Download solving equilibrium problems with the ti-92

SOLVING EQUILIBRIUM PROBLEMS WITH THE TI-nspire
Chemical equilibrium is one of the most important topics you will study this year. An understanding of the basic
principles of equilibrium will allow you to calculate the quantities of reactants and products at equilibrium for a wide
variety of chemical systems. However, even simple systems, such as the Haber process to produce ammonia,
3 H2 (g) + N2 (g)
2 NH3 (g)
can require the solution of a cubic equation, and many reactions will require even higher orders. Fortunately, the
TI-nspire makes the exact mathematical solution of even these complicated equations quite simple. However, you
must use your chemical knowledge to set up the appropriate equations, and interpret the results, if you are to obtain
the correct answer. Remember the computer maxim, Garbage in, garbage out!
The key to correctly solving equilibrium problems is in setting up the ICE table, or reaction stoichiometry matrix. This
will be covered in detail by your instructor, and the textbook; here, we are only interested in solving the equations that
result. An example can be found in the following exercise, featuring an industrial process known as the water-gas shift
reaction. What would be the equilibrium concentrations if 1.000 mol of each component were mixed in a 1.000 L
container at 700 K, where the equilibrium constant has a value of 5.10? The reaction and ICE table for this system are
shown below:
CO (g)
Initial
Change
Equilibrium
1.000M
-x
1.000-x
+ H2O (g)
CO2 (g) + H2 (g)
1.000M
-x
1.000-x
1.000M
1.000+x
1.000M
+x
1.000+x
+x
<-- follows balanced reaction
The equilibrium constant expression is :
[CO2][H2]
K = ---------------- = 5.10
[CO][H2O]
K=
(1.000+x) (1.000+x)
---------------------------- =
(1.000-x) (1.000-x)
(1.000+x)2
-------------- =
(1.000-x)2
5.10
If you recognize that the right side of the equation is a perfect square, you can simplify the expression by taking the
square root of both sides, yielding:
(1.000+x)
-------------- = 2.26
(1.000-x)
This equation is first order in x, and can be solved directly on the TI-nspire by the following steps:
1. Turn on the calculator. Make sure that it is in the “approximate” calculation mode by selecting “Settings”,
“settings”, and “general”. Set the Calculation Mode to “Approximate”.
2. Return to the home screen and select “Calculate”. Press the “menu” button and select “Algebra” and “Solve”.
Type the equation inside the parentheses as follows:
solve((1.000+x)/(1.000-x)=2.26,x)
note the comma
which says to take the equation (1.000+x)/(1.000-x)=2.26 and solve for x. You must abide by the format
solve( equation , variable sought )
and have matched parentheses, in order to avoid error messages. Press <Enter>. After a brief interval, the display
will read x = 0.3865. Note that this is just the solution to the equation, NOT the answer to the problem, which asked for
final concentrations of the components at equilibrium. These can be determined by substituting the value obtained for
x into the expression for the equilibrium concentration for each species, e.g.,
[CO]eq = 1.000-x = 1.000 - 0.3865 = 0.613 M.
What is the equilibrium concentration of [CO2]eq?
Note also that the calculator may return more significant figures than is appropriate. Your answer should be corrected
to show only the number of significant figures justified by the data.
If you did not recognize the equation as a perfect square, you could still have solved the problem by working with the
full equilibrium expression
[CO2][H2]
K = ---------------- = 5.10
[CO][H2O]
K=
(1.000+x) (1.000+x)
---------------------------- = 5.10
(1.000-x) (1.000-x)
This would be entered into the calculator as:
solve(((1.000+x)*(1.000+x))/((1.000-x)*(1.000-x))=5.10,x)
which gives the results
x = 2.589
or
x = 0.3862
Because this is a quadratic equation, it has two solutions. However, only one is physically meaningful. You will have to
use your understanding of the chemistry to choose the correct one. (Which would you choose? Can you have
negative concentration values?)
Note, also, that you must be VERY careful to ensure that the calculator is solving the problem that you want it to solve.
For the last case, if you had mistakenly typed:
solve((1.000+x)*(1.000+x)/(1.000-x)*(1.000-x)=5.10,x)
(
)(
) <-- missing these parentheses above
which differs from the earlier statement by the absence of a few parentheses (indicated), the result would have been:
x = 1.258 or x = -3.258
<-- incorrect values
Neither of these will give you the correct answer to the problem! Here, because the calculator used its internal
hierarchy of operations to interpret your entry, the numerator and denominator are not correctly grouped. Thus, the
calculator solved a different equation. In this case, it solved:
(1+x)
(1+x) * --------- * (1-x) = 5.10
(1-x)
which reduces to
(1+x)2 = 5.10
<-- incorrect
Take the time to examine the expression on the display screen, to be sure that it is what you want it to be. It should
look just like your equilibrium constant expression, substituted with the algebraic version of the equilibrium
concentrations.
As a final example, consider the problem below, which is typical of a chemical system with a small equilibrium
constant. Here, you start with a 0.50 M concentration of the NOCl reactant, and seek equilibrium concentrations for all
species. The reaction, ICE table, and equilibrium expressions are:
2 NOCl (g)
Initial
Change
Equilibrium
0.50 M
- 2x
0.50-2x
K = 1.6 x 105 =
Typing in
2 NO (g) + Cl2 (g)
0M
+ 2x
2x
[NO]2 [Cl2]
--------------[NOCl]2
solve(((2x)^2*(x))/(.5-2x)^2=1.6E-5,x),
0M
+x
x
=
<-- follows balanced reaction
(2x)2 (x)
-------------(0.50 - 2x)2
and then <Enter>, gives the solution
x=0.0097
yielding, e.g., [NO]equil = (2x) = 0.019 M. The equilibrium constant expression was solved exactly, WITHOUT making
the simplifying approximation 0.50 - 2x  0.50 discussed in your text (in a later chapter). Such approximations are
often useful to simplify the mathematical problem. While that may be unnecessary when solving the equations with the
TI-nspire, you should still be aware of the chemical situation on which the validity of the approximation rests. (Here, K
is so small that the reaction does not go very far to the right, so x is negligibly small.) This will enable you to make
important qualitative predictions about the system.