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.

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Presentation transcript:

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 M2 x M2.70 x M/s 25 x M4 x M5.40 x M/s 31 x M2 x M5.40 x M/s

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

Zero-Order GraphZero-Order Graph

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

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

1 st Order Graph1 st Order Graph

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

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

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) x x x x

Example 1: continuedExample 1: continued

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) x x x x

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

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

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

2 nd Order Graph2 nd Order Graph

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

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

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 ] x x x

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 ] x x x

Example 3: continuedExample 3: continued

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?

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

Homework  Read over lab procedure