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Firdavs [7]
3 years ago
12

A certain reaction has the form aA → bB. At a particular temperature and [A]0 = 2.80 × 10-3 molar, data was collected of concent

ration versus time for this reaction. A plot of 1/[A]t versus time resulted in a straight line with a slope value of 3.60 × 10-2 M-1s-1. What is the concentration of A after 300 seconds?
a.2.80 × 10-3 molar
b.2.72 × 10-3 molar
c.2.60 × 10-3 molar
d.3.60 × 10-2 molar
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
7 0
If the plot of 1/[A] versus time shows a constant slope, this means it is a second-order reaction. The equation would be:
1/[A] - 1/[A]0 = kt
1/[A] - 1/0.0028 = 0.036(300)
1/[A] = 0.00272 = 2.72 x 10^-3 M
This is equivalent to choice B.
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A reaction is second order respect to the reactant A. If the reaction is 85% complete in 12 minutes, how long would it take for
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Answer:

e. 22s

Explanation:

Initial Concentration = 100

Final concentration = 15 ( Since the reaction is 85% complete)

Time = 12 minutes = 720 seconds

The integrated rate law for second order reactions is given as;

1 / [A] = (1 / [A]o) + kt

Making k subject of interest, we have;

kt = (1 / [A] ) - (1 / [A]o )

k = (1 / [A] ) - (1 / [A]o ) / t

k = [(1/ 15) - (1 / 100) ] / 720

k = (0.0667 - 0.01 ) / 720

k = 0.0567 / 720

k = 7.875e^-5

How long would it take for the reaction to be 15% complete?

Final concentration = 85 (Since the reaction would be 15% complete)

The integrated rate law for second order reactions is given as;

1 / [A] = (1 / [A]o) + kt

Making t subject of interest, we have;

kt = (1 / [A] ) - (1 / [A]o )

t = (1 / [A] ) - (1 / [A]o ) / k

t = [(1 / 85 ) - (1 / 100) ] / 7.875e^-5

t = (0.0117 - 0.01 ) / 7.875e^-5

t = 0.001765 / 7.875e^-5

t = 22.41 seconds

Corect option = E.

3 0
4 years ago
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