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Use the References to access important values if needed for this question. A stu

ID: 562716 • Letter: U

Question

Use the References to access important values if needed for this question. A student obtained the following data for the decomposition of nitrous oxide at 565 °C. N20(g)--N2(g) + ½ O2(g) IN20], M 1.42 0.710 0.355 0.178 seconds 862 2.58x103 6.01x103 (1) What is the half-life for the reaction starting at t-0s? 862 What is the half-life for the reaction starting at t-862 s? 862 Does the half-life increase, decrease or remain constant as the reaction proceeds? increase (2) Is the reaction zero, first, or second order? second (3) Based on these data, what is the rate constant for the reaction? 8.17e-4 M's1

Explanation / Answer

the order of reaction (n) need to be determined.

The order is related to rate (-dN2O/dr) as -dN2O/dt= K[N2O]n, K is rate constant

for n=0, the reaction is zeor order and for zero order- dN2O/dt=K , when integrated, the equation becomes

[N2O]= [N2O]0-Kt, where [N2O]0 is initial concentration and [N2O] is concentration at any time. for zero order, the plot of [N2O] vs t need to be straight line.

for 1st order reaction, n=1 and the integrated rate expression becomes ln[N2O]= ln[N2O]0-Kt. So a plot of ln[N2O] vs t need to be straight line.

for second order, n=2 and the integrated rate expression becomes 1/[N2O]= 1/[N2O]0+ Kt, so a plot of 1/[N2O] vs time is straight line. All the three plots are verified in the following graphs.

For convenicem a plot of respective concentration terms vs t/100 is drawn. They are shown below.

the plots are

best fir is obtained for second order reaction with R2=1. The equation of best fit is

1/[N2O]= 1/[N2O]+0.081/100 * t

K= 0.081/100 =0.00081/M.sec

half life is the time required for concentratin to drop to 50% of the intial value.

i.e [N2O]= [N2O]o/2

hence 1/[N2O]= 1/[N2O]0+ Kt becomes

2/[N2O]00 =1/[N2O]0 + K* half life

half life = 1/[N2O]o*K

at t=0, [N2O]0= 1.42M

half lufe= 1/(1.42* 0.00081)=869 sec

at t= 862, [N2O]=0.71

half lfie = 1/(0.710* 0.00081)= 1738 seconds

as the reactin proceeds, half life increases.