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Unformatted text preview: 3 , calculate the value of K eq if the following concentration are present at equilibrium: [N 2 ]=.22M; [H 2 ]=.14M; [NH 3 ]=.055M Ans) K eq =[NH 3 ] 2 (.055) 2 [N 2 ] [H 2 ]= (.22)(.14) 3 = 5 Ex) Write the K eq expression for the following reaction: a) SnO 2 (s)+2CO(g)Sn(s)+2CO 2 (g) b) Ti(s) +2Cl 2 (g)TiCl 4 (l). Ans) a) K eq [CO 2 ] 2 /[CO] 2 b) K eq 1/[Cl 2 ] 2 Reaction Order= 0= no change; 1=doubles; 2=quadruples; 3= 8x; Dissociation of water= K w 2H 2 O(l)H 3 O + +OH-H 2 O H + +OH-H 2 OH + +OH-K eq = [H + ][OH-]/ [H 2 O] K w =[H][OH-]=10-14 M=K eq x H 2 O Buffers : Solution that resists change to pH. To make buffer, combine weak acid w/conjugant base....
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This note was uploaded on 02/20/2008 for the course CHEM 1560 taught by Professor Baird, b during the Fall '07 term at Cornell University (Engineering School).
- Fall '07
- BAIRD, B