At a certain temperature the first-order rate reaction 2N 2 O 5 4NO 2 + O 2 had

Question

At a certain temperature the first-order rate reaction

2N2O5 4NO2 + O2

had an initial concentration of N2O5 of 0.422 mol/L. After time t = 150.0 sec, theconcentration of N2O5 was found to be 0.211 mol/L.
1). What is the rate constant for this reaction (1/sec)?

For the reaction in the above question, at time = 0 sec the concentration of NO2O5 is 0.507 mol/L
2). How many half lifes will it take to reduce theconcentration
to 1 / 32 of that of the initial value ?

3). How many minutes will it take to reduce the concentration to 1 / 32 that of the of initial concentration givenabove (in question 2) ?

Explanation / Answer

(1)For the first order reaction K = ( 2.303 / t )* log ( a/ (a-x ) ) Where K = rate constant = ? t = time taken = 150 s a = initial concentretion = 0.422 mol / L (a-x) = concentration left after time t = 0.211 mol / s Plug the values we get , K = 4.62 * 10^-3 s^-1 (2)For the first order reaction K = 0.693 / T where T = half life =? T = 0.693 / K    = 149.9 s     1T               2T               3T                  4T                 5T   1 -------> 1/2 -------> 1/4  --------> 1/8 --------> 1/16 --------> 1/32 So for the reduction of the original amount to 1/32 , the no.of half lifes required are 5 So , the total time for this is 5 * half life                                       = 5 * 149.9                                       = 749.7 s

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