Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of...

Ch6.1 The Nature of EnergyEnergy – the capacity to do work or to produce heat.

Law of Conservation of Energy – energy can be converted from one form to another, but can be neither created or destroyed.

1st Law of thermodynamics – the energy of the universe is constant.

Potential energy – stored energy due to position or composition

Kinetic energy – energy of motion KE = ½mv2

Heat – (a form of energy) transfer of energy due to a temp difference

Temperature – property that reflects the randomness of the particles(NOT a form of energy)

Chemical energy

Exothermic reaction –

CH4(g) + O2(g)

Chemical energy

Exothermic reaction – heat flows out of the system (feel hot)

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy

Chemical energy

Exothermic reaction – heat flows out of the system (feel hot)

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy

In any exothermic reaction, some of the potential energy stored in chemical bonds is converted to thermal energy.

Endothermic reaction –

N2(g) + O2(g)

Endothermic reaction – absorb heat from their surroundings (feel cold)- heat flows into the system

N2(g) + O2(g) + energy 2NO(g)

1st Law of Thermodynamics – the energy of the universe is constant.

ΔE = q + w

Internal heat work energy energy

“All energy movements reflect the system’s point of view.”

ΔE = + Internal energy of the system increasedΔE = – Internal energy of the system decreased q = + Heat flowed into the system q = – Heat flowed out of the system w = + The surroundings do work on the system w = – The system does work on the surroundings

Ex1) Calculate ΔE for a system undergoing an endothermic processin which 15.6kJ of heat flows and where 1.4kJ of work is done on the system.

Work associated with gases:Pressure: Work:

Ex2) Calc the work associated with the expansion of a gas from 46L to 64L at a constant pressure of 15 atm.

Ex3) A balloon is being inflated to its full extent by heating the air inside it.

The volume changes from 4.00x106L to 4.50x106L when

1.3x108J of heat are added. Atmospheric pressure is 1.0 atm.Calculate ΔE for the process.

Ch6 HW#1 p283 21,23,25,26

Ch6 HW#1 p283 21,23,25,2721. Calculate ΔE for each of the following cases. a. q = +51 kJ, w = -15kJ b. q = +100 kJ, w = -65kJ c. q = -65kJ, w = -20kJ d. In which of these does the system do work on the surroundings?

Ch6 HW#1 p283 21,23,25,2621. Calculate ΔE for each of the following cases. a. q = +51 kJ, w = -15kJ b. q = +100 kJ, w = -65kJ c. q = -65kJ, w = -20kJ d. In which of these does the system do work on the surroundings?

ΔE = q + w

a. ΔE = +51 kJ + (-15kJ) = +36kJ (Internal energy increases)

b. ΔE = +100 kJ + (-65kJ) = +35kJ (Internal energy increases)

c. ΔE = -65kJ + (-20kJ) = -85kJ (Internal energy decreases)

d. All 3 have w = (-) so all systems did work on the surroundings

23. A gas absorbs 45kJ of heat and does 29kJ of work. Calculate ΔE.

ΔE = q + w

23. A gas absorbs 45kJ of heat and does 29kJ of work. Calculate ΔE.

ΔE = q + w

= (+45kJ) + (-29kJ)

= +16kJ (increase)

25. The volume of an ideal gas is decreased from 5.0 L to 5.0 mL at a constant pressure of 2.0 atm. Calculate the work associated with this process.

ΔV = (0.005L – 5L) = -4.995L

kPa 202.6atm 1

101.3kPa atm 2 P

25. The volume of an ideal gas is decreased from 5.0 L to 5.0 mL at a constant pressure of 2.0 atm. Calculate the work associated with this process.

ΔV = (0.005L – 5L) = -4.995L

W = –P. ΔV

= –(202.6kPa). (-4.995L) = +1012 J

= +1.012kJ (work done on the gas!)

kPa 202.6atm 1

101.3kPa atm 2 P

26. Consider the mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650. torr if all the energy of combustion is conserved into work to push back the piston.

ΔV = (V2 – 0.040L)

q W =

kPa 6.68760torr

101.3kPa torr 650 P

26. Consider the mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650. torr if all the energy of combustion is conserved into work to push back the piston.

ΔV = (V2 – 0.040L)

q W = –950J (expands – gas doing work)

W = –P. ΔV

–950J = –(86.6kPa). ΔV ΔV = 10.97L

10.97L = (V2 – 0.040L) V2 = 11.0L

kPa 6.68760torr

101.3kPa torr 650 P

Ch6.1B – PV Diagrams‘A closed system at equilibrium has no tendency to undergo spontaneous

macroscopic change.”4 basic models where a closed system is changed:

Constant Can change Equations Implications Thermo Eqn1. Isothermal

2. Isobaric

3. Isovolumetric

4. Adiabatic (no heat transferred to or from a system.)

Work equation:(Units: ______, or ______, or ______)

Ch6.1B – PV Diagrams‘A closed system at equilibrium has no tendency to undergo spontaneous

macroscopic change.”4 basic models where a closed system is changed:

Constant Can change Equations Implications Thermo Eqn1. Isothermal T V, P V1P1 = V2P2 If ∆T = 0, q = W

then ∆E=02. Isobaric P V, T ∆E = q + W

3. Isovolumetric V P, T No work ∆E = q done.

4. Adiabatic nothing V, P, T Q = 0 ∆E = W(no heat transferred to or from a system.)

Work equation: W = –P . ∆V(Units: Joules, or Pa.m3, or kPa.L)

