Chemical_Kinetics.pptx - Chemical Kinetics 1 Chemical...

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1 Chemical Kinetics
2 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time ( M /s). A B rate = - D [A] D t rate = D [B] D t D [A] = change in concentration of A over time period D t D [B] = change in concentration of B over time period D t Because [A] decreases with time, D [A] is negative .
3 A B rate = - D [A] D t rate = D [ B ] D t
4 Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time 393 nm light Detector D [Br 2 ] a D Absorption red-brown t 1 < t 2 < t 3
5 Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) average rate = - D [Br 2 ] D t = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time
6 rate a [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x 10 -3 s -1
7 2H 2 O 2 ( aq ) 2H 2 O ( l ) + O 2 ( g ) PV = nRT P = RT = [O 2 ] RT n V [O 2 ] = P RT 1 rate = D [O 2 ] D t RT 1 D P D t = measure D P over time
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9 2A B Two moles of A disappear for each mole of B that is formed. rate = D [B] D t rate = - D [A] D t 1 2 a A + b B c C + d D rate = - D [A] D t 1 a = - D [B] D t 1 b = D [C] D t 1 c = D [D] D t 1 d Reaction Rates and Stoichiometry
Example 13.1 Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products:
Example 13.1 Strategy To express the rate of the reaction in terms of the change in concentration of a reactant or product with time, we need to use the proper sign (minus or plus) and the reciprocal of the stoichiometric coefficient. Solution (a) Because each of the stoichiometric coefficients equals 1, (b) Here the coefficients are 4, 5, 4, and 6, so
Example 13.2 Consider the reaction Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M /s. (a) At what rate is N 2 O 5 being formed? (b) At what rate is NO 2 reacting?
Example Strategy To calculate the rate of formation of N 2 O 5 and disappearance of NO 2 , we need to express the rate of the reaction in terms of the stoichiometric coefficients as in Example 13.1: We are given where the minus sign shows that the concentration of O 2 is decreasing with time. 13.2
Example 13.2 Solution (a) From the preceding rate expression we have Therefore
Example 13.2 (b) Here we have so
16 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. a A + b B c C + d D Rate = k [A] x [B] y Reaction is x th order in A Reaction is y th order in B Reaction is (x +y)th order overall
17 F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g ) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ]
18 F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g ) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally.