Unformatted text preview: stoichiometry of the reactants (reactants & products in an equilibrium step) Rate = k 1 [I-(aq)][HOCl(aq)] (from the slow step) the rate of formation of [HOCl(aq)] is determined from the pre-equilibrium in the first, fast step [HOCl(aq)] = K [ClO-(aq)] [OH-(aq)] H 2 O, the solvent, is not included; substitute this expression into the first rate law we derived $ % & ’ ( ) Rate = k 1 [I-(aq)] * K [ClO-(aq)] [OH-(aq)] = k 1 K[I-(aq)][ClO-(aq)][OH-(aq)]-1 (3) This reaction is given in Box 13.3 on pg 518-519 Cl " + O 3 # ClO " + O 2 ClO " + " O " # Cl " + O 2 (4) ! H = -114 kJ·mol-1 E a (rev) = 225 kJ·mol-1 E a (fwd) So, for this specific reaction E a (fwd) = 225 kJ " mol-1- 114 kJ " mol-1 = 111 kJ " mol-1...
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- Spring '08
- Equilibrium, Chemical reaction, Trigraph, Rate equation