Question
In: Chemistry
Consider the reaction A -> products at 312 K. For this reaction it was observed that...
Consider the reaction A -> products at 312 K.
For this reaction it was observed that the first three half lives were 10.7 h, 5.35 h, and 2.675 h when [A] 2.873M. How long will it take for [A] to decrease by 47%?
Time for [A] to decrease by 47% (in hours) = Anwser is 10.1h need help on how to get this anwser.
Solutions
Expert Solution
From the given data we can analyse that after each half-life period, concentration of [A] drops in half when we plot [A] versus half life time it gives a curved (exponential) line that shows it is a first-order reaction.
For first order reaction the half life formula is
From here we can calculate the value of k
The equation for the first order reaction that relates the
remaining concentration of A to the time
Here the remaining concentration A is 100-47 = 53% of initial concentration.
Initial concentration A0 is = 2.873 M
Therefore from the formula
The correct answer should be 9.8 hours.
queen_honey_blossom answered 6 months agoRelated Solutions
Consider the reaction A → products at 312 K. For this reaction it was observed that...
Consider the reaction A → products at 312 K. For this reaction
it was observed that the first three half-lives were 11.3 h, 22.6
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second order
Consider the reaction A → products at 312 K. For this reaction it was observed that...
Consider the reaction A → products at 312 K. For this reaction
it was observed that the first three half-lives were 11.3 h, 22.6
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A⟶products
T (K)
k (s–1)
275
0.383
875
0.659
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point 1: ?=
point 1: ?=
For the reaction observed at 25oC:  
For the reaction observed at
25oC:
2CHCl3 (l) + O2
(g) à 2COCl2 (g) + 2HCl (g)
use the data in the table below.
ΔHof kJ/mol
ΔGof kJ/mol
CHCl3 (l)
COCl2 (g)
HCl (g)
O2
(g)
-134
-220
-92.3
0
-73.7
-206
-95.3
0
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