JOHN J. ALEXANDER
University of Cincinnati Cincinnati. 45221
The Temperature Dependence of the Equilibrium Constant
Answer 1) The potential energy diagram far the reaction would be:
James H. Burness The Pennsyluania State Uniuersity York Campus York, Pennsyluania 17403 Students in a first-year, college-level chemistry course often spend a great deal of time studying equilihrium constants and using them to perform calculations for chemical systems such a s weak acid-base equilibria, solubility product equilibria, equilibria in buffer solutions, homogeneous gas phase equilibria, etc. In addition to examining the effect of changing concentrations on the position of equilibrium a t a given temperature, students also learn that the value of K,, for a given reaction is dependent on temperature. Most freshman chemistry texts point out that the temperature dependence of the equilibrium constant can be predicted by use of Le Chatelier's Princiole: if heat is added to an equilihrium system a t constant pessure, the equilihrium shifts in the direction which absorbs heat. This is memorized by the student as a qualitative description of what happens, but often it is not clear to the student that the result can be derived rigorously. This exam question tests the student's ability to derive the temperature dependence of an equilibrium constant not by qualitatively applying Le Chatelier's Principle, hut by understanding the relationship between the kinetics of the reaction and the value of K,,. This question requires cognitive response a t the level of synthesis. Question You have learned that if heat is added to an equilibrium system a t constant pressure, the equilibrium shifts in the direction which absorbs heat, and K,, changes. For example, for the equilihrium,
2COdg) = 2CO(g) + Ozk) the forward reaction is endothermic ( U o = 566 kJ) so an increase in temperature increases the equilibrium constant. Nuw,suppose that you are told that the general reacrion. ~
2) The equilibrium constant K,, for the reaction is derived as follows:
Suppose the rate constant for the forward and reverse reactions are kt and h,, respectively. Then we have
At equilibrium, the rates ofthe forward and reverse reactions are equal, and so k/ [A21 IBd = h, IAB12
Defining the equilibrium constant K,, = kflk,, it is obvious that K,, = [ABI2/[A21IB21. The effect of temperature changes on K,, depend on how temperature changes affect kt and k,. If, for example, a 10°C rise in temperature doubles kt but also doubles h,, then K,, will not change. 3) The temperature dependence of a rate constant k is given by the Arrhenius equation, which is expressed in its logarithmic form as Ink = InA - E , I R T . Ifwe are comparing the relative values of a rate constant (say kt) at two different temperatures, T z > TI, we have In h,,, = In A - EJRT,
~
uccurs in one step, with the same actkated complex formed whether tht: reaction proceeds in rhe t'orward or reverse directitm ri.r., the reaction is an elementarv reartionl. Civen that the forward rmc.tion is endothermic, show m n l h e ~ m o f ~ c nwh) l l ~ the equilibrium n m t a n r fur the reaction must Increase if the temperature increases. useful to draw a potential energ). Onrionnl H i n t . I t will diagram for the reaction, indicating the relative magnitude of the activation energies for the forward and reverse reactions.
and
ink,,^ = In A - E,IRT2 Subtracting the first equation from the second and rearranging, we have . .. . An analogous equation exists for the reverse reaction,
.
Exam Question Exchange offersteachers an opportunity to share prize exam questions with others. Guidelines for preparing exam questions for submission were outlined on page 608 of the October 1977 issue. All questions submitted become the property of the Journal of Chemical Education and will not be returned. Questions should be submitted to the column editor.
But far a given increase in temperature from T I to T z , we see that
It is obvious from the potential energy diagram that E. is a consequence of AH > 0. Thus,
> E.'; this
Volume 56, Number 6, June 1979 1 395
tially, however, student performance on the exercise since no lab lectures were given during the 197576 academic year. This question requires application of knowledge. Question
and therefore
Since all values of k are positive, it follows that
.
which means that K,, at T7is greater than K.,, at T 1Q.E.D. .
Below are pairs of organic compounds. Based on your experience and studv. decide which of the indicated method or methods are be; suited to achieve a good separation. Place the letter which corres~ondsto the method of separation in the blank provided. methanol and ethanol capric acid and salicylic acid ~henoland cmrie acid mesityl tetrafluoroborate and nheptad 5) n-heptanol and ethanol 61 alkaline sdution ofcapric acid and phenol 1) 2) 31 4)
gas chromatography b) simple distillation C) extraction of aaueous layer with benzene d) recrystallization from water e) steam distillation a)
Selected Data on Organic Compounds
Separation of O r g a n i c Compounds
Solubiiitv
Paul R. Loconto Dutchess Community College Poughkeepsie, New York 12601 One final exam exercise that I have found useful directly pertains to the various separation and purification techniques commonly studied by students in the first semester of an introductory course in organic chemistry. The exercise seems to be an accurate test af whether the student can choose the most suitable technique based on his or her knowledge of the theory behind each technique. The instructor normally provides lectures on the theory of these techniques during the course in addition to the laboratory modules which also provide theory as well as practice. Laboratory lectures in class during the 1977-78 academic year have increased substan-
396 1 Journal of Chemical Education
Camoaund
mp lDCl
methanol ethanol n-heptanol capric acid salicylic acid phenol rnesitvl tetrafluoroborate
-97 -115 34 31 159 41 20
bp l0C1
64.5 78.3 176 269 182
H.0
H,O'"
EIOH
V V S S S S V
V V S S V
V V V V V V S
v V
Answer 1) a ; 2) e
and d ; 3) d ;
4) e; 5) o
and b; 6 ) e
Benzene V V
v V V
v S
