rated Rate Laws first-, and second- ch that they t line, Graph slope [A] vs. t -

Question

rated Rate Laws first-, and second- ch that they t line, Graph slope [A] vs. t -k n A vs. t -k vs. t k signed in as dana jagarna Belp Co previous l 7 of 21 nex Part A The reactant concentration in a zero order eaction was 0.100 Mafter 195 s and 1.50x10 2 Mafter 320 s What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. Value Units Submit Hints My Answers Give Up Review Part Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. Value Units

Explanation / Answer

Part A :

The rate law for a zero-order reaction is
[A] = -kt + [A]o.
We have [A] at 195s = 0.1 M and [A] at 320s = 1.5 x 10-2 M

0.1 = -k x 195 + [A]o   and 1.5 x 10-2 M = -k x 320 + [A]o

0.1 - 1.5 x 10-2 = (-k x 195) - (-k x 320 )
0.085 = 125k
k = 6.8 x 10-4 M/s

Part B:

[A] = -kt + [A]o.
0.1 = - 6.8 x 10-4 x 195 + [A]o

[A]o = 0.233 M

Part C:
at 20 s [A] = 8.10 x10-2 M and at 60s [A] = 6.60 x10-3 <
The rate law for a first-order reaction is:

ln [A] = -kt + ln [A]o.

Therefore;

ln(8.10 x10-2 ) = -k x 20 + ln [A]o and ln [6.60 x10-3] = -k x 60 + ln [A]o.

ln(8.10 x10-2 ) - ln [6.60 x10-3] = -k x 20 - (-k x 60 )
k = 0.0626 s-1

Part D:

The rate law for a Second-order reaction is:

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

0.76 M at 180s and 3.30 x 10-2 M at 880s

(1/ln [0.76 ]) = k x 180 + (1/ln [A]o) and (1/ln [3.30 x 10-2]) = k x 880 + (1/ln [A]o)

(1/ln [0.76 ])- (1/ln [3.30 x 10-2]) = k x 180 - k x 880

-3.35 = -700k
k = 0.0048 M-1 s-1 = 4.8 x 10-3 M-1 s-1

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