rate law determination and activation energy Rate Law Determination and Activati
ID: 579578 • Letter: R
Question
rate law determination and activation energy Rate Law Determination and Activation Energy Pre-Lab Assignment 1. For this question, to ton reter to the table of soutions you woll make on page 2 of the experimental procedure a. Calculate the moles of S O that are consumed in Trial 1 b. Calculate the moles of ls that are produced in Trial 1 c. If the reaction takes 83 seconds to turn blue, what is the rate for Trial 1 2. Using the Method of Initial Rates, determine the order of the reaction and write the rate law for the tollowing A products A] (M) 0.100 0.200 0.300 Initial Rate (M/s) 0.053 0.210 0.473 3. If the slope of a line from a plot of 1/T versus In k is tound to be -11000 K, what is the activation energy, E. for the reaction in the unit kJ/mol?Explanation / Answer
1) Data required; please supply the data that you recorded for the experiment.
2) The given reaction is
A --------> Products
The rate of the reaction is given as Rate = k*[A]n where k is the rate constant for the reaction and n is the order of the reaction with respect to A.
Using the method of initial rates, we must have,
0.053 M/s = k*(0.100 M)n ……..(1)
0.210 M/s = k*(0.200 M)n ……..(2)
Therefore, (0.210 M/s)/(0.053 M/s) = [k*(0.200 M)n]/[k*(0.100 M)n]
====> 3.96 = (0.200/0.100)n
4.0 = (2)n
====> (2)2 = (2)n
Therefore, n = 2.
Put n = 2 in the first equation and obtain
0.053 M/s = k*(0.100 M)2
=====> 0.053 M/s = k*(0.01 M2)
=====> k = (0.053 M/s)/(0.01 M2) = 5.3 M-1s-1
The rate law for the reaction is Rate = (5.3 M-1s-1)[A]2 (ans).
3) The integrated form of the Arrhenius equation is given as
ln k = -Ea/R*1/T + c where k is the rate constant at temperature T; Ea is the activation energy for the reaction and c is the constant of intergration.
The above expression is of the form y = mx + b where m is the slope of the plot of y and x and b is the constant. Comparing the two equations, we obtain,
m = -Ea/R
It is given that m = -11000 K; therefore,
-11000 K = -Ea/(8.314 J/mol.K)
====> Ea = (11000 K)*(8.314 J/mol.K) = 91454 J/mol = (91454 J/mol)*(1 kJ/1000 J) = 91.454 kJ/mol (ans).
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