prac 2 soln 1-4

prac 2 soln 1-4 - stoichiometry of the reactants (reactants...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
(1) the rate law is determined by the rate-determining (slow) step if the slow step is preceded by a fast step, this pre-equilibrium condition must also be accounted for NO 2 (g) + F 2 (g) " NO 2 F(g) + F(g) k 1 , slow F(g) + NO 2 (g) " NO 2 F(g) k 2 , fast when, and ONLY when, using elementary reaction mechanisms do you write rate laws using the stoichiometry of the reactants Rate = k 1 [NO 2 (g)][F 2 (g)] (2) the rate law is determined by the rate-determining (slow) step if the slow step is preceded by a fast step, this pre-equilibrium condition must also be accounted for ClO - (aq) + H 2 O(l) " HOCl(aq) + OH - (aq) K, fast (equilibrium) I - (aq) + HOCl(aq) # HOI(aq) + Cl - (aq) k 1 , slow HOI(aq) + OH - (aq) # OI - (aq) + H2O(l) k 2 , fast when, and ONLY when, using elementary reaction mechanisms do you write rate laws using the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 kJmol-1 E a (rev) = 225 kJmol-1 E a (fwd) So, for this specific reaction E a (fwd) = 225 kJ " mol-1- 114 kJ " mol-1 = 111 kJ " mol-1...
View Full Document

Ask a homework question - tutors are online