The first order rate constant for the decomposition of N2O5, 2N2O5--> 2NO2 + O2

Question

The first order rate constant for the decomposition of N2O5, 2N2O5--> 2NO2 + O2 at 70 degrees C is 6.82 x 10^-3 s^-1. Suppose westart with .0250 mol of N2O5 in a volume of 2.0 L.
a. how many moles of N2O5 will remain after 5.0 min?
b. how many minutes will it take for the quantity of N2O5 todrop to .010 mol?
c. What is the half-life of N2O5 at 70 degrees?
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I don't understand how to find out how many moles will remainafter the time period, or to find how long it will take for it todrop to .010.
Also, does temperature make a difference for half lifes?
a. how many moles of N2O5 will remain after 5.0 min?
b. how many minutes will it take for the quantity of N2O5 todrop to .010 mol?
c. What is the half-life of N2O5 at 70 degrees?
-----
I don't understand how to find out how many moles will remainafter the time period, or to find how long it will take for it todrop to .010.
Also, does temperature make a difference for half lifes?

Explanation / Answer

a )    k = 6.82 x 10-3s-1     [A]0 = 0.0250 mol / 2.0L            = 0.0125 M    [A]   = ?     t      = 5min            =5 * 60 s            =300 s Formula :                                   kt = ln [A]0/ [A]   6.82 x 10-3 s-1*  300 s = ln ( 0.0125 M) / [A]            2.046                   =ln ( 0.0125 M) / [A]                                [A]  = 0.00161 M Number ofmoles            = 0.00161 mol / L * 2 L                                       = 0.00322 mols b )     t   = 1/6.82 x10-3 s-1 ln ( 0.0250 mols / 0.010 mols) ( As the volume is same , mols are considered)             = 133.43 s             = 2.23 min c )     t1/2 = 0.693 /k              = 0.693 / 6.82 x 10-3 s-1               =101.61 s

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