Chapter_12_-_Kinetics_-_PowerPoint

Chapter_12_-_Kinetics_-_PowerPoint - Chapter 12 Chemical...

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Chapter 12 Chemical Kinetics
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Reaction Rates Reaction Rate Change in concentration of a reactant or product per unit time. (how fast a reaction occurs) Rate = Δ[A] Δt [A] = concentration of A in mol/L or M
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Reaction Rates 2NO 2 (g) → 2 NO (g) + O 2 (g) Rate = Δ[A] Δt
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Reaction Rates Rates are always positive (+), so: For reactants, Rate = - Δ[NO 2 ] (reactant is lost) Δt For products, Rate = + Δ[NO] (product is gained) Δt
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Reaction Rates Rates are proportional to stoichiometry: 2NO 2 (g) → 2 NO (g) + O 2 (g)
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Reaction Rates Calculate the average rate of the reaction for the first 50 sec. Rate = - Δ [NO 2 ] Δt = - (0.0079 M– 0.0100M) 50 s – 0 s = 4.2 x 10 -5 M / s
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Reaction Rates Rates change over time.
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Reaction Rates Reactions occur as molecules interact with each other. Less molecules (lower concentration) = less interactions (reactions) As reaction progresses, concentration of reactants decrease, so rate slows down.
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Reaction Rates
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Rate Laws 2NO 2 ( g ) → 2NO( g ) + O 2 ( g ) Rate = - Δ [NO 2 ] = k [NO 2 ] n Δt k = rate constant n = order of the reactant (zero order, first order, second order, …)
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Examples What happens when you double the reactant [A] when reactant order is: 0 order: Rate = k [A] 0 Rate doesn’t change 1 order: Rate = k [A] 1 Double [A], rate doubles 2 order: Rate = k [A] 2 Double [A], rate quadruples (x4) Rate = k
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Rate Laws Rate = k [NO 2 ] n n must be determined by experiment; NOT stoichiometry from the balanced equation.
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Rate Laws Determining the order of the reactant, n 2 N 2 O 5 → 4 NO 2 + O 2 Rate = k [N 2 O 5 ] n Rate 1 = k [N 2 O 5 ] 1 n Rate 2 = k [N 2 O 5 ] 2 n
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Rate Laws Determining the order of the reactant, n 2 N 2 O 5 → 4 NO 2 + O 2 Rate = k [N 2 O 5 ] n Rate 1 k [N 2 O 5 ] 1 n Rate 2 k [N 2 O 5 ] 2 n Set up as ratio. =
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Rate Laws Determining the order of the reactant, n 2 N 2 O 5 → 4 NO 2 + O 2 Find ratio: Rate 1 = k [N 2 O 5 ] 1 n Rate 2 k [N 2 O 5 ] 2 n 5.4 x 10 -4 M/s = k(0.90M) n 2.7 x 10 -4 M/s k(0.45M) n 2.0 = (2.0) n n = 1, first order
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Rate Laws Determining the rate constant, k 2 N 2 O 5 → 4 NO 2 + O 2 Rate = k [N 2 O 5 ] 1 5.4 x 10 -4 M/s= k(0.90M) 1 k = 6.0 x 10 -4 s -1
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Rate Laws If more than one reactant, the rate law becomes: Rate = k [A] n [B] m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Reaction order The sum of the exponents in the rate equation ( n + m) .
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Rate Laws Example 12.1 BrO 3 - + 5Br - + 6H + → 3Br 2 + 3 H 2 O Rate = k [BrO 3 - ] n [Br - ] m [H + ] p Find n , m , p , k, and the overall reaction order
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Example 12.1 - Solving n, order of reactant BrO 3 - For Exp. 1 and 2, ONLY BrO 3 - changes so: Rate 2 = k
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