Average KE same Heat (q) transfer thermal energy from hot to cool due to temp diff 2 .. 2 1 v m KE Average translational energy/KE per particle Heat Temperature Heat vs Temperature Symbol Q Unit - Joule Form of Energy Symbol T Unit – C/K Not Energy At 80C Distribution of molecular speed, Xe, Ar, Ne, He at same temp 2 . 2 1 v m KE 80 o C Diff gas have same average KE per particle. Click here Heat vs Temperature Click here specific heat capacity He – mass low ↓ - speed v high ↑ Xe – mass high ↑ - speed v low ↓ Temp ᾳ Average KE Higher temp - Higher average KE 2 .. 2 1 v m KE Movement of particle, KE. Heat energy (energy in transfer) 80 o C 50 o C degree of hotness/coldness Total KE/PE energy of particles in motion 1 liter water 80C 2 liter water 80C Same Temp Same Average kinetic energy per particle Same Average speed Same temp Diff amt heat More heat energy Less heat energy Heat energy (energy in transfer)
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IB Chemistry on Energetics experiment, Thermodynamics and Hess's Law
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Average KE same
Heat (q) transfer thermal energy from hot to cool due to temp diff
2..2
1vmKE
Average translational energy/KE per particle
Heat Temperature
Heat vs Temperature
Symbol Q Unit - Joule
Form of Energy Symbol T Unit – C/K
Not Energy
At 80C
Distribution of molecular speed, Xe, Ar, Ne, He at same temp
2.2
1vmKE
80oC
Diff gas have same average KE per particle.
Click here Heat vs Temperature Click here specific heat capacity
He – mass low ↓ - speed v high ↑ Xe – mass high ↑ - speed v low ↓
Temp ᾳ Average KE Higher temp - Higher average KE
2..2
1vmKE
Movement of particle, KE.
Heat energy (energy in transfer)
80oC 50oC
degree of hotness/coldness
Total KE/PE energy of particles in motion
1 liter water 80C 2 liter water 80C
Same Temp Same Average kinetic energy per particle Same Average speed
Specific heat capacity Amount of heat needed to increase
temp of 1 g of substance by 1C
Q = Heat transfer
Q = mcθ m = mass c = specific
heat capacity θ = Temp diff
Coffee-cup calorimeter
constant pressure – no gas
Calorimetry - techniques used to measure enthalpy changes during chemical processes.
Bomb calorimeter
Constant vol – gas released
Cup calorimeter Determine specific heat capacity of X
m = 1000 g
Heated 200 C
5000 ml water
m = 5000g
c = 4.18
Ti = 20 C
Tf = 21.8 C
Heat lost by X = Heat gain by water mc∆T = mc∆T
X
1000 x c (200 – 21.8) = 5000 x 4.18 x (21.8 – 20) c = 37620/ 178200 c = 0.211J/g/K
Benzoic acid – used std – combustion 1g release 26.38 kJ Combustion 0.579 g benzoic acid cause a 2.08°C increase in temp. 1.732g glucose combusted, temp increase of 3.64°C. Cal ΔHcomb glucose.
Bomb calorimeter (combustion) Find heat capacity of bomb and ∆Hc X
- Energy neither created nor destroyed - Converted from one form into another. - Amt heat lost by system equal to amt heat gain by its surrounding. - Total energy system plus its surrounding is constant, if energy is conserved.
