2. The data below was collected and three graphs were plotted to determine the r

Question

2. The data below was collected and three graphs were plotted to determine the rate law for the generic, hypothetical reaction below: (5 pts) Time5 (min) 10 15 20 25 30 35 40 X 0.099 0.0497 0.0332 0.0249 0.02 0.0166 0.0143 0.0125 Lx1vsTime a) What is the overall reaction order? 0.1 y -0-001993x + 0.07863 R-0.706691 b) Determine the rate constant (k) for the reaction (including units). 0.06 0.02 Time (min) c) Determine the half-life of the reaction. 1/X] vs Time 90 I y = 1.997x + 0.1624 R2 0.999986 80 , 40 d) Determine [K] after 3.00 hours 10 Time (min) In[X] vs Time 50 e) Determine the rate after 3.00 hours. J' =-0.05479x-2.399 R½ 0.918968 2 -2 Time (min)

Explanation / Answer

order of reactant(X) and rate( -dX/dt ) is releated as -dX/dt= K[X]n, n = order of reaction and K is rate constant.

if X=Xo at t=0 and X=Xat t=t

for n=0, the reaction is zero order and -dX/dt=K, X= X0-Kt, so a plot of X vs t is straight line with a slope of -K

for n=1, the reactino is 1st order and -dX/dt= KX, when integrated, lnX= lnXo-Kt, hence for 1st order reaction to be obeyed, the plot of lnX vs t has to be straight line.

n=2 gives second order reaction and -dX/dt= K[X]2 , when integrated, 1/X= 1/Xo+Kt, for second order reaction to be obeyed, the plot of 1/X vs t has to be straight line whose slope is K.

from the data of curves given, best plot is obtained ( high R2 value) from 1/X vs t. The reaction is second order and slope = K= rate constant = 1.997/M.min

the equation of best fit is 1/X= 1.997*t+0.1624(1/Xo) (1)

XO= 1/0.1624= 6.16M

for X= Xo/2 gives half ifie (t1/2) from Eq,1, 1/Xo/2= 1.997*t1/2+1/Xo

1/Xo= 1.997* t1/2

t1/2= 0.1624/1.997 min=0.0813 min

after 3 hrs= 3*60 min =180min, from Eq.1 1/X= 1.997*180+0.1624

X= 0.0028M

after 3 hrs, -dX/dt= K[X]2= 1.997*(0.0028)2= 1.56*10-5 M/sec

1

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