edned by
GRACE FISCHER McGUFFlE NOrthwestern High School Hyansville. MD 20782
Teaching Hess's Law John Davik York Communitv Hizh School Elmhurst, IL 60126 The Law of Constant Heat Summation or Hess's Law states that the heat change in a chemical reaction is independent of any intermediate reactions; that is, i t is the same whether it takes place in one or several stages.' Problems of Hess's Law are a part of the high school chemistry course, but students sometimes have difficultv with such ~rohlems.Students can s if they master be taught to deal successkdly with ~ k s s ' Law a few simplified rules. These rules might help to end the confusion of many students in attempting to decide which equations are to he added or subtracted. T h e following procedure will work best if it is orally presented hv the teacher and aiven to the students via the blackboard with several examples worked in detail. If this procedure is reproduced and given to the class, disaster will bccur. students need the reminder of key words such as coefficients, terms, and phases. An additional caveat would caution the teacher to he certain that the students understand endothermic and exothermic reactions along with the AH notations that are routinely employed. Simplified rules 1) Balance the equation. 2) Select the reactions. 3) Multiply the reactions. 4) Transpose if necessary. 5 ) Add the reactions. f simplified rules, the teacher Expanding upon each t ~those explains to the students the hllowing.
Rule I. Balance the Equation. Write the equation that is to he solved so that all ofthe coefficients are whole numbers in a balanced equation. Rule 2. Select the Reactions. From achart of Heats of Reaction2 select a reactian for each term in the equation that is to be solved. No reaction is needed for any term that is an element. Rule 3. Multiply the Reactions. Multiply each reaction by a factor so thnt the coefficient of the term in thereaction matches the coeffi-
the reaction appears with the products, the reaction must be reversed. -~ ~.. ~
~
Rule 5 Add t h r R m t r o n c Add i h c eqoaticm. nnd cam 11 rhuie termr uhirh arr nor in theequztiun l h u t is tc,l,esolvtd Suhrrncr the hrnr term from thnt side of the C ~ U ~ I I OnI hI e w the hear term ir the smaller amount of energy, and of course, subtract the samequantity from the other side of the equation.
Example of Each Rule Problem What is the heat of reaction of the oxidation of ammonia to yield nitrogen dioxide and water vapor as the products.
-
NH~ISI + 0 2 ( g 1 NOzig~+ H20lg) Rule I . Balance the equation using whole number coefficients. 4 NHw
+ 7 021g1
-
4 NOzigl + 6 H&I
(1)
Rule 2. From Table 1 select a reaction for NHu, another reactian for NO2,and another reaction for HzOl,l.No equation for OXis needed since it is an element.
% Nzlg)+ 312 33.9 kJ
-
NHx,,
-
+ 46.1 kJ
+ % Nx,) + 0 2 1 ~ 1 NOzig~
Hzci + '12 Ox8)- H20ip)+ 241.8 kJ
(2) (3) (4)
Rule 3. Multiply the reactions so that the coefficients This feature investigates aspects of the secondary school chemistry curriculum and related topics. New developments will be introduced and established techniques critiqued. Contributions are welcome.
match. Match the coefficients of the terms in eqn. (1) with those in eqns. (2), (3),and (4). Equation (2) must he multiplied by 4 to match the 4 NH3; eqn. (3) must be multiplied by 4 to match the 4 NO*; and eqn. (4) must he multiplied by 6 to match the 6 H20. This means that
'Moore, Walter J., "Physical Chemistry," 4th ed., Prentice-Hall, Inc.,Englewood Cliffs,N.J., 1972, p. 55. 2Parry, et al., "Chemistry, Experimental Foundations," 2nd ed., Prentice-Hall,Englewood Cliffs, N.J., 1975, p. 207. (S.I. units are used here.) Volume 57. Number 12, December 1980 1 895
135.6 kJ 6 Hzigl
- +
+ 2 Nsig1+ 4 02ig) 4
(6)
+ 3 02(.)
(7)
6 H 2 0 1 ~ ) 1450.8 kJ
Rule 4. In eqn. (I),the 4 N H 3 appears as a reactant, hut in eqn. (5) the N H 3 appears as a product. Reverse (or transpose) eqn. (5) to obtain eqn. (8). 184.4 kJ
+ 4 NH31,)
+
+
2 Nzig) 6 Hz(.)
320.0kJ+4NH~+2Nz+70~+6H,-4NO~+ 6 H z 0 + 2 Na + 6 Hz
+ 1450.8 kJ
(8)
Since in eqn. (11, the 4 NO2 appears as a product, and in eqn. (6), the 4 N O 2 appears as a product, the equation is not reversed. Likewise, in equation one, the 6 H z 0 appears as a product, and in equation seven, the 6 H z 0 appears as a product, this equation is also not reversed. Rule 5. Adding eqns. (61, (7), and (8) the summation is ohtained.
896 1 Journal of Chemical Education
Cancelling those terms that are not a part of the equation that is to be solved. eonation one is obtained with the enerev -.terms in the equation. Thus
and subtracting the smaller amount of energy from each side of the equation provides the solution to the problem. 4 N H x p ) + 7 Ozig)
-
4 N02ig)
+ 6 H z 0 + 1130.8 kJ
or AH = -1130.8 kJl4 mol N H 3 = -282.7 kJ/mol N H 3 If the students would have an easier time with using Kcal rather than the S.I. unit, kJ, the conversion can be made by dividing kJ values by 4.18, since one calorie equals 4.18 joules.
