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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).