REVERSABLE REACTIONS (EQUILIBRIUM)
This calculator can do some useful calculations with respect to
equilibriums of reversable reactions. When we enter <=> instead of
=> for an equation, it goes into this mode. It can calculate the
equilibrium point for up to 10 reagents.
In general, if we have the reaction
aA + bB <=> cC + dD
it will
solve the equation
K = ( ([C]+cx)^c * ([D]+dx)^d )/( ([A]-ax)^a * ([B]-bx)^b )
for x.
Here [A] is the initial concentration of A and [A]-ax is
the final concentration of A.
"a" is its stoichiometric coefficient.
A similar relationship holds for B,C,
and D. It may actually compute the equilibrium for as many reactants
and products as the spreadsheet will hold (10).
K is the equilibrium constant.
1. Consider the reaction H2+I2<=>2HI. The molar
concentrations are H2=0.00625, I2=0.01414, HI=0.0224 before the reaction.
The equilibrium constant, Kc, is 54.4 at a specified temperature. What
will be the new concentrations?
Solution:
-We enter the equation H2+I2<=>HI and SUBMIT it.
-Be sure to enter the sign <=> instead of =>.
-If it is not balanced, it will be
balanced automatically.
-We then enter the initial concentration values
under the "Initial" column and the Kc value under the "K" column.
-We
hit the STEPPER (CALCULATE) button and note the "Final" concentration values given.
(0.00184, 0.00973, 0.0312)
-Note that Q (5.6776) is also given. This is the equivalent of the equilibrium
constant calculated for the initial concentrations. It can help determine
the nature of the reaction. Another value given is lower case "x" which is
the unknown that is determined in the polynomial equation that must be
solved for these type of problems.
It is useful when you check the calculation by hand. Both Q and x
are shown in green boxes because you can read them but not set their
values.
2. A weak solution of Acetic acid (0.9M) ionizes in water to form H3O
according to the reaction:
CH3COOH + H2O <=> CH3COO[-] + H3O[+]
If K=0.8
find the equilibrium concentration of CH3COOH.
Solution:
-
We enter the above formula and SUBMIT it.
-In this case as with many
solvents, water does not appreciably change its concentration and it is
treated as a constant in the determination of K.
-We indicate "N" in the
"Calculate" column for H2O. We do not need to enter a value for it.
-In the "Initial" column we enter 0.9 for CH3COOH, 0 for CH3COO and 0
for H3O.
-We are assuming no reaction has occured as of yet.
-Finally,
we enter 0.8 in the K column and hit the STEPPER (CALCULATE) button.
-It should
quickly converge and indicate the "Final" or equilibrium concentration
of CH3COOH (0.3619 moles/liter).
3. In the above two examples, we dealt with Kc, the concentration
constant. We can also use Kp for partial pressures. Consider the
reaction: N2O4 <=> 2NO2 The equilibrium relationship holds with the
calculator where the initial partial pressures of N2O4 and NO2 are
given in the "Initial" column and Kp is supplied in the "K" column.
The solution yields the "Final" partial pressures after equilibrium
is reached. Note that it is assumed the volume and temperature of
the container remains constant and we are dealing with ideal gases.
4. We can also use the acid ionization constant Ka in a way similar to
Kc or Kp. Calculate the pH of a 0.050 M Nitrous acid solution HNO2 at
25 degC.
Solution:
-Water is a solvent and its concentration is assumed to
be constant.
-We have the reaction: HNO2 <=> H[+] + NO2[-] and we know
the inital concentration of HNO2.
-We assume the initial concentrations
of H[+] and NO2[-] are both zero.
-We look up Ka in a table and find it
is 4.5e-4 at 25 degC.
-We enter the above data in the "Initial" column and the "K" column
and hit the STEPPER (CALCULATE) button.
-We find that the final concentration of
H[+] is 0.0045237 M or about 4.6e-3 M.
-By definition, the pH is thus
pH = -log(4.6e-3) = 2.34. We use a pocket calculator to do the log
calculation.
5. The solubility constant Ksp can also be used. For example if we
have the reaction
Ca3(PO4)2 <=> 3Ca[2+] + 2PO4[3-]
we are interested in the
Molar Solubility of this compound in water. That is called s and it equals
the number of moles of a solute that is present in a saturated solution.
Solution:
-
The calculator can be manipulated to calculate this constant.
-We simply
enter N in the "Calculate" column for all the reactant entries, enter
0 for all the "Initial" Product entries and enter Ksp or 1.2e-26 in this case
for K.
-We then hit the STEPPER (CALCULATE) button and the value of s will be displayed
(2.6e-6 moles/liter).
6. One important and useful feature that should not be overlooked is that
for equilibrium calculations, you can give the initial
concentration
values and one final concentration value and it will calculate the value of K.
Note that all the other final concentration values must be blank for this
mode to work.
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