LabQuest 25 The Rate and Order of a Chemical Reaction A basic kinetic study of a
ID: 953820 • Letter: L
Question
LabQuest 25 The Rate and Order of a Chemical Reaction A basic kinetic study of a chemical reaction ofen involves conducting the reaction at varying concentrations of reactants. In this way, you can determine the order of the reaction in each species, and determine a rate law expression, Once you select a reaction to examine, you must decide how to follow the reaction by measuring some parameter that changes regularly as time passes, such as temperature, pH, pressure, conductance, or absorbance of light. f a chemical reaction often involves conducting the reaction at varying In this experiment you will conduct the reaction between solutions of potassium iodide and iron (Ill) chloride. The reaction equation is shown below, in ionic form. 2T (aq)+ 2 Fe" (aq)12 (aq)+2 Fe (aq) As this reaction proceeds, it undergoes a color change that can be precisely measured by a Vernier Colorimeter or a Vernier Spectrometer. By carefully varying the concentrations of the reactants, you will determine the effect each reactant has on the rate of the reaction, and consequently the order of the reaction. From this information, you will write a rate law expression for the reaction. OBJECTIVES In this experiment, you will Conduct the reaction of KI and FeCls using various concentrations of reactants Determine the order of the reaction in KI and FeCls Determine the rate law expression for the reaction. MATERIALS 0.020 M potassium iodide, KI, solution distilled water 0.020 M iron (IlI) chloride, FeCl. Vernier LabQuest LabQuest App Vernier Colorimeter or Spectrometer three 250 mL beakers two 100 mL beakers plastic cuvettes solution, in 0.10 M HC three 25 mL graduated cylinders five plastic Beral pipets 25-1 Advanced Chemistry with VernierExplanation / Answer
a. concentration is calculated alright for both FeCl3 and KI
An example would be,
Say for Trial 1:
[FeCl3] = 0.02 M x 20 ml/40 ml = 0.01 M
[KI] = 0.02 M x 20 ml/40 ml = 0.01
Similarly others can be calculated
b. Let the rate law be,
rate = k[FeCl3]^x.[KI]^y
with,
k = rate constant
x and y are order with respect to FeCl3 and KI
From Trial 1 and 3, [KI] is constant
rate 1/rate 3 = 0.00094855/0.0019259 = (0.01/0.005)^x
Taking log of both sides,
x = -1
From Trial 1 and 2, [FeCl3] is constant,
rate 1/rate 2 = 0.00094855/0.0036239 = (0.01/0.005)^y
taking log of both sides,
y = -2
The order with respect to [FeCl3] is -1 and with respect to [KI] is -2
c. The rate law becomes, rate = k/[FeCl3].[KI]^2
d. rate constant k = 0.00094855/(0.01).(0.01)^2 = 948.5 M^-2.s^-1
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