2

2

1

1

T

V

T

V

2

2

1

1

T

P

T

P

2

22

1

11

T

PV

T

PV

50 Work = (area under the PV curve.) 40 Decide +/- based on direction P 30(Pa) 20 10 1 2 3 V (m3)

101.3 P(kPa)

1 2 3 V (L) 150

P 100(Mpa) 50 1 2 3 V(cm3)

50 Work = –(area under the PV curve.) 40 Decide +/- based on direction P 30 W = ½(3m3)(40N/m2) (Pa) 20 = –60J 10 1 2 3 V (m3)

101.3 W = (101.3kPa)(3L) P (1kPa 1000Pa)(kPa) = –303.9J (1L 0.001m3)

1 2 3 V (L) 150

No work done P 100 (Isovolumetric)(Mpa) (1 Mpa 1,000,000Pa) 50 (1cm3 0.000001m3) 1 2 3 V(cm3)

Ex3) A cylinder contains 264.5 moles of a monatomic gas that is initially at state A at standard temperature, and a pressure of 6x105Pa. a. What volume does the gas occupy? 6 b. Please graph state A. 5c. A certain amount of heat is added, P 4 bringing the gas isobarically to state B, (x105 Pa) 3 at a volume of 3m3, while its internal energy 2 is increased by 400kJ. How much heat was added? 1 Please graph state B. 1 2 3 4 5 6 7 8 V(m3)d. The gas is brought isothermally to state C, where the pressure was reduced to 3x105Pa. What is the new volume? Please graph state C.

Ch6 HW#2 1 – 4

Ex3) A cylinder contains 264.5 moles of a monatomic gas that is initially at state A at standard temperature, and a pressure of 6x105Pa. a. What volume does the gas occupy? 6 b. Please graph state A. 5c. A certain amount of heat is added, P 4 bringing the gas isobarically to state B, (x105 Pa) 3 at a volume of 3m3, while its internal energy 2 is increased by 400kJ. How much heat was added? 1 Please graph state B. 1 2 3 4 5 6 7 8 V(m3)d. The gas is brought isothermally to state C, where the pressure was reduced to 3x105Pa. What is the new volume? Please graph state C. a. PV=nRT

c. ∆E = q + W W = –(area) = (+2m3)(6x105Pa) = –1,200,000J (+400,000J) = q + (–1,200,000J) + q = +1,600,000J d. VB = 3m3 VC = ? VB

. PB = VC. PC

PB = 6x105Pa PC = 3x105Pa (3)(6) = (VC)(3) TB = ? same TC = ? VC =6m3

36

3

1 106

)273)(31.8)(5.264( m

Pax

KKmolmPa

mol

P

nRTV

Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heatas it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure?

2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas?What do you think happens to the temp of the gas as it is compressed?

Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heatas it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure?

q = + 4186J ∆E = q + w ∆V = + 15L 0.015m3 w = (+286J) – (+4186J) ∆U = +286J w = – 3890J

2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas?What do you think happens to the temp of the gas as it is compressed?

kPaL

J

V

WP

259 15

3890

Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heatas it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure? q = + 4186J ∆E = q + w ∆V = + 15L 0.015m3 w = (+286J) – (+4186J) ∆U = +286J w = – 3890J

2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas?What do you think happens to the temp of the gas as it is compressed? ∆V = –20.0m3 W = – P. ∆V P = 101,300Pa = –(101,300Pa)(– 20m3) W = ? = +2,026,000J

(work done on the gas)Compression of a gas increases temp.

kPaL

J

V

WP

259 15

3890

3. A gas experiences a change from state I to state F. Determine the work done.

4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinderat a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy? A = 0.50m2 h = 6.5cm

6 5 F P 4(Mpa) 3 2 I 1

1 2 3 4 5 6 V (cm3)

3. A gas experiences a change from state I to state F. Determine the work done.65 F Work = 1 + 2 + 3

P 4 = 12J(Mpa) 3

2 I1 1 2 3 4 5 6 V (cm3)

4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinderat a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy?

12

3

A = 0.50m2

h = 6.5cm

3. A gas experiences a change from state I to state F. Determine the work done.65 F Work = 1 + 2 + 3

P 4 = 12J(Mpa) 3

2 I1 1 2 3 4 5 6 V (cm3)

4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinderat a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy? A = 0.50m2 ∆V=A. ∆h=(.50m2)(.065m) = +0.0325m3

h = 6.5cmW = – P. ∆V = (0.3x106Pa)(+0.0325m3)

= – 9750J

∆E = q + W = (+10,000) + (– 9750) = +250J

12

3

Ch6.1C – PV DiagramsEx1) The graph shows P vs V for 2 moles of gas cycling from state A to B.The internal energy increases by 400J. a. Calc work done. 1000b. Is heat added or removed? How much? P 800 c. The pressure is reduced to 200Pa (Pa) 600 A B

isovolumetrically as gas goes from state B 400 to state C. Please graph state C. 200d. The graph is compressed isothermally to state A. 1 2 3 4 5 6 7 8 9 10 Draw the lines to represent the cycle. V (m3)e. Is heat added or removed from C A? How much?

Ex1) The graph shows P vs V for 2 moles of gas cycling from state A to B.The internal energy increases by 400J. a. Calc work done. 1000b. Is heat added or removed? How much? P 800 c. The pressure is reduced to 200Pa (Pa) 600 A B

isovolumetrically as gas goes from state B 400 to state C. Please graph state C. 200 C

d. The graph is compressed isothermally to state A. 1 2 3 4 5 6 7 8 9 10 Draw the lines to represent the cycle. V (m3)e. Is heat added or removed from C A? How much? a. W = P. ∆V = (600Pa)(+6m3) = +3600J b. Added: Why else would the gas expand? Q = W + ∆U = +3600J + (+400J) = +4000J c. isovolumetric d. isothermal – P and V can change, while ∆T = 0, so ∆U = 0. e. Heat is removed. Q = W + ∆U

= W + 0Q = WorkCA = (Area) = -2400J

(Work is (-) becuz the gas is compressed by outside force. Work done on the gas.)

