Thank you for downloading my chemical equilibrium program!!
This program was originally published on May 25, 2000.
Gas chemical equilibrium problems are some of the most tedious, commonly assigned problems in modern chemistry courses. These problems often require the manipulation of large chemical equations, the solving of high degree algebraic equations, and a thorough understanding of stoichiometry in order to be solved. I assume that if you chose to download this chemical equilibrium program that you are taking either AP Chemistry in high school or college chemistry. You are probably also familiar with equilibrium problems and how tedious and outrageously difficult they can be to solve. Well, consider your chemical equilibrium troubles annihilated with this Gas Chemical Equilibrium Problem Solver. This program can solve just about any gas equilibrium problem with various numbers of reactants and products, and varying values for K. Simply enter the coefficients of every reactant and product in the balanced chemical equation, along with the initial concentrations of every reactant and product and the value for the equilibrium constant (K) and let the calculator perform all the tedious operations necessary to find the equilibrium concentrations.
In this explanation I am going to assume that you have a basic understanding of gas chemical equilibrium problems, that is that you know what such problems are, what reactants, products, initial concentrations, equilibrium concentrations, equilibrium constants, shifts, and x-values are. If you do not recall what all these things are, consult your chemistry book.
Let us first of all examine your standard gas chemical equilibrium problem. A common problem that you are likely to encounter in a problem set or on an exam involves the decomposition of dinitrogen tetroxide to nitrogen dioxide by the equation:
N2O4(g)↔2NO2(g).
An equilibrium problem involving this decomposition may be stated like this:
Dinitrogen tetroxide decomposes to nitrogen dioxide by the equation:
N2O4(g)↔2NO2(g).
Calculate the equilibrium concentrations of both species in this reaction if 3 moles of N2O4 is placed in a 1-liter flask at a temperature such that the equilibrium constant for this reaction is 5.1.
To solve this problem, like any problem, you must know what information is needed and where to find that information. This problem, like any standard gas chemical equilibrium problem, is asking you to find the equilibrium concentrations of the species in a reaction given the balanced chemical equation for their reaction, their initial concentrations before the reaction takes place, and a value for the equilibrium constant. To solve this problem, or any standard gas chemical equilibrium problem for that matter, all you must know is the number of reactants and products in the reaction, the coefficients of every reactant and product in the balanced equation, the initial concentrations of every reactant and product, and the value for the equilibrium constant. My gas chemical equilibrium program asks for these and only these pieces of information.
Let us now obtain the information needed from the prompt. From the balanced equation we find that the coefficient of the single reactant is 1 and the coefficient of the product is 2. The prompt says that 3 moles of reactant are placed in a 1-liter flask, so the initial concentration of the reactant (N2O4) must be 3 molar. The prompt says nothing about any NO2 being present before the reaction, so we can assume that the initial concentration of the product NO2 is zero. The prompt, like most standard equilibrium prompts, flat out gives the value for the equilibrium constant.
Now that we have all the information required to solve the problem, we are ready to run the calculator program, which will find the problem’s solution. When we start the program we find that it first asks for the number of reactants in the reaction. In this case there is only one reactant, so we enter 1 at the first prompt. The calculator then prompts us for the number of products. In this case there is also only one product, so we also enter 1 at the second prompt. The calculator now asks us for the coefficient of our reactant. Its coefficient is one, so we enter 1 once again. It now asks for our reactant’s initial concentration. This is three molar, as we determined earlier, and we now enter 3. If we had more than one reactant, the calculator would now ask us for the coefficient and initial concentration of our second reactant. It would then ask us for these same values for our third reactant and so on until all the reactant information had been obtained. The calculator will now ask us for the same information, coefficients and initial concentrations of our products. Since our product’s coefficient is two, we enter 2 at the third prompt. Its initial concentration is assumed to be zero, so we enter 0 at the fourth prompt. Once the calculator has all the product information, it asks for the value of the equilibrium constant (K). We enter 5.1 for K.
We can now sit back and let the calculator do its work. DO NOT BE SURPRISED IF IT TAKES THE CALCULATOR SEVERAL SECONDS TO SOLVE THE PROBLEM. This program utilizes three processing-hungry routines. The first of these routines expands binomials. Binomials are crucial to equilibrium problems. An equilibrium problem has as many binomials as it does reactants and products. This means that the more reactants and products there are, the longer it will take the calculator to solve the problem. The second processing-hungry routine is one that multiplies polynomials. To solve equilibrium problems an equation must always be solved. To transform this equation into one the calculator can solve, its expanded binomials, which are polynomials, must be multiplied together to yield a single polynomial. The time that it takes to complete this multiplication depends on the size of the coefficients you entered earlier. Larger coefficients mean longer processing time. Once the calculator has obtained a single polynomial it can now set this polynomial equal to zero and solve it for x with its built-in polynomial root finder. The larger this polynomial is, the longer it will take to solve the equation. Once the calculator has solved the equation for x it can use that value of x to find the equilibrium concentrations.
The calculator will now display the equilibrium concentrations of our reactants and products. If there are more than seven species in the reaction, the calculator will display the equilibrium concentrations of the first seven species and pause so that you may write them down. You must press enter to view the remaining equilibrium concentrations. The calculator displays the x-value and tells you the direction of the equilibrium shift, right or left. For our problem the calculator displays x=1.4195, and says that the equilibrium “Shifts Right.” It tells us that our reactant’s equilibrium concentration is 1.580 molar and that our product’s equilibrium concentration is 2.839 molar.
Now you know how to use my gas chemical equilibrium problem-solving program. I hope you find it useful for problem sets, exams, and any other applications you may find for it. I apologize if this explanation is lengthy and confusing. I believe this program is fairly easy to use and even if I confused with this text, I am sure you’ll have no problem learning to use this piece of software on your own. I wish you happy computing.
Jerry Vigil
TI Programmer