prac 2 soln 1-4

# prac 2 soln 1-4 - stoichiometry of the...

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(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
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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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