Exp. 4 Rate Law Report PDF

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Experiment 4 Rate Law General Chemistry 1112-98L The University of Texas -Rio Grande Valley Fall 2020 Rickey M. Perez Graded By:

Objective: The objective of this experiment is to determine the rate law for a specific chemical reaction, while utilizing graphical data analysis of the experimental data to; determine the order for each reactant in the reaction and determine the activation energy for the same reaction. As well as observe and describe the factors that influence reaction kinetics.

Introduction: The factors affecting the rate of a chemical reaction depend on the reaction mechanism. Depending on the nature of the reactants being consumed or the products being formed, the reaction mechanism may differ in a number of ways. The reaction rate may be followed as a change in concentration of one of the reactants or products per unit of time, the volume of gas produced per unit for time, or the change in color per unit of time. A general rule of thumb for the reaction mechanism is that the overall reaction rate should not be greater than the rate of the slowest step. In this experiment, five different solutions will be placed in a Spectrophotometer to determine the amount of light absorbed and contrasted with a blank solution. The Rate Laws of each solution will be determined as well. The Rate Law can be described as the relationship between the rate of a chemical reaction and the concentration of the reactants. The Rate constant, k, is a proportionality constant and term for the concentration of each reactant raised to the power of its order. The reaction orders are determined experimentally and are as follows; Zero order, where n = 0. First order, where n = 1, and Second order, where n = 2.

Data Sheet:

Graphs:

Reaction order with Respect to Iodate ion -6.85E+00 f(x) = 0.417663169621516 x − 6.59752200471635 R² = 0.991025369761876

log(rate)

-7.00E+00

-7.15E+00

-7.30E+00

-1.6

-1.5

-1.4

-1.3

-1.2

-1.1

-1.0

-0.9

log[I- ]0

Reaction Order with Respect to HSO3 -6.95E+00

log(Δ(mol I3- )/Δt)

-7.00E+00 -7.05E+00 -7.10E+00

f(x) = 0.414217087726613 x − 6.59464190810914 R² = 0.988783520673049

-7.15E+00 -7.20E+00 -7.25E+00 -7.30E+00 -7.35E+00 -1.8

-1.7

-1.6

-1.5

-1.4

log[H2O2]0

-1.3

-1.2

-1.1

-1.0

-0.8

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