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Example 5:Example 5: Determine the rate law for the following reaction---- NH 4 + (aq) + NO 2 - (aq) N 2(g) + 2H 2 O (l) Experiment[NH 4 + ] initial [NO 2 - ] initial Rate initial 15 x 10 -2 M2 x 10 -2 M2.70 x 10 -7 M/s 25 x 10 -2 M4 x 10 -2 M5.40 x 10 -7 M/s 31 x 10 -1 M2 x 10 -2 M5.40 x 10 -7 M/s

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Zero-Order ReactionsZero-Order Reactions Rate is NOT dependent on reactant concentration Graph of [A] vs. time gives STRAIGHT LINE If no straight line, reaction is NOT zero order Slope = -k

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Zero-Order GraphZero-Order Graph

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Integrated Rate LawIntegrated Rate Law Enables the determination a reactant’s concentration at any moment in time Enables the determination of the time it takes to reach a certain reactant concentration Enables the determination of the rate constant or reaction order

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1 st Order Reactions1 st Order Reactions Integrated Rate law ln[A] t – ln[A] 0 = - kt ln[A] vs. time graph yields STRAIGHT LINE If no straight line, reaction is NOT 1st order Slope = -k

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1 st Order Graph1 st Order Graph

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1 st Order Integrated Rate Law Only used with 1 st order reactions Focus on initial concentration and Δ C for one reactant Initial concentration of reactant known---- can determine reactant concentration at any time Initial and final reactant concentrations known---can determine rate constant

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1 st Order Integrated Rate Law Rate = - Δ [A] = k [A] Δ t -take equation and integrate with calculus to get…. ln[A] t – ln[A] 0 = - kt [A] 0 = initial concentration (t = 0) [A] t = concentration after a period of time

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Example 1: A B + 2DExample 1: A B + 2D Using the data provided for a 1 st order reaction, determine the rate constant and [A] at time = 5.0 x 10 2 s. Time (s)[A] (M) 00.020 5.0 x 100.017 1.0 x 10 2 0.014 1.5 x 10 2 0.012 2.0 x 10 2 0.010

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Example 1: continuedExample 1: continued

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Example 1: A B + 2DExample 1: A B + 2D Using the data and graph provided, determine the rate constant and [A] at time = 5.0 x 10 2 s. Time (s)[A] (M) 00.020 5.0 x 100.017 1.0 x 10 2 0.014 1.5 x 10 2 0.012 2.0 x 10 2 0.010

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Half-life Radioactive decay is a 1 st order process Half-life (t 1/2 )— Time it takes for half of a chemical compound to decay or turn into products Focus on reactant Constant, not dependent on [ ] Rate changes with temperature so half-life varies based on temperature

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Example 2:Example 2: Find the half-life for the following reaction with a rate constant (k) of 1.70 x 10 -3 s -1

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2 nd Order Reactions2 nd Order Reactions Integrated Rate Law 1___ - 1__ = kt [A] t [A] 0 1/[A] vs. time graph yields STRAIGHT LINE If no straight line, reaction is NOT 2nd order Slope = k

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2 nd Order Graph2 nd Order Graph

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2 nd Order Integrated Rate Law Used only for second order reactions Focus on initial concentration and Δ C for one reactant with reaction 2 nd order with respect to it. Initial concentration of reactant known---- can determine reactant concentration at any time Initial and final reactant concentrations known---can determine rate constant

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2nd Order Integrated Rate Law Rate = - Δ [A] = k [A] 2 Δ t -take equation and integrate with calculus to get…. 1__ - 1__ = kt [A] t [A] 0 [A] 0 = initial concentration (t = 0) [A] t = concentration after a period of time

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Example 3: 2NO 2(g) 2NO (g) + O 2(g) Using the data provided, find the rate constant if the rate law = k[NO 2 ] 2. Time (s)[NO 2 ] 0.00.070 1.0 x 10 2 0.0150 2.0 x 10 2 0.0082 3.0 x 10 2 0.0057

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Example 3: 2NO 2(g) 2NO (g) + O 2(g) Using the data and graphs provided, find the rate law and rate constant. Time (s)[NO 2 ] 0.00.070 1.0 x 10 2 0.0150 2.0 x 10 2 0.0082 3.0 x 10 2 0.0057

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Example 3: continuedExample 3: continued

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Example 4:Example 4: NO 2 reacts to form NO and O 2 by second-order kinetics with a rate constant = 32.6 L/mol min. What is the [NO 2 ] after 1 minute if the initial [NO 2 ] = 0.15M?

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Concentration and Time DataConcentration and Time Data Use data to construct all graphs for zero, 1 st, and 2 nd reaction orders Determine which graph yields a straight line. [A] vs. TimeZero Order ln[A] vs. Time1 st Order 1/[A] vs. Time2 nd Order

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Homework Read over lab procedure

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