Heating and Cooling Curves

1
Heating and Cooling Curves
2
Phase Changes

• When a substance undergoes a phase change, the
  TEMPERATURE REMAINS CONSTANT.
• All energy absorbed or released is due to
  rearrangement of intermolecular forces NOT a
  change in molecular motion.

3
Calculating Q

• There is no temperature change so we can not use
• Q m Cp ?T
• ?H (enthalpy) is used instead of specific heat.
• Q m ?H
• There are different ?H values for each phase
  change (?Hf enthalpy of fusion is for
  melting/freezing ?Hv is for boiling/condensing.)

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VALUES FOR WATER

• Cp of water 4.18 J/gC
• Cp of ice 2.06 J/gC
• Cp of steam 2.03 J/gC
• ?Hf 334 J/g
• ?Hv 2261 J/g

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Solid
Liquid
Gas
Kinetic energy ? (molecules move faster) and
temperature ? until melting point is
reached THEN
Energy ? (molecules move faster) and temperature
? until boiling point is reached THEN
Energy ? (molecules move slower) and temperature
? until condensation point is reached THEN
Substance melts but temperature remains the same
until it is completely melted (all energy goes
into breaking bonds (changing PE) not making
molecules move faster).
Substance boils but temperature remains the same
until it is completely vapor.
Substance condenses but temperature remains the
same until completely liquid.
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Heating Cooling Curve
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Section 1

• This is ice, below the melting point. As energy
  is added, the temperature rises because the
  molecules are moving faster (change in Kinetic
  Energy (?KE)).
• Since energy is ADDED, this is an ENDOTHERMIC
  process.

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Section 2

• This begins with ice at the melting point. As
  energy is added, the ice melts and becomes liquid
  (at the very end of the line it is now completely
  liquid at the melting point). The bonds are
  breaking (change in Potential Energy (?PE)).

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Section 3

• This is water, below the boiling point. As
  energy is added, the temperature rises because
  the molecules are moving faster (?KE).

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Section 4

• This begins with water at the boiling point. As
  energy is added, the water boils and becomes gas
  (at the very end of the line it is now completely
  gas at the boiling point). The bonds are
  breaking (?PE).

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Section 5

• This is steam (gas), above the boiling point. As
  energy is added, the temperature rises because
  the molecules are moving faster (?KE).

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BOILING POINT
MELTING POINT
13
Another Diagram
This is an EXOTHERMIC process. The substance must
give off energy to cool, condense and freeze.
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• Enthalpy, Entrophy, and Hesss Law

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Enthalpy

• The heat absorbed or released by a reaction at
  constant pressure is the change in enthalpy (?H).
• q ?H
• This value (?H) is often calculated for one mole
  of product formed. Multiply or divide as
  necessary if more or less than one mole is
  formed.

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RULES

• ?H is directly proportional to the quantity of a
  substance that reacts or is produced by a
  reaction.
• Example
• 1 H2 ½ O2 ? 1 H2O ?H -285.8 kJ
• 2 H2 1 O2 ? 2 H2O ?H -571.6 kJ

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• ?H for a reaction is equal in magnitude but
  opposite in sign for the reverse reaction.
• Example
• HgO ? Hg ½ O2 ?H 90.7 kJ
• Hg ½ O2 ? HgO ?H -90.7 KJ

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Endothermic or Exothermic?

• If ?H is negative (-) the
• reaction is exothermic and
• heat is given off.
• If ?H is positive the reaction is
• endothermic and heat is
• absorbed.

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Entropy (S)

• A measure of the disorder of a system.
• The universe tends towards a higher entropy.
  Natural processes create more disorder.
• Gases have the highest entropy, solids the
  lowest.

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Chemical Reactions

• Chemical reactions can be classified as
• Spontaneous once started, will continue on
  their own.
• Non-spontaneous must be pushed along
  for the entire
  reaction.

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How can we tell?

• SPONTANEOUS

• ?H is negative exothermic
• ?S is positive entropy increases

• NON-SPONTANEOUS

• ?H is positive endothermic
• ?S is negative entropy decreases

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Gibbs Free Energy

• Used to determine spontaneity
• ?G ?H - T ?S
• If ?G is negative, the reaction is spontaneous.
• WE WILL NOT BE CALCULATING THESE VALUES!
• Temperature is important in determining
  spontaneity of a reaction where either entropy
  or enthalpy is not favorable.

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Temperature

• Temperature is important in determining
  spontaneity of a reaction where either entropy
  or enthalpy is not favorable
• Example Think of an ice cube placed on a warm
  stove vs. an ice cube in a freezer.
• The ice cube on the stove spontaneously melts
  (even though its an endothermic process) while
  the ice cube in the freezer does not (even though
  it decreases entropy) The temperature makes the
  difference!

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• Hesss Law

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Hesss Law

• ?H of a reaction ? (? Hs of the parts)
• If Rxn 1 Rxn 2 Rxn 3, then ?H1 ?H2 ?H3
• Adding reactions species that exist on both
  sides of the equation cancel each other out.
  A B 2C ? 4D
  4D ? 2C G
  . E F ? A 3H .
• B E F ? G 3H

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Example 1

• Determine the enthalpy change for the reaction
• C(s, diamond) ? C(s, graphite)
• Use the following information
• 1. C(s, diamond) O2(g) ? CO2(g) ?H
  -395.40 kJ/mol
• 2. C(s, graphite) O2(g) ? CO2(g) ?H
  -393.51 kJ/mol

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• 1. C(s, diamond) O2(g) ? CO2(g) ?H
  -395.40 kJ/mol
• 2. C(s, graphite) O2(g) ? CO2(g) ?H
  -393.51 kJ/mol (Reverse this reaction)
• CO2(g) ? C(s, graphite) O2(g) ?H 393.51
  kJ/mol
• __________________________________________________
  ____________________________
• C(s, diamond) ? C(s, graphite) ?H
  -395.40 393.51-1.89 kJ

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• Example 2

• Calculate the ?H for C2H4(g) 6F2 ? 2CF4(g)
  4HF(g)
• Use the three reactions below.
• 1 C(s) 2F2 ? CF4(g) ?H -680
  kJ/mol
• 2 2HF(g) ? H2(g) F2(g) ?H 537 kJ/mol
• 3 2C(s) 2H2(g) ? C2H4(g) ?H 52.3 kJ/mol
•  

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• Reaction 3 reverse to put C2H4 on the left
  (change sign of ?H)
• Reaction 1 multiply by 2 (including ?H) to
  cancel 2C
• Reaction 2 reverse and multiply by 2 to get
  4HF on the right
• C2H4(g) ? 2C(s) 2H2(g) ?H -52.3 kJ
• 2C(s) 4F2 ? 2CF4(g) ?H
  -1360 kJ
• 2H2(g) 2F2(g) ? 4HF(g) ?H
  -1074 kJ .
• C2H4(g) 6F2 ? 2CF4(g) 4HF(g) ?H -
  2486 kJ