The concentration vs. time data below was collected for the decomposition of N_2

Question

The concentration vs. time data below was collected for the decomposition of N_2O_5 at 55 degree C. 2 N_2O_5(g) rightarrow 4 NO_2(g) + O_2(g) What quantity should be plotted vs. time to find out if the order with respect to N_2O_5 is: 0^th 1^st 2^nd Make a plot to show that the decomposition of N_2O_5 os is first order. Write the rate equation and find the value of the rate constant, k? For first order reactions the half-life, t_1/2 = 0.693/k, is independent of concentration. How could you show that the reaction is first order with respect to N_2O_4by simply inspecting the data given in Number 1? What is the half-life and the value of k?

Explanation / Answer

The relation between order and concentration is given by

-dCA/dt= KCAn where n is order and K = rate constant . CA= Concentration at any time and CAO= initial conentration.

When n= 0, the reaction is zero order. Hence CA= CAO-Kt. So a plot of CA vs t gives a straight line.

For n=1, -dCA/dt =KCA when integrated we get, lnCA= lnCAO-Kt, so a plot of lnCA vs t gives a straight line whose slope is -K

For n=2, --dCA/dt= KCA2, when integrated 1/CA= 1/CAO+ kt, a plot of 1/CA vs t gives a straight line whose slope is K.

from the plot of lnCA vs t slope is =0.001 /sec

half life is the time required for the concentration to drop to 50% of initial value. From the data point it is 400 sec for concentration to reach 0.010M from 0.020M. hence half life =400 sec and K=0.693/400 =0.001733/sec

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