Transcript HESS’S LAW
HESS’S LAW
ALSO CALLED…
Law of constant heat summation First law of thermodynamics
HESS’S LAW
States that: The total enthalpy of a reaction is independent of the reaction pathway.
PARAPHRASE OF HESS’S LAW
The heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps.
Mathematically, ΔH overall = sum of ΔH’s of individual reactions.
ENTHALPY DIAGRAM FOR HESS’S LAW
IMPORTANCE OF HESS’ S LAW
Allows us to calculate the enthalpy changes of reactions that cannot be measured directly in the laboratory.
EXAMPLE
In the production of CO 2 , there are 2 reaction paths that can be taken before final product is obtained. When the enthalpy change of these 2 reactions are added, the enthalpy change of the overall reaction is obtained.
EXAMPLE
C + CO + ½O 2 ½O 2 CO CO 2 C + O 2 CO 2 Using Hess’s law, H 3 = H 1 + H 2 H 3 = Y kJ + Z kJ H 3 = X kJ H 1 H 2 = Y kJ = Z kJ H 3 = XkJ
MORE QUESTIONS
Two reactions occurring in the manufacture of sulfuric acid are shown below: S(s) +O 2 (g) SO 2 SO 2 (g) + ½O 2 (g) (g) Δ SO 3
H
(g)
1 Ө = –297 kJ
Δ
H 2 Ө = –92 kJ
Deduce the Δ
H Ө value of this reaction:
S(s) + 1½O
2
(g)
SO
3
(g)
(November 2005)
RULES FOR SOLVING QUESTIONS ON HESS’S LAW.
When it is necessary to reverse a chemical equation, change the sign of ΔH for that reaction. When multiply equation coefficients, multiply values of ΔH by the same factor. Make sure to rearrange the given equations so that reactants and products are on the appropriate sides of the arrows.
QUESTION 1
Calculate the enthalpy change, Δ
H 4 reaction for the
C + 2H 2 + ½O 2 CH 3 OH Δ
H 4
Using Hess’s Law, and the following information.
CH 3 OH + 1½O 2
mol -1
CO 2 + 2H 2 O Δ
H 1 = -676 kJ
C + O 2 H 2 + ½O 2 CO 2 H 2 O Δ
H 2
Δ
H 3 = -394 kJ mol -1 = -242 kJ mol -1 (May 2006)
QUESTION 2
Using the equations below Cu(s) + ½ O 2 (g) → CuO(s)∆
H ο = –156 kJmol -1
2Cu(s) + ½ O 2 (g) → Cu 2 O(s)∆
H ο = –170 kJmol -1
What is the value of ∆
H ο (in kJmol -1 ) for the following reaction?
2CuO(s) → Cu 2 O(s) + ½ O 2 (g) A. 142 B. 15 C. –15 D. –142 (May 2004)
QUESTION 3
Consider the following equations.
Mg(s) + ½O 2 (g) → MgO(s) ∆
H ο H 2 (g) +
½
O 2 (g) → H 2 O(g) ∆H ο = –602 kJ = –242 kJ
What is the ∆
H° value (in kJ) for the following reaction?
MgO(s) + H 2 (g) → Mg(s) + H 2 O(g) A. –844 B. –360 C. +360 D. +844 (November 2004)
HESS’S LAW ENERGY CYCLE
C CO + C + ½O 2 ½O 2 + O 2 CO CO 2 CO 2 Reactants H H = – 110.5 kJ = – 283.0 kJ H = – 393.5 kJ C + O 2 H CO = – 110.5 kJ + ½O 2 Intermediate H = – 393.5 kJ H = – 283.0 kJ Products CO 2
STEPS IN DRAWING ENTHALPY DIAGRAMS 1.
2.
3.
4.
5.
6.
7.
Balance the equation(s).
Sketch a rough draft based on H values.
Draw the overall chemical reaction as an enthalpy diagram with the reactants on one line, and the products on the other line.
Draw a reaction representing the intermediate step by placing the
relevant reactants
on a line.
Check arrows. Look at equations to help complete balancing (all levels must have the same number of all atoms).
Add axes and H values.
MORE EXAMPLES
GeO(s) Ge(s) + O 2 (g) GeO(s) + ½ O 2 (g) Ge(s) + ½ O 2 (g) GeO 2 (s) GeO 2 (s) Intermediate Reactants H = + 255 kJ H = – 534.7 kJ H = – 279.7 kJ Ge(s) + O 2 (g) H = + 255 kJ GeO(s) + ½ O 2 (g) Products GeO 2 (s) H = –280 kJ
TO TRY
Calculate the enthalpy change for the reaction below
NO(g) + ½ O 2 (g)
NO 2 (g)
Using Hess’s law and the following information: NO(g) ½ N 2 (g) + O 2 (g) ½ N NO 2 2 (g) + ½ O (g) 2 (g) H = – 90.37 kJ H = + 33.8 kJ With the above information, draw an enthalpy diagram for the reaction using the knowledge acquired from the above examples.
REFERENCE
www.chalkbored.com/lessons/chemistry-12/hess-law.ppt
http://ibchem.com/root_htm/defn/IBgloss.htm#Hess’ law http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.ht
ml http://www.sparknotes.com/testprep/books/sat2/chemistry/c hapter9section5.rhtml
http://www.bbc.co.uk/scotland/learning/bitesize/higher/chem istry/calculations_3/hess_rev1.shtml
http://chemactive.com/articulate/IB/Energetics_2010_Elearni ng_IB/player.html