Unformatted text preview: 2 ] which reacts. [H 2 ] [I 2 ] [HI] initial 2.0 2.0 5.0 changexx + 2 x* equilibrium 2.0  x 2.0  x 5.0 + 2x * the 2 comes from the reaction stoichiometry. [HI] 2 (5.0 + 2x) 2 K c =  =  = 46 [H 2 ][I 2 ] (2.0 x)(2.0 x) Since both the numerator and denominator of this fraction are perfect squares, this can be solved by taking the square root of both sides of the equation. Thus, the equation reduces to 5.0 + 2x K c =  = (46)0.5 = 6.78 2.0  x Solving for x gives 5.0 + 2x = 6.78(2.0  x) 5.0 + 2x = 13.56  6.78 x (2 + 6.78) x = 13.56  5 8.78 x = 8.56 x = 0.975 Substituting back to get the equilibrium concentrations gives [H 2 ] = 2  x = 2  0.975 = 1.025 [I 2 ] = 2  x = 2  0.975 = 1.025 [HI] = 5.0 + 2x = 5.0 + 2 (0.975) = 6.95 Check: (6.95)2/(1.025)(1.025) = 46 (Your number may vary slightly depending on how you rounded off your answer.)...
View
Full
Document
This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.
 Spring '08
 Martin
 Equilibrium

Click to edit the document details