per degree of freedom per mole formula

A

E = 1/2 RT
energy = 1/2 gas constant x temperature

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

gas constant

A

8.314 J K-1 mol-1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

kinetic and potential energy comparison

A

kinetic energy is in the 10’s of kJ and potential energy is in the 100’s, to potential > kinetic.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

most important type of work in system

A

work of expansion - since it will have the largest value, so will contribute to most of the work.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
work of expansion requirement
there must be a change in the number of moles of a gas in a reaction
26
properties of gasses (basic)
p ∝ 1/V at constant temperature V ∝ T at constant pressure p ∝ T at constant volume
27
how should you understand constant pressure and volume
constant pressure is like a piston pushing down on a chamber at consistent force constant volume is like an indestructible box with something inside of it.
28
ideal gas law
pV = nRT pressure x volume = number of moles of gas x gas constant x temperature.
29
ideal gas law assumptions
the gas particles are infinitely small points the gas particles move in straight lines the gas particles have no intermolecular forces of interaction between each other
30
mixture of gasses pressure formula (Daltons law)
p = p1 + p2 + p3 +... the pressure of a mixture of gasses = the pressure of all of the different components of the gas mixture added together. where p1 = x1p where x1 is the percentage composition of the mixture of gasses.
31
work of expansion formula
w = -p(ΔV) work of expansion = minus pressure x change in volume this can then be expanded using the ideal gas law w = -ΔnRT work of expansion = minus change in moles x gas constant x temp
32
work of expansion + or -
work of expansion of a system will be -ve since w = -ΔnRT and Δn is +ve.
33
work of contraction + or -
work of contraction of a system will be +ve since w = -ΔnRT and Δn will be -ve.
34
specific heat capacity definition
the energy required for 1g of a substance to have its temperature raised by 1K
35
molar heat capacity definition
the energy required for one mole of a substance to have its temperature raised by 1K
36
specific heat capacity formula
Cs = q/mΔT
37
molar heat capacity formula
Cp = q/nΔT
38
enthalpy change formula relating to internal energy, pressure and volume
ΔH = ΔU + pΔV
39
enthalpy at constant pressure
ΔH = ΔU + pΔV pΔV = -w ΔH = ΔU - w ΔU = q + w ΔH = q + w -w ΔH = q
40
enthalpy at constant volume
ΔH = ΔU + pΔV pΔV = 0 at constant volume ΔH = ΔU = qv at constant volume
41
exothermic reactions definition
a chemical reaction where there is heat energy transferred from the reaction system to its surroundings, meaning that heat is released and there is a -ve enthalpy change.
42
endothermic reaction definition
a chemical reaction where there is heat energy transferred from the surroundings to the reaction system meaning heat is absorbed and there will be a +ve enthalpy change.
43
forming bonds type of reaction
forming bonds will mean electrons will go from a high energy to low energy state, which will mean that heat energy is transferred from the system to the surroundings, which will mean that it is an exothermic reactions and the enthalpy change will be -ve
44
breaking bonds enthalpy and type of reaction
breaking bonds will mean electrons go from a low energy state to high energy state, which will result in heat energy being absorbed by the system from surroundings, which will mean a +ve enthalpy change and an endothermic reaction.
45
Hess law equation
Σ v Δf H°(products) - Σ v Δf H°(react) note the v is just for stoichiometry
46
Kirchhoff's law equation
Δr H (T2) = Δr H (T1) + Δr Cp ΔT change in enthalpy at T2 = change in enthalpy at T1 + molar heat x change in temperature.
47
rule for enthalpy relating to temperature
as the temperature increases the enthalpy of the products and reactants will increase, and the change in enthalpy will increase since the difference between products and reactants will increase.
48
important assumptions of Kirchhoff's law
there cant be a state change the molar heat capacity is assumed to be constant.
49
latent heat definition
the enthalpy change required to change a substance from one state to another.
50
fusion
solid to liquid
51
vaporization
liquid to gas
52
sublimation
solid to gas
53
freezing
liquid to solid
54
condensation
gas to liquid
55
deposition
gas to solid
56
entropy colloquial definition
the measure of disorder within a system
57
entropy better definition
the measure of how energy is dispersed between different arrangements in a system whilst the systems energy remains constant.
58
arrangements
W
59
rule of entropy and number of atoms
As the number of atoms increases the number of different arrangements that energy can be dispersed between will increase, which will result in an increase in entropy.
60
rule of entropy and state change
As a substance goes from a solid to a liquid to a gas the intermolecular forces holding the molecules in place break apart allowing the molecules to move more freely, which will result in a greater number of arrangements for energy to be dispersed between, meaning the entropy of the system will increase.
61
which is bigger: the change in entropy between solids and liquids, or liquids and gasses?
