(She posted it early this morning and asked us to return it tomorrow morning in
ID: 926396 • Letter: #
Question
(She posted it early this morning and asked us to return it tomorrow morning in class. I already have 2 final exams tomorrow, my brain is not on it right now. Its part of my final grades! PLEASE HELP !!! Save my life! )
P.S: Show your work please!
Name:………………………………… section:…………………….. Grade: (………/30 points)
8.23 ×10-3 mol of InCl (s) is placed in 1.00 of 0.010 M HCl (aq) at 25C. The InCl (s) dissolves quite quickly. and then the following reaction occurs:
3In+ (aq) 2 In (s) + In3+ (aq)
As this disproportionation proceeds, the solution is analyzed at intervals to determine the concentration on In+ (aq) that remains.
Time (s) [In+] (mol. L-1)
0 8.23 × 10-3
240 6.41 × 10-3
480 5.00 × 10-3
720 3.89 × 10-3
1000 3.03 × 10-3
1200 3.03 × 10-3
10000 3.03 × 10-3
a) Determine the rate constant and the half- life of this reaction and predict the concentration of [In+] after 600 seconds.
b) Determine the equilibrium constant K for the reaction under the experimental conditions.
c) Determine the standard Gibbs energy (Grxn) under these conditions.
d) What is the activation energy for this reaction under these conditions if the frequency factor for this reaction under these conditions is 5.6 ×10-5 ?
e) What will be the standard cell potential for this reaction?
Explanation / Answer
a) we know that
ln[A]=kt+ln[A]o
ln (6.41 × 10-3) = -kX 240 + ln(8.23 × 10-3)
-5.05 = -k X 240 - 4.79
0.26 = k X 240
So rate constant = 0.00108 sec^-1
(ii) t1/2 = 0.693 / k = 641.66 seconds
(iii) ln[A]=kt+ln[A]o
ln[A] = -0.00108 X 600 + ln (8.23 × 10-3)
ln[A] = -0.648 - 4.79 = -5.438
[A] = 0.00434 = 4.34 X 10^-3
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