Question
Asked Feb 15, 2020
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A: Consider the following reaction:

N2H5+  +  4 Fe(CN)63-  -->  N2(g)  +  5 H+  +  4 Fe(CN)64-

If the starting [N2H5+] is 0.18 M and the reaction half-life is 45.0 s, what is the N2H5+ concentration after 3.00 minutes?

Group of answer choices
0.18 M
0.045 M
0.090 M
0.023 M
 
B:

If the half-life for a first-order reaction is 2.54 hrs, what is the rate constant for this reaction?

Group of answer choices

1.76 hr-1

0.273 hr-1

126 hr-1

0.693 hr-1

 
 
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Expert Answer

Step 1

Half-life of a reaction is the time at which the initial concentration of the reactant becomes half.

The half-life of a reaction can be calculated as using the following formula :

Chemistry homework question answer, step 1, image 1

Again, for a 1st order reaction, half-life is expressed as:

Chemistry homework question answer, step 1, image 2

Step 2

Part A :

The given reaction is :

N2H5+  +  4 Fe(CN)63-  →  N2(g)  +  5 H+  +  4 Fe(CN)64-

Initial concentration, [N2H5+]o = 0.18 M

Reaction half-life = 45.0 s

Time, t = 3.00 min = 3×60 s = 180 s

 

The N2H5+ concentration after 3.00 minutes, [N2H5+]t can be calculated using equation (1) as :

Chemistry homework question answer, step 2, image 1

The answer can be confirmed with help of the definition of half-life also.

Half-life of a reaction is the time at which the initial concentration of the reactant becomes half.

As per the definition, the concentration can be calculated as :

Chemistry homework question answer, step 2, image 2

The correct answer as per the values given in part A will be = 0.01125 M  (not any of the answer choices given)

Step 3

Part B :

Given,

Half-life for a first-order reaction = 2.54 hrs

The rate constant for this reaction can be calculated using equation (2) as :

Chemistry homework question answer, step 3, image 1

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