Chapter_14

# Chapter_14 - Chapter 14 Equivalent Rates 4 NH 3 (g) + 5 O 2...

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Unformatted text preview: Chapter 14 Equivalent Rates 4 NH 3 (g) + 5 O 2 (g) 4 ( 29 + 6 2 ( 29-1 [ 3 ]-1 [ 2 ] + 1 + 1 [ 2 ] 4 5 4 6 Equivalent reaction rate for reaction like this one defined as speed of formation of a product with coefficient of 1. To calculate equivalent rate multiply slope of reactant or product by 1/coefficient and change sign for reactants (reactant slopes are negative). This effectively gives reaction rate in change of equivalents per unit of time. Integrated Rate Laws First-Order: ln(c t /c ) = -kt Second-Order: (1/c t ) - (1/c ) = +kt Rate Law From Kinetic Data 2 NO(g) + O 2 (g) 2 2 ( 29 Expt. [NO] [O 2 ] Init. Rate 1 0.0126 0.0125 1.41 x 10-2 2 0.0252 0.0250 1.13 x 10-1 3 0.0252 0.0125 5.64 x 10-2 (a) Calculate rate law. (c) Calculate rate constant. Arrhenius Equation Activation energy measured by comparing rates at different temperatures k = Ae-Ea/RT k 2 /k 1 = (Ae-Ea/RT 2 )/(Ae-Ea/RT 1 ) = e-Ea/RT 2-(-Ea/RT 1 ) k 2 /k 1 = e-(Ea/R)(1/T 2 - 1/T 1 ) ln(k 2 /k 1 ) = -(Ea/R)(1/T 2- 1/T 1 ) Bimolecular Reaction Mechanism...
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## This note was uploaded on 12/28/2011 for the course CHEM 1202 taught by Professor Staff during the Fall '08 term at LSU.

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Chapter_14 - Chapter 14 Equivalent Rates 4 NH 3 (g) + 5 O 2...

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