Ex2) A piston with a radius of 0.05m seals a cylinder that contains a gas at a pressure of 101.3kPa. 100J of heat is added to the cylinder, and the piston rises 0.04m isobarically.a. How much work is done by the gas? hb. What is its change in internal energy?The total pressure exerted on the gas increases to 107.7kPa. c. If the piston seals 1m3, when the pressure increased, and the piston lowers isothermally, what is the new volume?

Ch6 HW#3 5 – 8

Ex2) A piston with a radius of 0.05m seals a cylinder that contains a gas at a pressure of 101.3kPa. 100J of heat is added to the cylinder, and the piston rises 0.04m isobarically.a. How much work is done by the gas? hb. What is its change in internal energy?The total pressure exerted on the gas increases to 107.7kPa. c. If the piston seals 1m3, when the pressure increased, and the piston lowers isothermally, what is the new volume? a. ∆V = A. ∆h = π(.05m)2(+.04m) = + 0.0003m3

W = – P. ∆V = – (101,300Pa)(+0.0003m3) = – 31.8J b. ∆U = q + W = (+100J) + (– 31.8J) = 68J

c. V1 = 1 m3 V2 = ___m3 V1. P1 = V2

. P2 P1 = 101,300Pa P2 = 107,700Pa (1m3)(101,300Pa) = V2(107,700Pa) T1 = ? same T2 = ? V2 = .9m3

Ch6 HW#3 5 – 8 1. Given that we have 1200 cm3 of helium at 15°C and 99 kPa, what will be its volume at STP?

2. Three moles of a gas are injected into an evacuated container,with a fixed volume, and maintained at a temp of 60°C. If the pressure increases to 103,200 kPa what volume is the container?

PV = nRT



PV = nRT

2

22

1

11

T

PV

T

PV 3

22

3

1112 )273(

)3.101(

)288(

)99)(1200(cmV

K

kPaV

K

kPacm



PV = nRT

2

22

1

11

T

PV

T

PV 3

22

3

1112 )273(

)3.101(

)288(

)99)(1200(cmV

K

kPaV

K

kPacm

3

3

008.0

700,103

)333(31.8)3(

m

Pa

KKmol

mPamol

P

nRTV

3. A 0.05 m3 volume of gas absorbs 10 kJ of heat and expands to a volume of 0.15 m3, while remaining at a constant pressure. During the process, the internal energy increases by 4500 J. What was the pressure? ∆V = +.10m3

q = +10kJ ∆E = +4.5kJ P = ?

3. A 0.05 m3 volume of gas absorbs 10 kJ of heat and expands to a volume of 0.15 m3, while remaining at a constant pressure. During the process, the internal energy increases by 4500 J. What was the pressure? ∆V = .10m3 W = ∆U – q = (+4.5kJ) – (+10kJ) = – 5.5kJ = – 5500J Q = +10kJ ∆U = +4.5kJ P = ?4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas?

b. What is the change in internal energy? c. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

Pam

J

V

WP 000,55

10.0

55003

4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3

W = – P. ∆V = b. What is the change in internal energy?

∆E = q + W = c. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?


W = – P. ∆V = –(100,000Pa)(– 0.005m3) = +500J b. What is the change in internal energy?

∆E = q + W = c. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?



∆E = q + W = (-200J) + (+500J) = +300J c. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?



∆E = q + W = (-200J) + (+500J) = +300J

c. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

32

23

04.

2

)02(.

mV

T

V

T

m

Ch6.2A – Enthalpy

Enthalpy, H – the ‘heat content’ of a substance

defined as: H = E + PVH is related to q. Officially: at constant pressure q = H

For practical purposes: q is the heat measured in an experimentH is the heat related to a chem rxn.


ΔH = Hproducts – Hreactants

Exothermic reaction:

Enthalpy

time


ΔH = Hproducts – Hreactants

Endothermic reaction:

Enthalpy

time

Ex1) When 1 mol of methane is burned at constant pressure, 890kJ of energy is released as heat. Calculate ΔH for a processin which a 5.8g sample is burned.

Ex2) a. Write an eqn for the combustion of ethanol, C2H5OH.

1235 kJ of heat is given off. b. Draw an energy diagram. c. How much heat is produced when 12.5g of ethanol reacts?

Ch6 HW#4 p283 29,30,31,33,34

Ch6 HW#4 p283 29,30,31,33,3429. The equation for the fermentation of glucose to alcohol

and carbon dioxide is

C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)

The enthalpy change for the reaction is -67kJ. Is the reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

Ch6 HW#4 p283 29,30,31,33,3429. The equation for the fermentation of glucose to alcohol

and carbon dioxide is

C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) + 67kJ

exothermic heat is evolved as the reaction occurs?

30. The reaction SO3(g) + H2O(l) → H2SO4(aq) is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is -227 kJ. In the designing a sulfuric acid plant, is it necessary to provide

for heating or cooling of the reaction mixture? Explain.

30. The reaction SO3(g) + H2O(l) → H2SO4(aq) is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is -227 kJ. In the designing a sulfuric acid plant, is it necessary to provide

for heating or cooling of the reaction mixture? Explain.

SO3(g) + H2O(l) → H2SO4(aq) + 227 kJ

Cool the plant!

