06 - Chem 204 Lecture6 Time to use our tools and...

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Chem 204–Lecture 6 Time to use our tools and relationships to solve equilibrium problems. We have relationships between K p or K c and the amounts, both relative and absolute, of reactants and products. We need to relate: I nitial conditions, C hange with reaction and final E quilibrium. The I.C.E. method. There are many variations. We will use this repeatedly in the course.

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Lecture 6 –page 2 Consider the % decomposition of HBr, with an initial concentration of 1.2 mM. At 500 K, K c =7.7 x 10 -11 . 2HBr(g) H 2 (g) + Br 2 (g) Initial: 1.2 x10 -3 M 0 0 Change -2x +x +x ( note stoichiometry !) Equilib 1.2 x10 -3 -2x xx K c =[H 2 ][Br 2 ]/[HBr] 2 = x 2 /(1.2 x10 -3 -2x) 2 =7.7 x 10 -11 . We could solve using the quadratic equation…
Lecture 6 –page 3 This is the crude brute force method. If you do: x 2 + 3.67x10 -13 x -1.01x10 -16 = 0 x = 1.05x10 -8 M, the positive root. The amount decomposed is 2x 2.1x10 -8 M. But this is very small compared to 1.2x10 -3 M, the initial concentration, so 2.1x10 -8 /1.2x10 -3 = 1.75x10 -5 or % decomposition is 1.75x10 -3 %.

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06 - Chem 204 Lecture6 Time to use our tools and...

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