ࡱ> +-* ?bjbjqq yBee$5555<q,5:RTTTTTTgTQTRR`X5X6>0%:%%<TT:% : 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)  EMBED ISISServer  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) + H2O (g)  EMBED ISISServer  CO2 (g) + H2 (g) Initial 1.000M 1.000M 1.000M 1.000M Change -x -x +x +x <-- follows balanced reaction Equilibrium 1.000-x 1.000-x 1.000+x 1.000+x The equilibrium constant expression is : [CO2][H2] (1.000+x) (1.000+x) (1.000+x)2 K = ---------------- = 5.10 K = ---------------------------- = -------------- = 5.10 [CO][H2O] (1.000-x) (1.000-x) (1.000-x)2 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) 0note 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 . 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] (1.000+x) (1.000+x) K = ---------------- = 5.10 K = ---------------------------- = 5.10 [CO][H2O] (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 which reduces to (1+x)2 = 5.10 <-- incorrect (1-x) 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)  EMBED ISISServer  2 NO (g) + Cl2 (g) Initial 0.50 M 0 M 0 M Change - 2x + 2x + x <-- follows balanced reaction Equilibrium 0.50-2x 2x x [NO]2 [Cl2] (2x)2 (x) K = 1.6 x 10!5 = --------------- = -------------- [NOCl]2 (0.50 - 2x)2 Typing in solve(((2x)^2*(x))/(.5-2x)^2=1.6E-5,x), and then <Enter>, gives the solution x=0.0097 yielding, e.g., [NO]equil =(/01  = ǺǫǺǫǟǺǫ}ǫǫǺsnaj9A heC>H*UV heC>H*jheC>H*UhYAOJQJ^JaJj9A heC>CJH*UV heC>CJH*jheC>CJH*UheC>6OJQJ]^JaJheC>H*OJQJ^JaJheC>OJQJ^JaJheC>5OJQJ\^JaJhYA5CJOJQJ\^JheC>5CJOJQJ\^J#01 3 d e [$$ `P^`Pa$$ ` ^ `a$$ `^`a$$P`Pa$$ `^`a$$@ `@ a$$a$$a$ [\hl*+./TU @CK_l}rhYAhYA5OJQJ^JaJhYAOJQJ^JaJh[OJQJ^JaJheC>H*OJQJ^JaJ heC>6CJOJQJ]^JaJhHOJQJ^JaJheC>5OJQJ\^JaJheC>H*OJQJ^JaJheC>OJQJ^JaJjheC>H*UjheC>H*U'$'V&'T/$`a$$@ `@ a$$`a$$a$$ ` ^` a$$ `^`a$$ `0^`0a$.:BN\^t Rln07?BQS幧ɛ}lZZ"heC>56OJQJ\]^JaJ heC>6CJOJQJ]^JaJheC>6CJOJQJ]aJheC>5OJQJ\^JaJh'FOJQJ^JaJ"hYAhYA5OJQJ\^JaJhYAhYAOJQJ\^JaJhYAOJQJ^JaJhYAhYA5OJQJ^JaJheC>OJQJ^JaJhYA5OJQJ\^JaJ#/0TU  uv456Fhi$p`pa$$ `P^`Pa$$ ` ^` a$$ `0^`0a$$@ `@ a$$a$ ,3opqsLM 9<RZ04KOr%'."+jkuheC>CJH*OJQJ^J heC>6CJOJQJ]^JaJheC>5>*OJQJ\^JaJheC>>*OJQJ^JaJheC>H*OJQJ^JaJheC>6OJQJ]^JaJheC>5OJQJ\^JaJheC>OJQJ^JaJ5  st&'PQW$ `^`a$$`a$$ `P^P`a$$p`pa$$a$$ `^`a$uvw+8      % & ' * , - f ۤەۋyoەbەSheC>5OJQJ\^JaJheC>H*OJQJ^JaJjBheC>H*Uj9A heC>H*UV heC>H*jheC>H*UheC>6OJQJ]^JaJhSPtOJQJ^JaJheC>>*OJQJ^JaJ heC>6CJOJQJ]^JaJheC>H*OJQJ^JaJheC>OJQJ^JaJheC>CJH*OJQJ^JheC>CJOJQJ^J + , f *!|"""##~??$ ` ^` a$$ ` ^ `a$$`a$$@ `@ a$$a$f g !!!!$!%!"""""""#H#L#R#p#~#####$<<<==N>W>d>i>z>>M?X?zkheC>6OJQJ]^JaJh[OJQJ^JaJheC>OJQJaJheC>>*OJQJ^JaJUheC>5H*OJQJ\^JaJheC>H*OJQJaJheC>H*OJQJ^JaJheC>H*OJQJ^JaJ heC>6CJOJQJ]^JaJheC>OJQJ^JaJheC>5OJQJ\^JaJ* (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. 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