A.) The rate law for the combination of myoglobin, Mgb, with oxygen O 2, in homo

Question

A.) The rate law for the combination of myoglobin, Mgb, with oxygen O2, in homogenous aqueous solution to form oxymyoglobin, MgbO2 , is first order in each of these two reactants and therefore is second order overall.

i.) Calculate the rate constant k2 given that the time, t1/2, needed for the same initial concentrations of [Mgb]=[O2]=1.82*10-6M to be reduced to half of this value is 15.7ms at 25oC.

ii.) Hence, using your result for k2, obtain the time needed for the 90% conversion of these initial concentrations to MgbO2

iii.) what is the activation energy Ea if the rate constant increases by a factor of 1.4 from 298 to 310 K?

Explanation / Answer

for 2nd order reaction and equal concentratinos of reactants

KCAOt= Damkohler number = XA/(1-XA)

half life ( t1/2)is the time required for the concentraton to drop to 50% of its initial value.

XA= 0. 5, t= t1/2 =15.7*10-3 seconds   CAO= initial concentration of reactant = 1.82*10-6 M

K*1.82*10-6*t1/2 =1

K*1.82*10-6* 15.7*10-3 =1

K= 3.5*107/M.sec at 25 deg.c

b. Time required for 90% conversion, 3.5*107 * 1.82*10-6*t= 0.9/(1-0.9) =9

t= 0.142 sec=142 ms

c) let K1= 3.5*107 /M.sec , T1=298K

K2= 1.4*3.5*107/M.sec , T2= 310

K2/K1= 1.4

From Arhenius equation , ln(K2/K1)= (Ea/R)*(1/T1-1/T2), R= 8.314 J/mole.K, Ea= activatino energy

ln(1.4)= (Ea/R)*(1/298-1/310)=0.0001298* (Ea/R)

Ea= 22741 Joules/mole

B)

when concentratinon of each reactor are different

the conversion and time are related as

-ln({1-XA)/(M*(M-XA)= KCAO*(M-1)t,

M= CBO/CAO= 12/8= 1.5 = ratio of initlal concentrations of reactants

XA= conversion of [OH-]

XB= conversion of nitroethane =0.5

from CAOXA= CBOXB

XA = 12*0.5/8= 0.75

-ln({1-XA)/(M*(M-XA)= KCAO*(M-1)t, K= rate constant = 39.1/M.sec, t =time in seconds

XA=0.75, M=1.5 and CAO= 8mM=8*10-3 M

1.50= 39.1*8*10-3*0.5*t

t= 21.3 seconds time taken for 50% conversion of   B.

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