The kinetics of the following second-order reaction were studied as a function o

Question

The kinetics of the following second-order reaction were studied as a function of temperature:
C2H5Br(aq)+OH(aq)C2H5OH(l)+Br(aq)

Temperature (C)

k (L/mols)

25

8.81×105

35

0.000285

45

0.000854

55

0.00239

65

0.00633

Determine the activation energy for the reaction.

Determine the frequency factor for the reaction.

Determine the rate constant at 10 C.

If a reaction mixture is 0.155 M in C2H5Br, and 0.260 M in OH, what is the initial rate of the reaction at 90 C?

Temperature (C)

k (L/mols)

25

8.81×105

35

0.000285

45

0.000854

55

0.00239

65

0.00633

Explanation / Answer

Arhenius Equation can be written as

K= Koe(-Ea/RT)

ko is frequency factor, Ea =activation energy and R= gas constant

lnK= lnK0 -Ea/RT

So a plot of lnK Vs 1/T gives a straight line having slope of -Ea/R and intercept K0, the frequeny factor

The slope is -Ea/R= -10770

and Ea= 10770*8.314 Joules/mole.K=89542 Joules/mole.K

lnKo= 26.78 ( intercept)

K0= Frequecny factor = e(26.78)= 4.3*1011

at 10 deg.c= 10+273.15 K

T= 283.15K and 1/T= 1/283.15=0.003532

from the equation

lnK(10 deg.c)= -10770*1/283.15+26.78=-11.26

K =1.29*10-5

at 90 deg.c= 90+273.15 K= 363.15K

lnK(363.15)= -10770/363.15+ 26.78

K= 0.056294 /M.Sec The rate constant at 90 deg.c

Rate for second order reacton R= K*[C2H5Br] [OH-] =0.056294*0.155*0.260=0.002269 M/Sec

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