Rate Law Equation

Rate Law Equation - 0.018 4 0.004 0.001 0.008 5 0.004 0.002...

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Rate Law Equation (Expression) Since For the reaction a X + b Y products The Rate Law The rate of a reaction will always be proportional to the initial  concentration of the reactants raised to some exponents. Rate Law Equation Examples: For the reaction:
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What happens to the rate if the [H 2 O 2 ] is doubled? For the reaction: What happens to the rate if [NO 2 ] is doubled? 
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Rate Law Equations These are equations show the quantitative relationship between the concentration of reactants and the rate of reaction. They are derived from experimental data. For example, using the following data derive the rate law equation for the following reaction. 2 NO (g) + 2 H 2 (g) N 2 (g) + 2 H 2 O (g) Trial [NO] (mol/L) [H 2 ] (mol/L) Rate (mol/(L∙s)) 1 0.001 0.004 0.002 2 0.002 0.004 0.008 3 0.003 0.004
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Unformatted text preview: 0.018 4 0.004 0.001 0.008 5 0.004 0.002 0.016 6 0.004 0.003 0.024 Compare trials 1,2 and 3: [NO] is changed (independent variable) Rate is measured (dependent variable) [H 2 ] is constant (controlled variable) Compare Trials 4,5 and 6: [H 2 ] is the independent variable. Rate is the dependent variable. [NO] is the controlled variable. Another Example, using the previous reaction and data chart, calculate the rate of the reaction if the NO has a concentration of 0.018 mol/L and H 2 has a concentration of 0.046 mol/L. Method: 1) Use any of the data sets in the chart to calculate k. This is a constant as long as the temp and pressure remain constant. . 2) Use the constant and the new concentration values to calculate the new rate....
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This note was uploaded on 07/06/2011 for the course SCH 4U taught by Professor White during the Spring '10 term at Beacon FL.

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Rate Law Equation - 0.018 4 0.004 0.001 0.008 5 0.004 0.002...

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