change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal energy
q = heat transfer
w = work done by/on system
Thermodynamics Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK Gain energy from surrounding as heat or work
Heat add , q = + 100 J Work done by gas, w = - 20 J ∆E = + 100 – 20 = + 80 J
Q = Heat gain
+ 100J
w = work done by system = -20 J
w = work done on system = +20 J
Q = Heat lost
- 100J Heat lost , q = - 100 J Work done on gas, w = + 20 J ∆E = - 100 + 20 = - 80 J
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat and work Heat only
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J No work done = 0 ∆E = q + w ∆E = + 100 = + 100 J
Q = Heat lost
- 100J
Heat lost, q = - 100 J No work done = 0 ∆E = q + w ∆E = - 100 = - 100 J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Enthalpy change w = work done by/on gas
1st Law Thermodynamics
∆E = q + w ∆E = q
∆E = q + w + q = sys gain heat - q = sys lose heat + w = work done on sys - w = work done by sys
∆E = q + w
Work done by gas
No gas – No work
change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal energy
q = heat transfer
w = work done by/on system
Thermodynamics Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK Gain energy from surrounding as heat or work
No work done by/on system ∆E = q + w w = 0 ∆E = + q (heat flow into/out system) ∆H = ∆E = q (heat gain/lost)
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat only – Exothermic and Endothermic reaction
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J No work done = 0 ∆E = q + w ∆E = + 100 J ∆E = ∆H = + 100 J
Q = Heat lost
- 100J Heat lost, q = - 100 J No work done = 0 ∆E = q + w ∆E = - 100 J ∆E = ∆H = - 100J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Constant pressure
Enthalpy change w = work done by/on gas
1st Law Thermodynamics
P∆V = 0
∆E = q + w ∆H = ∆E + P∆V
∆E = q + 0 ↓
∆E = q
No gas produced V = 0
∆H = ∆E + 0 ↓
∆H = ∆E
At constant pressure/no gas produced
∆H = q ∆Enthalpy change = Heat gain or lost
No work done w = 0
H
E
E
∆H = + 100J
H ∆H = - 100J
Enthalpy Change
Heat(q) transfer from system to surrounding ↓
Exothermic ∆H < 0 ↓
HOT
Heat
energy ∆H = - ve
Enthalpy Change = Heat of rxn = -∆H
Mg(s) + ½ O2(g) → MgO(s) ∆H = -1200kJ mol-1
Mg + ½ O2
MgO
∆H= -1200
- Energy neither created nor destroyed - Converted from one form into another. - Amt heat lost by system equal to amt heat gain by its surrounding. - Total energy system plus its surrounding is constant, if energy is conserved.
Reactant (Higher energy - Less stable/weaker bond)
- Energy neither created nor destroyed - Converted from one form into another. - Amt heat lost by system equal to amt heat gain by its surrounding. - Total energy system plus its surrounding is constant, if energy is conserved.
Std Enthalpy Changes ∆Hθ
Std condition
Pressure 100kPa
Conc 1M All substance at std states
Temp 298K
Bond Breaking Heat energy absorbed – break bond
Bond Making Heat energy released – make bond
Std ∆Hcθ combustion
Std Enthalpy Changes ∆Hθ
∆H for complete combustion 1 mol sub in std state in excess O2
∆H when 1 mol solute dissolved form infinitely dilute sol
∆H when 1 mol form from its element under std condition
∆H = - ve
Std ∆Hf θ formation
Na(s) + ½CI2 (g)→ NaCI (s)
∆H = - 50 – (+12)
= - 62 kJ mol -1
∆H = - 50 kJ mol -1
Cold water = 50g
CuSO4 .100H2O
CuSO4 (s) + 95H2O
→ CuSO4 .100H2O
CuSO4 (s) + 100H2O
→ CuSO4 .100H2O
Mass cold water add = 50 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final Temp mixture = 31 C
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
Slow rxn – heat lost huge – flask used. Heat capacity flask must be determined.
CuSO4 (s) + 5H2O → CuSO4 .5H2O
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask CuSO4
Heat capacity flask, c = 63.5JK-1
Mass CuSO4 = 3.99 g (0.025mol) Mass water = 45 g T initial flask/water = 22.5 C T final = 27.5 C
Mass CuSO4 5H2O = 6.24 g (0.025mol) Mass water = 42.75 g T initial flask/water = 23 C T final = 21.8 C
Temp correction – using cooling curve for last 5 m
Measure temp CuSO4, every 0.5m interval then add zinc in excess
Zinc powder (excess)
Min/Max gradient
∆H = - 113 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50 g
MgSO4 .100H2O
MgSO4.7H2O + 93H2O
→ MgSO4 .100H2O
MgSO4 + 100H2O
→ MgSO4 .100H2O
Mass cold water add = 50 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final Temp flask/mixture = 31 C
MgSO4 (s) + 7H2O → MgSO4 .7H2O ∆H = ?
Slow rxn – heat lost huge – flask used. Heat capacity flask must be determined.
MgSO4(s) + 7H2O → MgSO4 .7H2O
1. Find heat capacity flask
Ti = 23.1 C
Hot water = 50 g T i = 41.3 C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask MgSO4
Heat capacity flask, c = 63.5JK-1
Mass MgSO4 = 3.01 g (0.025mol) Mass water = 45 g T initial mix = 24.1 C T final = 35.4 C
Mass MgSO4 7H2O = 6.16 g (0.025mol) Mass water = 41.8 g T initial mix = 24.8 C T final = 23.4 C
Assumption No heat lost from system ∆H = 0 Water has density = 1.00g ml-1
Sol diluted Vol MgSO4 = Vol H2O All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized Dont need to Plot Temp/time
No extrapolation
Error Analysis Heat loss to surrounding
Heat capacity sol is not 4.18 Mass MgSO4 ignored
Impurity present MgSo4 already hydrated
limiting
∆H = - 286 kJ mol -1
∆H = - 442 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50g
Mass cold water add = 50g Mass warm water add = 50g Initial Temp flask/cold water = 23.1C Initial Temp warm water = 41.3C Final Temp flask/mixture = 31C
Mg(s) + ½ O2 → MgO ∆H = ?
