Chern 12 Notes , 2.8 Notes -Equilibrium Calculations (ICETablesl A chemical system can be thought of as being either: 1. At Equilibrium or 2. Not At Equilibrium (Initial) A system which is not at equilibrium will move spontaneously to a position of being at e uilibrium. I IYPe 1 Calculations - System is AT EQUILIBRIUM • If the system is at equilibrium, there are no changes in concentrations. • Keq can be easily calculated by "plugging" in the values for equilibrium concentrations into the Keq expression. • Whenever a question says something like " ...the equilibrium concentrations of the following are ..." or something to that effect, it is a Type 1 Calculation. Type 1 - Example: Given the equilibrium system: \ PCls cg )~ \ PCbcg) +\ Chcg) The system is analyzed at a certain temperature and the equilibrium concentrations are as follows: [PCl s ] = 0.32 M, [PCh] = 0.40 M and the [Cl-] = 0.40 M. Calculate the Keq for this reaction: L 9~\~~ [ C.AL1 ____ , •• ",. "" ----...--'"'"' ... .7.-.•.•-.•..-.--~- L v (.,\ S"~ \~ (O\4lJ -)CO,'1b) CO,"6:L) 0.50 1 I I I
Transcript
Chern 12 Notes,
2.8 Notes -Equilibrium Calculations (ICETablesl
A chemical system can be thought of as being either:
1. At Equilibrium
or 2. Not At Equilibrium (Initial)
A system which is not at equilibrium will move spontaneously to a position ofbeing at e uilibrium. I
IYPe 1 Calculations - System is AT EQUILIBRIUM
• If the system is at equilibrium, there are no changes in concentrations.• Keq can be easily calculated by "plugging" in the values for equilibrium
concentrations into the Keq expression.• Whenever a question says something like " ...the equilibrium
concentrations of the following are ..." or something to that effect, it is aType 1 Calculation.
Type 1 - Example:Given the equilibrium system: \ PClscg) ~ \ PCbcg) +\ Chcg)
The system is analyzed at a certain temperature and the equilibriumconcentrations areas follows:
[PCls] = 0.32 M, [PCh] = 0.40 M and the [Cl-] = 0.40 M.
** Notice that there are no units given in the answer. Even though Keq technicallyhas units, they are fairly meaningless and they are just dropped. So don'tinclude any units when you state the value for Keq .
Type 1 - Example 2:
At 200°C, the Keq for the reaction: N2(g) + 3H2(g) ~ 2NH3(g) is 625
If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate theequilibrium [H2]. '~\~ '\ ("'\ \t0hCt\ jou \<.f)QW ~
\<,~ = L t,n\.=.J 2- := CO. \ 2.. ') 2- f"'<~?fo 2- s:C N w~ [. \-\J.,j:' l 0,0:::'0') C \-\)=~/>C-~ \
- [\-\L1~ UALJ'~ OIOq~Do Worksheet 2.8: Type 1 Calculations -= lO . Ol)LM
Type 2 Calculations - System has MOVED to Equilibrium ~
Problems).:.
l( 0,000) (iolS)]
0 .. boolloi>
Given initial concentrations of all species and equilibrium concentration of onespecies. You are asked to calculate the equilibrium concentrations of all species orthe Keq.
E=
" (\ \ -\\ C\. \ [j
G'f\C\ns~ \:\ C ~eq,U\ \\b~u~ C1
Q\\ \ {\U\'\~ \--SO-\-
" ~il
We set up an ICEtable: 1=
C=
I
I
Type 2 -- Example:
Given the reaction:
Some H2 and N2 are added to a container so that initially, the [N2] = 0.32 Mand [H2] =O.66M. I Q.r\\~(~C\L~~~A(\~6 ose acldtd, f\SSlAme- \~,'\,'t\\
I L N'~bj \~ 0.tJ\·At a certain temperature and pressure, the equilibrium [H2lis found to be 0.30M.
G) Wfi te ,-~he ()(. n·a) Find the equilibrium [N21 and, INH3]. ~ <:L . . \. r- c: .1 L I' ,;
~f A and .60 moles fB are placed in 2.0 L container. 'equilibrium is reached, the [A] is found to be 0.08 M. Calculate theequilibrium IB] and the equilibrium Ie]:
'I. G,u'Y\ 8''/ en Cl~ O\"\f\\- \ f\ mo\4, ChC\'~e.
C - Oq02- ., o~oCo -\- 0" 0 L\-S o~o~ O. L~ o Q 04
Do Hebden Q's pg. 7- #47-49
Trial Keg Calculations
When we are given the initial concentrations of species (before equilibrium isestablished), we can calculate a Trial Keq. We can then compare the Trial Keq tothe actual Keq value and determine which way the system will shift to reach
equilibrium.
Recall how we write the equilibrium expression:
~eq = C PYO c\IA.C;\ 5)I [ ~-et\ e:\- CW"\\-S~
sCl W\ {. cx(Ak ~S i()()
-to I( tv; G\ \ t< ~J~ '
U~L I(\n--\ ct \ [J IS .
If Trial Keq <Keq the reaction will shift to the right. (a higher[products] is neededs ..,.
~la(A- i-tI"\\'S ~ ?(~c\.vtc\-~.If Trial Keq > Keq the reaction will shift to the left. (a lower [products]
and a higher [reactants] is needed.) <. D(Z c Cu:J -2 r f'ods
If Trial Keq = Keq , the reaction will not shift at all. (It is already atequilibrium).
W\' C\\ \\.e.~-::: .;(\,\) G\ \ L ?ft) c\~'f\\'~ l\ \ C K~~,t~j
Example 1:
At a particular temperature, for the reaction:
H2 + 12 ~ 2HI The Keq = 55.6Ab6\..\n'\-t c: I-\- Ij =- D ,,(\,\-\ 6\\\j
If the initial [H2l= 0.200 M and [12l= 0.200 M, calculate the trial Keq. Whichway will the reaction shift?
~ 0
TV" ~\ ~e.t: <,lo ')
~1-l5C;,1o)
( 0 ..1.00J lO \200 ')
~-eC\G~\OYl w\\' ~ \1 \fl"
(Z\~\\-\ -\u m a\<0\fY\ovt ~VDdu- Gt~ '
Example 2:
CO(g) + H20(g) ~ CO2(g) + H2(g)
The value Of~~~~~~...:.O\~ a temperature of600 °C.
A reaction mixture is analyzed and found to contain O.80M CO, O.050M H20,O.SOMCO2 and 0.40M H2., \'"\ \' ~ (A \C 1\~.Determine which dirlction (left, right or not at all) the reaction will have toshift in order to reactl equilibrium.
