What is the activation energy for this reaction in the presence of the catalyst?
ID: 482483 • Letter: W
Question
What is the activation energy for this reaction in the presence of the catalyst?
Express your answer using three significant figures.
Mastering chemistry. Homework chapter 13 Lectures EFGH Google Chrome Secure https:// ession masteringchemistry.com /myct/itemView? assignment ProblemID 78030294 CHEM 121 SP17 TALAGA Exercise 13.10 Resources e previous 24 of 24 I return to assignment Exercise 13.101 A certain compound, A, reacts to form products according to the reaction A P. The amount of A is measured as a function of time under a variety of different conditions. 25.0 °C 35.0 °C 45.0 °C Time (s) Al (M Al (M Al (M 1.000 1.000 1.000 0.779 10 0.662 0561 0,591 0461 0.312 0.453 0.338 0.100 0.259 0.057 0.136 0.200 0.093 0.032 The same reaction is conducted in the presence of a catalyst, and the following data are obtained: 25.0 °C 35.0 °C 45.0 °C Time (s) Al (M Al (M Al (M) 1.000 1.000 1.000 0.724 0,668 0.1 0,598 0.2 0.511 0.433 0.341 0.3 0.375 0.291 0.202 0.190 0.119 0,4 0.275 0.122 0.198 0.5 0,071 0.141 0.080 0.6 0.043 Part A 4:29 PM Cortana. Ask me an ENG 2/12/2017Explanation / Answer
For determining the activation energy, first we need to determine the rate constant and order in the absence of catalyst.
Order can be determined by methods of integral analysis. The plot of CA vs time gives some idea with respect to order of reaction. For 1st order, the concentration vs time graph is exponential.
The plot of CA vs time is shown below.
The equation for 1st order is CA= CAO*e(-Kt)
Where K is rate constant and CA is concentration at any time, t. The plots for all the three temperatures gives the rate constants.
The plots are shown below.
the rate constant (K) at three different temperatures ( in K) are
298 0.02/sec, 308 : 0.04/s and 318 : 0.05/s
From lnK= lnKo-Ea/RT, a plot of lnK vs 1/T gives a straight line whose slope can be used to calculate the activation energy.
the plot along with data points is shown below
Ea/R = 4364 and Ea= 4364*8.314 J/mole =36282 J/mole
similar calculations are done in the presence of catalyst and shown below
-2.9957323
the slope is Ea/R= 2422, Ea= 2422*8.314=20137 J/,mole
K T 1/T lnK 0.02 298 0.0033557 -3.912023 0.04 308 0.0032468 -3.2188758 0.05 318 0.0031447 -2.9957323Related Questions
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