31. Are the following processes exothermic or endothermic? a. When solid KBr is dissolved in water, the solution gets colder.

KBr(s) + _____ → K+(aq) + Br-

(aq) + _____

b. Natural gas (CH4) is burned in a furnace.

CH4(g) + O2(g) + _____ → CO2(g) + H2O(g) + _____

c. When concentrated H2SO4 is added to water, the solution gets very hot.

H2SO4(aq) + _____ → 2H+(aq) + SO4

2-(aq) + _____

d. Water is boiled in a teakettle.

H2O(l) + _____ → H2O(g) + _____


KBr(s) + heat → K+(aq) + Br-

(aq)


CH4(g) + O2(g) → CO2(g) + H2O(g) + heat


H2SO4(aq) + _____ → 2H+(aq) + SO4

2-(aq) + _____


H2O(l) + _____ → H2O(g) + _____


KBr(s) + heat → K+(aq) + Br-

(aq)


CH4(g) + O2(g) → CO2(g) + H2O(g) + heat


H2SO4(aq) → 2H+(aq) + SO4

2-(aq) + heat


H2O(l) + heat → H2O(g)

33. For the reaction

S(s) + O2(g) → SO2(g) ΔH = -296 kJ/mola. How much heat is evolved when 275g sulfur is burned in excess O2?

b. How much heat is evolved when 25mol sulfur is burned in excess O2?

c. How much heat is evolved when 150. g sulfur dioxide is produced?



q =



2536kJ1molS S 32.1g

296kJ S 1mol S 275g



q =


q =


2536kJ1molS S 32.1g

296kJ S 1mol S 275g

kJ4007S 1mol

296kJ S 25mol



q =


q =


q =

2536kJ1molS S 32.1g

296kJ S 1mol S 275g

kJ4007S 1mol

296kJ S 25mol

kJ936SO 1mol SO 64.1g

296kJ SO 1mol SO 150g

22

22

34. The overall reaction in commercial heat packs can be represented

as 4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔH = -1652kJ

a. About how much heat is released when 4.00 mol iron is reacted with excess O2?

b. How much heat is released when 1.00 mol Fe2O3 is produced?

c. How much heat is released when 1.00 g iron is reacted with excess O2?

d. How much heat is released when 10.0 g Fe and 2.00 g O2 are reacted?



a. About how much heat is released when 4.00 mol iron is reacted with excess O2?

b. How much heat is released when 1.00 mol Fe2O3 is produced?



kJ826OFe 2mol

1652kJ OFe 1mol

32

32

kJ1652Fe 4mol

1652kJ Fe 4.00mol





(excess Fe)

released

kJ40.7Fe 4mol Fe 55.8g

1652kJ Fe 1mol Fe 1.00g

kJ0.74Fe 4mol Fe 55.8g

1652kJ Fe 1mol Fe 10.00g

kJ4.34Fe 3mol O 32.0g

1652kJ O 1.00mol O 2.00g

2

22

Ch6.2B – Calorimetry

Calorimetry – the measure of heat flow

Heat Capacity (C) – of heat flow to a substance, compared to its temp change.

(In lab 6.2 you will find C for a foam cup)

in temp increase

absorbedheat C

T

H

C



Molar Heat Capacity – heat flow per mole, per temp change

units:

in temp increase

absorbedheat C

T

H

C

Tmol

H

molmol

C

mole

CapacityHeat TH

Kmol

J



Specific heat capacity (s or Cp) – measure of heat flow to 1 gram of a substance, compared to its temp change.

Exs:

(In lab 6.1 you will find the Cp for an unknown…per gram of it)

in temp increase

absorbedheat C

Cg

J

18.4C waterp Cg

J

45.0C Fe p

T

H

C

Ex1) A 46.2 g sample of copper is heated to 95.4˚C and then placed ina calorimeter containing 75.0 g water at 19.6˚C. The final temperature of the metal and water is 21.8˚C. Calculate the specific heat capacity ofcopper, assuming that all the heat lost by the copper is gained by water.

Ex2) 50ml of 1.00M HCl at 25.0oC is mixed with 50ml of 1.00M NaOHat 25.0oC in a calorimeter. The temp of the mix reaches 31.9oC.

If the Cp of the solution is ,

and the density of the final soln is 1.0 g/ml, calc the enthalpy changeper mole for this neutralization reaction.

Ch6 HW#5 p284 27,42,43

Cg

J

18.4

Ex2) 50ml of 1.00M HCl at 25.0oC is mixed with 50ml of 1.00M NaOHat 25.0oC in a calorimeter. The temp of the mix reaches 31.9oC.

If the Cp of the solution is ,

and the density of the final soln is 1.0 g/ml, calc the enthalpy changeper mole for this neutralization reaction.

HCl + NaOH NaCl + HOH + ______J 0.05mol 0.05mol

(100ml)(1.0g/ml) = 100g

q = (100g)(4.18 J/g.0C)(5.90C) = 2466J

ΔH = 2466J/0.10mol = 24,660J

Ch6 HW#5 p284 27,42,43

Cg

J

18.4

Ch6 HW#5 p284 27,42,4327. A balloon filled with 39.1 mol helium has a volume of 876L a 0.0˚C and 1.00 atm pressure. The temperature of the balloon is increased to 38.0˚C as it expands to a volume of 998 L, the remaining pressure remaining constant. Calculate q, w, ΔE for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 J/˚C·mol.)