Slow rxn – heat lost huge – flask is used. Heat capacity flask must be determined.
Mg + ½ O2 → MgO
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb Flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask Mg
Heat capacity flask, c = 63.5JK-1
Mass Mg = 0.5 g (0.02 mol) Vol/Conc HCI = 100 g, 0.1M T initial mix = 22 C T final = 41 C
Mass MgO = 1 g (o.o248 mol) Vol/Conc HCI = 100 g, 0.1M T initial mix = 22 C T final = 28.4 C
Assumption No heat lost from system ∆H = 0 Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2O All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized Dont need to Plot Temp/time
No extrapolation
Error Analysis Heat loss to surrounding
Heat capacity sol is not 4.18 Mass MgO ignored
Impurity present Effervescence cause loss Mg
+ 2HCI
MgCI2 + H2
HCI Flask MgO
+ ½ O2
MgCI2 + H2O
+ 2HCI
MgCI2 + H2O
∆H MgO + 2HCI → MgCI2 + H2O
Mg + ½ O2 → MgO
MgCI2 + H2
+ 2HCI
MgCI2 + H2O
+ ½ O2
limiting
MgCI2 + H2O
Data given
2KHCO3(s) → K2CO3 + CO2 + H2O
∆H = + 51.4 kJ mol -1
Enthalpy change ∆H for rxn using calorimeter
Cold water = 50g
Mass cold water add = 50 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final Temp flask/mixture = 31 C
2KHCO3(s) → K2CO3 + CO2 + H2O ∆H = ?
Slow rxn – heat lost huge – vacuum flask is used. Heat capacity of flask must be determined.
1. Find heat capacity vacuum
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask KHCO3
Heat capacity vacuum, c = 63.5JK-1
Mass KHCO3 = 3.5 g (0.035 mol) Vol/Conc HCI = 30 g, 2M T initial mix = 25 C T final = 20 C
Mass K2CO3 = 2.75 g (0.02 mol) Vol/Conc HCI = 30 g, 2M T initial mix = 25 C T final = 28 C
Strong acid vs Strong alkali Weak acid vs strong alkali
0.05 mol = - 2541 J 1 mol = - 51 kJ mol-1
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C 21 22. 24 26 27 28 28 28 27 27 26
∆H n always same (-57.3) regardless strong or weak acid H+ + OH- → H2O ∆H = 57.3 kJmol-1
∆H n weak acid vs strong alkali is lower (less –ve) Weak acid - molecule doesn’t dissociate completely Heat absorb to ionize/dissociate molecule – less heat released
∆H neutralization rxn
CH3COOH + NaOH → CH3COONa + H2O
1.9M CH3COOH – 5 ml added and temp recorded
∆H rxn = Heat absorb water = (mc∆θ)
= 76 x 4.18 x (29 – 21)
= 76 x 4.18 x 8
= - 2541J
V, NaOH = 50 ml M, = 1M T initial = 21 C T final = 29 C
Smaller size--↑ -ve ∆H hydration More heat released (-ve)
CI- Br- I-
CI-
Br-
I-
Ag+ + CI- → AgCI ∆H = -50 kJ mol -1
Bigger size--↑ - Weaker attraction bet I- with H2O ↓
Easier to release I- to form ppt
∆H hydration
C2H5OH + 3O2 → 2CO2 + 3H2O
Cold water = 50g
Mass cold water add = 50 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1C Initial Temp warm water = 41.3C Final Temp flask/mixture = 31C
1. Find heat capacity calorimeter
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O = Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Heat capacity flask, c = 63.5JK-1
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn= 250 x 4.18 x (58 – 30) + 63.5 x (58 – 30)
∆Hrxn= - 31 kJ
Lit value = - 1368 kJ mol -1
∆Hc combustion using calorimeter
Mass water = 250 g Initial Temp water = 30 C
Final Temp water/flask = 58 C
2. Find ∆Hc combustion data
Mass initial spirit lamp/alcohol = 218.0 g Mass final spirit lamp/alcohol = 216.6 g Mass alcohol combusted = 1.4 g