quiz/attempt.php?attempt = 502202 rong Trinh Liên Kêt Quóc Té Khoa Ky thuat Hoa

Question

quiz/attempt.php?attempt = 502202 rong Trinh Liên Kêt Quóc Té Khoa Ky thuat Hoa hoc> Hoa ly 2 (CH2031 n Quang Long (OT_HK171) » Chù dè 9 Group quiz 171 au hoi 1 hura duoc tra lo hám dióm cúa ,00 For a reaction A B, the following data were collected when a kinetic study was camed out at several temperatures between 25 and 45 oC. [A]. M t. min T-25 oC T-30 oC T-35 oC T-40 oC T-45 oC 0.750 0.750 0.750 0.750 0.750 15 0.648 0.622 0.590 0.556 0.520 0.562 0.530 0.490 0.440 0.400 0.514 0.467 0.410 0.365 0.324 60 0.460 0.410 0.365 0.315 0.270 75 0.4 0.378 0.315 0.275 0.235 9 0.385 0.336 0.290 0.243 0.205 (a) Use one of the data sets and make appropriate plots or perform linear regression to determine the order of the reaction (b) Having determined the rate constants at several temperatures, determine the activation energy

Explanation / Answer

for determining the order of reaction, one has to determine the order of reaction (n). The order of reaction is related to rate constant (K) as -dCA/dt= KCAn (1)-dCA/dt= rate of reaction. Integral method of analysis is used where ther order is assumed and Eq.1 is integrated. for n=0( Zero order) -dCA/dt=KCA , when integrated with boundary conditions at t=0, CA=CAO and at t=t, CA=CA

CA= CAO-Kt, So if zero order reaction is obeyed, the plot of CA vs t is a straight line.

for 1st order, -dCA/dt= KCA and the integrated equation -1 becomes lnCA= lnCAO-Kt, the plot of lnCA vs time is straight line.

for second order -dCA/dt= KCA2, the integrated equation-1 becomes 1/CA= 1/CAO+ Kt, so a plot of 1/CA vs time is straight line.

So all the three plots ( CA vs t, lnCA vs t and 1/CA vs t ) are generated at 25 deg.c are generated and shown below.

from the three plots, the plot of 1/CA vs t is straight line whose R2 value is maximum. Hence the reaction is second order.

Next the rate constant at all the temperatrues will have to determined by generating the plot of 1/CA vs time at all the temperatrues starting from 25 deg.c to 45 deg.c. slopes at each temperature, gives the rate constant. The plot is shown below.

now the data on rate constant K vs Temperature in K need to be generated K= t(deg.c)+ 273

Arhenius equation can be used to calculate the activation energy. The equation is

K= Ko*e-(E/RT)

K= rate constant, Ko= Frequency factor, E= activation energy, R= gas constant= 8.314 J/mole.K. T is temperature in K

Taking ln, the equation becomes

lnK= lnKo- E/RT

so a plot of lnK vs 1/T gives straight line whose slope is -E/R. The plot is generated and shown below.

from the plot, the slope is -E/R= -4943, E= 4943*8.314 J/mole =41096 J/mole= 41.096 KJ/mole

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.