n = 39.1 mol C/mol = 20.8 J/˚C·molVi = 876L Vf = 998 LTi = 0.0˚C Tf = 38.0˚CP = 1.00 atm = 101.3kPa (stays const)

q = ΔH =

W = –P.ΔV =

ΔE = q + W =

C11mol

J 20.8o

T

H

C

27. A balloon filled with 39.1 mol helium has a volume of 876L a 0.0˚C and 1.00 atm pressure. The temperature of the balloon is increased to 38.0˚C as it expands to a volume of 998 L, the remaining pressure remaining constant. Calculate q, w, ΔE for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 J/˚C·mol.)

n = 39.1 mol C/mol = 20.8 J/˚C·molVi = 876L Vf = 998 LTi = 0.0˚C Tf = 38.0˚CP = 1.00 atm = 101.3kPa (stays const)

q = ΔH =

W = –P.ΔV = –(101.3kPa)(122L) = –12,359J

ΔE = q + W = 30,905J + (-12,359J) = +18,546J

30,905J C11mol

C38 mol 39.1 J 20.8o

o

42. A 15.0 g sample of nickel metal is heated to 100˚C and dropped into 55.0 g of water, initially at 23˚C. Assuming that all the heat lost by the nickel is absorbed by the water, calculate the final temp of the nickel and water. (Specific heat of nickel = 0.444 J/˚C · g)

Heat lost by nickel = Heat gained by water

– m.cpNi.ΔT = + m.cpw

.ΔT

42. A 15.0 g sample of nickel metal is heated to 100˚C and dropped into 55.0 g of water, initially at 23˚C. Assuming that all the heat lost by the nickel is absorbed by the water, calculate the final temp of the nickel and water. (Specific heat of nickel = 0.444 J/˚C · g)



.ΔT

Cof

f

ff

o

of

o

of

2.25 T

5954 T 237

5288-230T666-6.66T

Cg

C23-T 55g J 4.18

Cg

C100-T 15g J 0.444

43. A 150.0-g sample of metal at 75.0˚C is added to 150.0 g of H2O at 15.0˚C. The temperature of water rises to 18.3˚C. Calculate the specific heat capacity of the metal, assuming that all the heat lost is gained by the water.



.ΔT

43. A 150.0-g sample of metal at 75.0˚C is added to 150.0 g of H2O at 15.0˚C. The temperature of water rises to 18.3˚C. Calculate the specific heat capacity of the metal, assuming that all the heat lost is gained by the water.



.ΔT

Cp

o

oo

o243.0C

Cg

C3.3 150.0g J 4.18C56.7 g 150.0

gJ

pC

Ch6.2C More Calorimetry and EnthalpyEx1) 1.00L of 0.10M Ba(NO3)2 solution at 25.0oC is mixed with 1.00L of 0.10M Na2SO4 soln at 25.0oC in a calorimeter. A ppt forms.The temp of the mix reaches 28.1oC.

If the Cp of the solution is , and the density of the final soln is 1.0 g/ml, calc the enthalpy change

per mole of ppt.

CgJ

18.4

Ex2) 15.00 g of CaCl2 at 21.1oC is dissolved in 50.0mL of water also at 21.1oC in a calorimeter. The temp of the mix reaches 42.5oC. Heat absorbed by the calorimeter is negligible.If the Cp of the solution is , and the density of the final soln is 1.0 g/ml, calc the enthalpy change per mole of ppt.

Ch6 HW#6 p284 37,45,47,48

CgJ

18.4

Lab6.1 Calorimetry

- unknown due @ end of period.

- lab write up due 2 days

- Ch6 HW#6 due tomorrow

Ch6 HW#6 p284 37,45,47,4837. The specific heat capacity of aluminum is 0.9 J/˚C·g. a. Calculate the energy needed to raise the temperature of an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.

cp = 0.9 J/˚C·gm = 8.5x102 gΔT = 71.8˚C

energy needed = ? q = m.cp.ΔT

b. Calculate the molar heat capacity of aluminum.

37. The specific heat capacity of aluminum is 0.9 J/˚C·g. a. Calculate the energy needed to raise the temperature of an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.

cp = 0.9 J/˚C·gm = 8.5x102 gΔT = 71.8˚Cenergy needed = ?


J927,54 C11g

C71.8 g 850 J 0.9o

o

?Cmol

J

o

37. The specific heat capacity of aluminum is 0.9 J/˚C·g. a. Calculate the energy needed to raise the temperature of an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.

cp = 0.9 J/˚C·gm = 8.5x102 gΔT = 71.8˚Cenergy needed = ?


J927,54 C11g

C71.8 g 850 J 0.9o

o

Cmol4.24

Al 1mol Cg

Al 27.1g J 0.9oo

J

45. In a coffee-cup calorimeter, 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed to yield the following reaction:

NIE: Ag+(aq) + Cl-(aq) →AgCl (s)

The two solutions were initially at 22.60˚C, and the final temperature is 23.40˚C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 g and has a specific heat capacity of 4.18 J/˚C·g.

ΔT = 0.8 ˚C Ag+(aq) + Cl-(aq)

water AgCl (s) + heat (ΔH)

cp = 4.18 J/˚C·g V=0.05L V=0.05L m = 100g

ΔH: kJ/mol = ? (q/n) M=0.100M M=0.100M

ΔT = 0.8 ˚C Ag+(aq) + Cl-(aq)

water AgCl (s) + ____J

cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product

ΔH: kJ/mol = ? (q/n) 100g total soln

q = + m.cp.ΔT

0.005 moles yielded 334J of energy to the 100g of solution,so how much heat would 1mol yield?

