Rising Above_ Dubai Floods and the Fortitude of Dubai International Airport.pdf
AP Chemistry Chapter 15 Outline
1. Chapter 15 Chemical Equilibrium John D. Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science , 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
29. What Do We Know? 1.87 x 10 -3 At equilibrium Change 0 2.000 x 10 -3 1.000 x 10 -3 Initially [HI], M [I 2 ], M [H 2 ], M
30. [HI] Increases by 1.87 x 10 -3 M 1.87 x 10 -3 At equilibrium +1.87 x 10 -3 Change 0 2.000 x 10 -3 1.000 x 10 -3 Initially [HI], M [I 2 ], M [H 2 ], M
31. Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much. 1.87 x 10 -3 At equilibrium +1.87 x 10 -3 -9.35 x 10 -4 -9.35 x 10 -4 Change 0 2.000 x 10 -3 1.000 x 10 -3 Initially [HI], M [I 2 ], M [H 2 ], M
32. We can now calculate the equilibrium concentrations of all three compounds… 1.87 x 10 -3 1.065 x 10 -3 6.5 x 10 -5 At equilibrium +1.87 x 10 -3 -9.35 x 10 -4 -9.35 x 10 -4 Change 0 2.000 x 10 -3 1.000 x 10 -3 Initially [HI], M [I 2 ], M [H 2 ], M
33. … and, therefore, the equilibrium constant. K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )