evaluating equilibrium expressions

# evaluating equilibrium expressions - EVALUATING EQUILIBRIUM...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EVALUATING EQUILIBRIUM EXPRESSIONS Most problems that deal with equilibrium concentrations can be solved by rearranging the equilibrium expression into the form of a quadratic polynomial. For the reaction aA + bB cC + dD init. [A] o [B] o [C ] o [D] o equil. [A] eq = [A] o- ax [B] eq = [B] o- bx [C] eq = [C] o + cx [D] eq = [D] o + dx With the initial amounts and equilibrium amounts on the initial and equilibrium line respectively, the equilibrium expression is K c = [C] o + cx ( 29 c [D] o + dx ( 29 d [A] o- ax ( 29 a [B] o- bx ( 29 b . Evaluation of the reaction quotient (Q c ) Q c = [C] o c [D] o d [A] o a [B] o b allows one to determine whether x is positive (Q c &lt;K c ) or negative (Q c &gt;K c ). For the case a=b=c=d=1 one has after rearranging K c ([A] o- x)([B] o- x)- ([C] o + x)([D] o + x) = and after grouping terms with a common power of x one obtains the quadratic equation x 2 + x + = where = K c- 1 = - K c [A] o + [B] o { } + [C] o + [D] o ( 29 and = K c [A] o [B] o- [C] o [D] o . The roots of this equation are the values of x which are given by x =- 2- 4 2 . As an example let [A] o =1.00M, [B] o =2.00M, [C] o =1.00M, [D] o =3.00M and K c =1.25. Then Q c =1.50&gt;K c so that x must be negative. Evaluating the coefficients gives 1 =0.250,=-7.75 and =-0.500. Then the physical root is x=-0.064 so that [A] eq =1.06M, [B] eq =2.06M, [C] eq =0.94M and [B] eq =2.94M. Calculators can be used to remove several undesirable features of the previous analysis. i) Calculators can solve for the roots of a polynomial greater than second order so that the restriction of a+b and c+d being less that or equal to two can be removed. ii) The form of the equation to be solved by the calculator is quite arbitrary so that rearranging equations can be avoided. The TI-83 Calculator: The equation to be solved is K c ([A] o- ax) a ([B] o- bx) b- ([C] o + cx) c ([D] o + dx) d = which was obtained by rearranging the equilibrium expression. This equation can be entered by pressing the Y= button. Since only single letter variables can be used the symbols in the above equation can be defined as follows. Symbol in Equation Symbol on Calculator K c K [A] o A a B [B] o C b D [C] o E c F [D] o G d H With this notation the equation to enter is \ Y 1 = K ((A- B X)^B) ((C- D X)^D)- ((E + F X)^F) ((G + H X)^H) . For the previous problem the data is stored as 1.25 K 1 A 1 B 2 C 1 D 1 E 1 F 3 G 1 H To make a graph of the function \Y 1 the equal sign must be highlighted. To highlight it put the cursor on the equal sign and push ENTER. To dehighlight the equal sign place 2 the cursor on it and press ENTER. the cursor on it and press ENTER....
View Full Document

## This note was uploaded on 04/29/2008 for the course CHE 1302 taught by Professor Young during the Spring '08 term at Baylor.

### Page1 / 8

evaluating equilibrium expressions - EVALUATING EQUILIBRIUM...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online