Jq 4.334 Cg

C(.8) 100.0g J 4.18o

o

ΔT = 0.8 ˚C Ag+(aq) + Cl-(aq)

water AgCl (s) + heat (q)

cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product

kJ/mol = ? (q/n) 100g total soln

q = + m.cp.ΔT

While the entire mass of solution had its temperature changed because of the rxn, only the moles of the rxn provided the heat! - 0.005 moles yielded 334J of energy to the 100g of solution, - 1 mol of reactants woulda provided 66,900J to the solution!

molkJ9.66

AgCl mol 0.005

J 334.4

Jq 4.334 Cg

C(.8) 100.0g J 4.18o

o

47.Consider the dissolution of CaCl2:

CaCl2(s) → Ca2+(aq) + 2Cl-(aq) ΔH = -81.5 kJ

An 11.0g sample of CaCl2 is dissolved in 125 g of water, with both substances at 25.0˚C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/˚C·g

CaCl2(s) water

Ca2+(aq) + 2Cl-(aq) + 81.5 kJ

m=11g m=125g Ti = 25C Ti = 25C per 1 mol CaCl2

47. CaCl2(s) water

Ca2+(aq) + 2Cl-(aq) + 81.5 kJ

m=11g m=125g Ti = 25C Ti = 25C per 1 mol CaCl2

q = – m.cp.ΔT

kJ 44.8CaCl mol 1 CaCl 111.1g

81.5kJ CaCl mol 1 CaCl 11.5g

22

22

C8.39 T

Cg

C25-T 136g J 4.188436

of

o

of

J

48. Consider the reaction

2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ

Calculate the heat when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.500 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0˚C and that the final mixture has a mass of 400.0 g and a specific heat capacity of 4.18 J/˚C · g, calculate the final temperature of the mixture.

2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2H2O(l) +118 kJ 0.10mol 0.15mol

Ti = 25.0˚C

this will run out q = – m.cp.ΔT

48. 2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2H2O(l) +118 kJ 0.10mol 0.15mol total mass = 400g

Ti = 25.0˚C

q = – m.cp.ΔT

Tf = 32.1˚C

Cg

C25-T 400g J 4.1811,800J

o

of

kJ 8.11HCl 1mol

kJ 118 HCl 0.10mol

Ch6.3 – Hess’s Law

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes placein one step or a series of steps.

1 step: N2(g) + 2O2(g) 2NO2(g) ΔH1 = 68 kJ

2steps: N2(g) + O2(g) NO(g) ΔH2 = 180 kJ2NO(g) + O2(g) 2NO2(g) ΔH3 = –112 kJ

Net rxn:

Ch6.3 – Hess’s Law

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes placein one step or a series of steps.

1 step: N2(g) + 2O2(g) 2NO2(g) ΔH1 = 68 kJ

2steps: N2(g) + O2(g) NO(g) ΔH2 = 180 kJ2NO(g) + O2(g) 2NO2(g) ΔH3 = –112 kJ

Net rxn: N2(g) + 2O2(g) 2NO2(g) ΔH2 + ΔH3 = 68 kJ

Ex1) Calculate ΔH for converting graphite to diamond,

C(s)graphite C(s)diamond using the enthalpies for combustion given:

C(s)graphite + O2(g) CO2(g) ΔH = -394kJ C(s)diamond + O2(g) CO2(g) ΔH = -396kJ

Ex2) Calculate ΔH for the synthesis of diborane (B2H6),given the following reactions:

2B(s) + 3/2O2(g) B2O3(s) ΔH = –1273 kJB2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) ΔH = –2035 kJ H2(g) + ½O2(g) H2O(l) ΔH = –286 kJ H2O(l) H2O(g) ΔH = 44 kJ

HW#55) Given the following data:

2O3(g) → 3O2(g) ΔH = -427 kJO2(g) → 2O(g) ΔH = +495 kJ

NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ

calculate ΔH for the reaction

NO(g) + O(g) → NO2(g)

Ch6 HW#7 p284 51,53,55,57

HW#55) Given the following data:

2O3(g) → 3O2(g) ΔH = -427 kJO2(g) → 2O(g) ΔH = +495 kJ

NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ


2NO(g) + 2O3(g) → 2NO2(g) +2O2(g) ΔH = -398 kJ 3O2(g) → 2O3(g) ΔH = +427 kJ

2O(g) → O2(g) ΔH = -495 kJ2NO(g) + 2O(g) → 2NO2(g) ΔH = -466 kJ

NO(g) + O(g) → NO2(g) ΔH = -233 kJ

Ch6 HW#7 p284 51,53,55,57

Ch6 HW#7 p284 51,53,55,5751. The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is -283.3 kJ/mol CO. Use these data to calculate ΔH for the reaction


2C(s) + O2(g) → 2CO(g)


2C(s) + O2(g) → 2CO(g)

2C(s) + 2O2(g) → 2CO2(g) ΔH = -787.4 kJ/mol

2CO2(g) → 2CO(g) + O2(g) ΔH = +566.6 kJ/mol

2C(s) + O2(g) → 2CO(g) ΔH = -220.8 kJ/mol

53. Given the following data:

S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ


S(s) + O2(g) → SO2(g)


S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ


S(s) + O2(g) → SO2(g)

2S(s) + 2(3/2O2(g)) → 2SO3(g) ΔH = -790.4 kJ 2SO3(g) → 2SO2(g) + O2(g) ΔH = +198.2 kJ

2S(s) + 2O2(g) → 2SO2(g) ΔH = -592.2 kJ

S(s) + O2(g) → SO2(g) ΔH = -296.1 kJ

55. In class57. Given the following data:

H2(g) + 1/2O2(g) → H2O(l) ΔH = -285.8 kJ N2O5(g) + H2O(l) → 2HNO3(l) ΔH = -76.6 kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) ΔH = -174.1 kJcalculate the ΔH for the reaction

