Cliffs AP Chemistry, 3rd Edition

4. The student used the ratios in Mixture I and ran the experiment at two different tempera-
   tures. Calculate the rate, the rate constant, log k and 1/T for each temperature studied.
   From the data, plot k versus 1/T and determine the activation energy. Given that the acti-
   vation energy for the reaction is 8.6 × 104 Joules, calculate the % error.
   25 °C 240 seconds
   40 °C 45 seconds

Begin by creating a table summarizing the known information and leaving room for calcu-

lated results:

Acetone H + (I 2 )o Temp °C Time (sec) rate k log k 1/T (K)

0.80 0.20 0.0010 25 240 4.2 × 10 –6 2.6 × 10 –5 −4.58 3.36 × 10 –3

0.80 0.20 0.0010 40 45 2.2 × 10 –5 1.4 × 10 –5 −3.86 3.19 × 10 –3

rate = k ⋅(acetone)^1 ⋅((I 2 )o)^0 ⋅(H+)^1

rate = (I 2 ) 0 ⋅time–1

At 25°C = 298 K:

..
rate sec
M
240
==0 0010 42 10# -^6

(. )(. )

k..

080 020

==42 10# -^6 26 10# -^5

At 40°C = 313 K:

..
rate sec
M
45
==0 0010 22 10# -^5

(. )(. )

k..

080 020

==22 10# -^5 14 10# -^4

Slope log k vs. 1/T

..

...

xx

yy

3 19 10 3 36 10

386 458 424 10

21

21

33

3

##

= - #

• =
  
  
  • 
    
  
  
  
  


• 
  
  
  
  • 
    

    =-

    
    

    . log.
    log
    EkRT
    T

    
    
  
  
  
  

k

a=-^230 ::::=-^230 - 1 R

Ea= –2.30 ×8.31 ×(−4240) = 8.10 × 104 Joules

% %%

.

..

exp

exp

error

obs

100 100

86 10

81 10 86 10

4

44

#

#

= - = ##- #

= −5.8%

Laboratory Experiments