Hour exam 106

1
Hour exam- 10/6

• -Please arrive at 830. Those who must arrive at
  915 due to a time conflict will sit in the front
  row of the auditorium.
• -Please bring two pens or pencils AND a
  calculator
• -You will be provided with a periodic table and a
  table illustrating the activity series as well as
  any important constants.
• -Sections 1-4 will take the exam in 201 White
  Gravenor
• -Everyone else will take the exam in 103 Reiss
  Science
• -Those who must take the exam on Tuesday should
  report to my office (601 B Reiss Science). You
  will be taking the exam next door in my
  laboratory.
• -The exam will end promptly at 1005. You will be
  asked at that time to put down your pencil.
  Working on your exam beyond that time will be
  considered an infraction of the Honor Code.

2
Enthalpies of Reaction
Recall from last time The change in enthalpy
for a reaction is ?H Hfinal Hinitial and is
equal to the heat transferred under conditions of
constant pressure. For a reaction, this means
?H enthalpy of reaction Hproducts
Hreactants Example 4 Fe(s) 3 O2 (g) ?
2 Fe2O3 (s) ?H -1644.52 kJ This an
exothermic reaction since the change is negative.
(The system loses energy to the surroundings.)
3
Enthalpies of reaction

• It makes sense that if twice as many mols of
  reactant were used that twice as much heat would
  be released. This is correct, ENTHALPY IS AN
  EXTENSIVE PROPERTY.
• It also makes sense that if the reaction were
  reversed that an equal quantity of energy would
  have to be transferred from the surroundings to
  the system in the re-formation of reactants. This
  is also correct. THE ENTHALPY CHANGE FOR THE
  REVERSE REACTION IS EQUAL IN MAGNITUDE, BUT OF
  THE OPPOSITE SIGN.
• Finally, since heat would have to be added to
  the reactants to form iron(II) oxide from liquid
  iron metal, that would mean that the change in
  enthalpy would be different. THE ENTHALPY CHANGE
  FOR A REACTION DEPENDS ON THE PHYSICAL STATE OF
  THE REACTANTS AND PRODUCTS.

4
Enthalpies of Reaction

• Enthalpy is an extensive property
• If the enthalpy change for
• 2 H2 (g) O2 (g) ? 2 H2O (g)
• is -483.5 kJ, then the enthalpy change for
• 4 H2 (g) 2 O2 (g) ? 4 H2O (g)
• is 2 x (-483.6 kJ) or -967.2 kJ

5
Enthalpies of Reaction

• The enthalpy change for the reverse of a reaction
  is equal in magnitude, but of the opposite sign.
• If the enthalpy change for
• 2 H2 (g) O2 (g) ? 2 H2O (g)
• is -483.6 kJ, then the enthalpy change for
• 2 H2O (g) ? 2 H2 (g) O2 (g)
• is (483.6 kJ) or 483.6 kJ

6
Enthalpies of Reaction

• The enthalpy change for a reaction depends on the
  physical state of the reactants and products.
• If the reaction between hydrogen and oxygen
  gas is written to give water as a liquid, there
  must have been additional heat lost by the system
  in order to condense the water. This shows up as
  an increase in the enthalpy of reaction.
• 2 H2 (g) O2 (g) ? 2 H2O (g) ? H -483.6
  kJ
• 2 H2 (g) O2 (g) ? 2 H2O (l) ? H -571.6 kJ

7
Hesss Law

• In the reactions describing the formation of
  water from hydrogen and oxygen gas, the ? H
  values must differ by the amount of heat
  associated with condensing 2 moles of water from
  the gaseous to the liquid state. We can find the
  amount of heat by looking at the chemical
  equation that describes the reaction producing
  H2O (l) as the sum of two equations
• 2 H2 (g) O2 (g) ? 2 H2O (g) ? H -483.6 kJ
• 2 H2O (g) ? 2 H2O (l) ? H ?
• 2 H2 (g) O2 (g) ? 2 H2O (l) ? H -571.6 kJ

8
Hesss Law

• 2 H2 (g) O2 (g) ? 2 H2O (g) ? H -483.6 kJ
• 2 H2O (g) ? 2 H2O (l) ? H ?
• 2 H2 (g) O2 (g) ? 2 H2O (l) ? H -571.6 kJ
• This means that the enthalpy change for the
  condensation step must be
• ? H 571.6 (483.6 kJ) 88.0 kJ
• This is an example of the application of Hesss
  Law
• If a reaction is carried out in a series of
  steps, ? H for the reaction will equal the sum of
  the enthalpy changes for the individual steps.

9
Hesss Law

• Example
• If the following enthalpy changes are given
• NO(g) O3 (g) ? NO2 (g) O2 (g) ? H -198.9
  kJ
• O3 (g) ? 3/2 O2 (g) ? H -142.3 kJ
  O2 (g) ? 2 O (g) ? H 495.0 kJ
• What is the change in enthalpy for the following
  reaction?
• NO(g) O (g) ? NO2 (g)

10
Hesss Law

• NO(g) O3 (g) ? NO2 (g) O2 (g) ? H -198.9
  kJ
• O3 (g) ? 3/2 O2 (g) ? H -142.3 kJ
  O2 (g) ? 2 O (g) ? H 495.0 kJ
• Which reactions have the same reactants as our
  reaction in which we are interested? The first
  and the reverse of the third. Since we are
  reversing the third, we must change the sign of
  the ? H.
• Will we need to multiply either reaction? Yes,
  the third reaction must be divided by 2 in order
  to yield 1 oxygen atom. This means we must divide
  the ? H for that reaction by 2.
• The second reaction can be reversed to get rid of
  the extra reactants and products.

11
Hesss Law

• OK as is NO(g)O3(g) ? NO2(g)O2(g) ?H -198.9
  kJ
• Reverse O3 (g) ? 3/2 O2 (g) ? H -142.3
  kJ
• Reverse x ½ O2 (g) ? 2 O (g) ? H 495.0
  kJ
• NO(g)O3(g) ? NO2(g)O2(g) ?H 198.9 kJ
• 3/2 O2 (g) ? O3 (g) ? H 142.3 kJ
• O (g) ? ½ O2 ? H 247.5 kJ
• NO(g) O (g) ? NO2 (g) ? H 304.1 kJ

12
Enthalpies of Formation

• If we could somehow have a value of ?H for the
  reactions that correspond to the enthalpy change
  for the formation of each reactant and product
  from the constituent elements, it would be less
  like a puzzle to find ?H. ? Enthalpies of
  formation, ?Hf
• These are defined as the enthalpy change for the
  reaction to form one mole of a given substance
  from the constituent elements in their standard
  state (the form in which they exist at 25ºC). The
  enthalpy of formation of elements in their
  standard states are defined as 0 kJ/mol.
  (Tabulated in Appendix C at the back of your
  book.)

13
Enthalpies of Formation

• Example
• NO(g) O (g) ? NO2 (g) can be broken down into
• Rxn 1 ½ N2(g) ½ O2(g) ? NO (g) (must
  reverse)
• Rxn 2 ½ O2(g) ? O(g) (must reverse)
• Rxn 3 ½ N2(g) O2(g) ? NO2(g) (use as is)
• ?Hoverall rxn -?H1 -?H2 ?H3
• Note the ?Hf values for reactants are all
  reversed in sign while those for products are as
  tabulated. In general
• ?Hoverall rxn S ?Hf,,products -S ?Hf,
  reactants