2N2(g) + 5O2(g) → 2N2O5(g)


4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ 10(H2(g) + 1/2O2(g) → H2O(l)) ΔH=10(-285.8)kJ 2(2HNO3(l) → N2O5(g) + H2O(l)) ΔH =2(+76.6)kJ



2N2(g) + 6O2(g) + 2H2(g)→ 4HNO3(l) ΔH=-696.4kJ 10H2(g) + 5O2(g) → 10H2O(l) ΔH =-2858kJ 4HNO3(l) → 2N2O5(g) + 2H2O(l) ΔH =+153.2kJ 12H2(g) + 6O2(g) → 12H2O(l)

+2N2(g) + 5O2(g) → 2N2O5(g) ΔH =-3401.2kJ total



2N2(g) + 6O2(g) + 2H2(g)→ 4HNO3(l) ΔH=-696.4kJ 10H2(g) + 5O2(g) → 10H2O(l) ΔH =-2858kJ 4HNO3(l) → 2N2O5(g) + 2H2O(l) ΔH =+153.2kJ 12H2(g) + 6O2(g) → 12H2O(l)

+2N2(g) + 5O2(g) → 2N2O5(g) ΔH =-3401.2kJ total 12(H2O(l) → H2(g) + ½O2(g)) ΔH=12(+285.8)kJ

2N2(g) + 5O2(g) → 2N2O5(g) ΔH =-28.4kJ

Ch6.4 – Standard Enthalpies of Formation

C(s)graphite C(s)diamond

Enthalpy

time

Standard Enthalpy of Formation (ΔHfo)

- change in enthalpy that accompanies the formationof 1 mole of a substance from its elements.(All substances in their standard states)


- change in enthalpy that accompanies the formationof 1 mole of a substance from its elements.(All substances in their standard states)

- use the state of the substance as it exists at 1 atm and 25oC.Exs) oxygen: _____ sodium: ______ mercury: ______

- for gases use a pressure of 1 atm. - for solutions use a concentration of 1M.


- change in enthalpy that accompanies the formationof 1 mole of a substance from its elements.

- use the state of the substance as it exists at 1 atm and 25oC. - for gases use a pressure of 1 atm. - for solutions use a concentration of 1M. - elements in their standard states = 0. - multiply # of moles to ΔH

ΔHoreaction = ΣnpΔHf

o(products) – Σnr Δhfo(reactants)

Ex1) Use the standard enthalpies of formation listed in Table6.2 to calc the standard enthalpy change for the given reaction:

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

Ex2) Use the standard enthalpies of formation listed in Table6.2 to calc the standard enthalpy change for the given reaction:

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)

Ch6 HW#8 p285 61,65,66Ch6 Rev p283 28,35(Go over B4 lab pt 1)

Ch6 HW#8 p285 61,65,66 Ch6 Rev p283 28,35 (Go over B4 lab pt 1)61. Use the values of ΔHf

o in Appendix 4 to the calculate ΔHo for the following reactions.

a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g) 2mol(-80kJ/mol)+ 3(0) + 2mol(-75kJ/mol) 2mol(135.1kJ/mol) + 6mol(-242kJ/mol)

ΔHoreaction = (products) – (reactants)

61. Use the values of ΔHf

o in Appendix 4 to the calculate ΔHo for the following reactions.

a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g) 2mol(-46kJ/mol)+ 3(0) + 2mol(-75kJ/mol) 2mol(135.1kJ/mol) + 6mol(-242kJ/mol)

(-92kJ) + (-150kJ) (270kJ) + (-1452kJ)

(-242kJ) (-1182kJ) ΔHo

reaction = (-1182kJ)– (-242kJ) = -940kJ

61 - b. Ca(PO4)2(s) + 3H2SO4(l) → 3CaSO4(s) + 2H3PO4(l) 1mol(-4126kJ/mol) + 3mol(-814kJ/mol) 3mol(1433kJ/mol)+2mol(-1267kJ/mol)


61 - b. Ca(PO4)2(s) + 3H2SO4(l) → 3CaSO4(s) + 2H3PO4(l) 1mol(-4126kJ/mol) + 3mol(-814kJ/mol) 3mol(-1433kJ/mol)+2mol(-1267kJ/mol)

(-4126kJ) + (-2442kJ) (-4299kJ) + (-2534kJ)

(-6568kJ) (-6833kJ) ΔHo

reaction = (products) – (reactants)

ΔHoreaction = (-6833kJ)– (-6568kJ) = -265kJ

61 - c. NH3(g) + HCl(g) → NH4Cl(s) 1mol(-46kJ/mol) + 1mol(-92kJ/mol) 1mol(-314kJ/mol)


61 - c. NH3(g) + HCl(g) → NH4Cl(s) 1mol(-46kJ/mol) + 1mol(-92kJ/mol) 1mol(-314kJ/mol)

(-46kJ) + (-92kJ) (-314kJ)

(-138kJ) (-314kJ) ΔHo


ΔHoreaction = (-314kJ)– (-138kJ) = -176kJ

65. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction isCalculate the ΔHo for this reaction.

3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g) 3(0) + 3mol(-295kJ/mol) 1mol(-1676)+1mol(-704)+1mol(90)+6(-242)

ΔHo


ΔHoreaction =

65. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction isCalculate the ΔHo for this reaction.