liquids to gasses will have a greater change in entropy.
62
how does temperature affect entropy
temperature will increase the average kinetic energy if the molecules, which will mean there will be more arrangements available for energy to be dispersed into.
63
molar entropy of a reaction equation
Δr S°m = Σ Δr S°m (products) - Σ Δr S°m (reactants) molar entropy at standard conditions = products molar entropy - reactants molar entropy
64
entropy related to heat energy and temperature
ΔS = q(rev)/T change in entropy = heat energy transfer of a reversible process/ temperature
65
entropy at a different temperature formula
Δr S°(T2) = Δr S°(T1) + Cp ln(T2/T1) entropy at temperature 2 = entropy at temperature 1 + molar heat capacity x ln T2/T1
66
reaction feasibility for entropy rule
a reaction will proceed spontaneously if there is a total increase in the entropy of the universe, meaning the reaction will always move to a more distributed configuration if available.
67
total entropy formula
Stot = Ssys +Ssur
68
Gibbs free energy formula
ΔrG° = ΔrH° - TΔrS°
69
rule for reaction feasibility in terms of free energy
if ΔG<0 then a reaction is feasible, since ΔG = -StotT
70
reaction spontaneity ΔH < 0 ΔS <0
the entropy change is -ve, so the entropy of the reaction system will be decreasing, the enthalpy is -ve so the reaction surroundings will be absorbing heat energy and the entropy of the surroundings will be increasing. This will mean that the reaction spontaneity will be temperature dependent.
71
reaction spontaneity at ΔH <0 ΔS >0
the entropy change will be positive, meaning that the reaction systems entropy will be increasing, and the reaction is exothermic which will mean that the surroundings entropy will be increasing, therefore the total entropy will always be greater than 0, so the reaction will always be feasible.
72
reaction spontaneity at ΔH>0 ΔS< 0
the entropy change is -ve which means the entropy of the system will be decreasing, the reactions enthalpy is endothermic which will mean that the surroundings will lose heat energy and therefore lose entropy, this will result in the reaction system never being feasible since the entropy of the reaction will always be decreasing.
73
reaction spontaneity at ΔH> 0 ΔS>0
the reaction systems entropy will be increasing because the change in entropy is +ve, however the enthalpy change is - +ve meaning heat energy is being absorbed by the reaction system, which means the energy and entropy of the reaction surroundings is decreasing, meaning that the temperature will affect the reaction feasibility.
74
formula for Δ G at standard condition
Δr G° = Σ v Δr G ° (products) - Σ v Δr G ° ( reactants)
75
what does the reaction quotient describe
the ratio of products to reactants
76
when does the reaction quotient equal the equilibrium constant
at equilibrium when the Gibbs free energy is at its minimum
77
reaction quotient equation
Q = [C]^c [D]^d [A]^a [B]^b
78
equilibrium equation
K = [C]^c [D]^d [A]^a [B]^b
79
requirements for equilibrium
a reaction must not proceed to completion there must be a minimum free energy within the mixture where there is not 100% products or reactants. the reaction will always proceed towards this minimum free energy once this minimum free energy is achieved the reaction rate of the forwards and reverse reaction will be constant, meaning that the concentrations of products and reactants will remain constant.
80
why does a reaction not proceed to completion in equilibrium
because the minimum free energy in the reaction occurs when there are products and reactants present, and since ΔG will always be negative going towards the minimum free energy, the reaction will proceed until this minimum free energy, then products and reactants concentrations will remain constant.
81
why do reactants and products remain constant at the minimum free energy in equilibrium
the reaction of the forwards and reverse reaction will occur at the exact same rate, so the reaction concentrations will remain constant
82
which way does a reversible reaction proceed in a relation to Gibbs free energy
towards the minimum in Gibbs free energy.
83
free energy at non standard conditions
Δr G = Δr G° + RT lnQ
84
free energy at standard states related to equilibrium (vant hoffs isotherem)
Δr G° = - RT lnK since Δr G = 0 at equilibrium and Q = K
85
calorimetry at constant volume pressure procedure
calibrate the calorimeter with hot and cold water measure the temperature changes of the hot and cold water use -CsMΔT2 = CsMΔT1 + C cal ΔT1 to find the calorimeter constant. then use ΔH = Cs M ΔT + cal ΔT
86
calorimeter constant equation
-CsMΔT2 = CsMΔT1 + C cal ΔT1 heat energy lost by hot water = heat energy gained by cold water + heat energy lost by calorimeter.
87
enthalpy change of a calorimeter at constant pressure equation
ΔH = Cs M ΔT + cal ΔT enthalpy change = specific heat capacity x mass x temperature change + calibration constant x temperature change
88
calorimetry at constant volume explanation (bomb calorimeter)
At constant volume no work of expansion can be lost, so the change in internal energy = the heat lost from the calorimeter chamber, which we can calculate using the calorimeter constant and the change in temperature.
89
calorimetry at constant volume equation
ΔU = Ccal ΔT

chemistry 1A (9 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2025 Bold Learning Solutions. Terms and Conditions