3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g) 3(0) + 3mol(-295kJ/mol) 1mol(-1676)+1mol(-704)+1mol(90)+6(-242)

(-885kJ) (-1676kJ) + (-704kJ) + (+90) + (-1452)

(-885kJ) (-3742kJ) ΔHo


ΔHoreaction = (-3742kJ)– (-885kJ) = -2857kJ exothermic

66. The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide for propulsion: Calculate the ΔHo for this reaction.

4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g) 4(54) + 5(-20) 12(-242) + 9(0) + 4(-393.5)


66. The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide for propulsion: Calculate the ΔHo for this reaction.

4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g) 4(54) + 5(-20) 12(-242) + 9(0) + 4(-393.5)

(+116kJ) (-4478kJ)


ΔHoreaction = (-4478kJ)– (+116kJ) = -4594kJ very exothermic!

Do u think that might bea very powerful andvery spontaneous reaction?!?

Where does all the energy come from?

Ch6 Rev p283 28,35,46,54,6428. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of 30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00atm and 100°C, 40.66kJ of heat is released. If the density of H2O(l) at the this temp and pressure is 0.996 g/cm3, calc ∆E for the condensation of one mole of of H2O.

H2O(g) H2O(l)

1mol d= 0.996g/cm3 V=30.6LP=1atmT=100°C

∆E =?

28. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of 30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00atm and 100°C, 40.66kJ of heat is released. If the density of H2O(l) at the this temp and pressure is 0.996 g/cm3, calc ∆E for the condensation of one mole of of H2O.

H2O(g) H2O(l) + 40.66kJ 1mol 1mol

V=30.6L d= 0.996g/cm3 ΔH qP=1atmT=100°C

∆E =?

∆E = q + W work done on system

∆E = q + (-P.∆V) ∆E = (-40.66kJ) + (101.3kPa) .∆V



V=30.6L d= 0.996g/cm3 qP=1atmT=100°C

∆E =? ∆V = 0.018L – 30.6L = -30.58L


∆E = q + (-P.∆V) ∆E = (-40.66kJ) + (101.3kPa) .∆V

0.0181L1.180.996g OH 1mol

1cm OH 18.0g OH 1mol 3

2

322 cm



V=30.6L d= 0.996g/cm3 qP=1atmT=100°C

∆E =? ∆V = 0.018L – 30.6L = -30.58L


∆E = q + (-P.∆V) ∆E = (-40.66kJ) + (-(101.3kPa) .∆V) ∆E = (-40.66kJ) + (-(101.3kPa) .(-30.58L)) ∆E = -40.66kJ + (+3.097kJ) = -37.56kJ

0.0181L1.180.996g OH 1mol

1cm OH 18.0g OH 1mol 3

2

322 cm

35. Consider the combustion of propane:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) + 1.3x108J

Assume that all the heat in sample Exercise 6.3 comes from the combustion of propane. What mass of propane must be burned to furnish this amount of energy assuming the heat transfer process is 60% efficient? 1.3x108J (from 6.3) 60% efficient means you gotta burn a whole lot more

1.3x108J / 0.60 = 2.17x108J will get you there.

35. Consider the combustion of propane:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) + 1.3x108J ?g

83

838

83838

HC g3.73HC 1mol J1.3x10

HC 44.0g HC 1mol J2.17x10

Rev – Day 2 (Lab6.2 Day 2)46. In a coffee-cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00oC. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34o C. Assuming the solution has a heat capacity of 4.18 J/oC.g and assuming that no heat loss to the calorimeter. Calculate the enthalpy charge for the dissolution of NH4NO3 in units of kJ/mol .

NH4NO3 water NH4+ + NO3

- m=1.60g m=75.0g ∆H = ? (kJ/mol)

Ti = 25.00C Ti = 25.00C

Tf = 23.34C Tf = 23.34C

Heat absorbed by ammonium nitrate, comes from the soln,and is measured in the temp change of the total soln:

q = – m.cp.ΔT

46. NH4NO3 water NH4+ + NO3

- m=1.60g m=75.0g

Ti = 25.00C Ti = 25.00C

Tf = 23.34C Tf = 23.34C

Heat absorbed by ammonium nitrate, comes from the soln,and is measured in the temp change of the total soln :

q = – m.cp.ΔT

J 532 q

Cg

C1.66 76.6g J 4.18o

o

q

46. NH4NO3 +____ water NH4+ + NO3

- m=1.60g m=75.0g

Ti = 25.00C Ti = 25.00C

Tf = 23.34C Tf = 23.34C


q = – m.cp.ΔT

When 1.60g NH4NO3 dissolved, 532 J (q) was released.How much heat (ΔH) would 1 mol produce?

J 532 q

Cg

C1.66 76.6g J 4.18o

o

q

46. NH4NO3 + 26.6kJ water NH4+ + NO3

- m=1.60g m=75.0g

Ti = 25.00C Ti = 25.00C

Tf = 23.34C Tf = 23.34C


q = – m.cp.ΔT

When 1.60g NH4NO3 dissolved, 532 J (q) was released.How much heat (ΔH) would 1 mol produce?

J 532 q

Cg

C1.66 76.6g J 4.18o

o

q

J 600,26 mol 1 NONH 1.60g

NONH 80.0g J 532

34

34


C2H2(g) + 5/2O2(g) 2CO2(g) + H20(l) ∆H= -1300kJ C(s) + ½O2(g) CO2(g) ∆H= -394kJ H2 (g) + ½O2(g) H2O(l) ∆H= -286kJCalculate ∆H for the reaction:

2C (s) + H2 (g) C2H2 (g)

64.Calculate ∆H˚ for each of the following reactions using the data provided:2Na(s)+2H2O(l) → 2NaOH(aq)+H2(g)

2Na(s)+CO2(g) → Na2O(s)+CO(g)

Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of...

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Transcript of Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of...