4. Consider two competing first order reactions A->B and A->C. One (B) has a pre

Question

4. Consider two competing first order reactions A->B and A->C. One (B) has a preexponential factor of 1010 units, the other (C) of 1013 units. At 144 deg F, both reactions proceed at a rate of 0.001 mol/(l s) in a 0.7 mol/l solution of A.

a. (4) What are the activation energies of each of the reactions

b. (4) At which temperature is the production of B twice as fast as that of C? At which temperature is the production of C twice as fast as that of B? If B is the desired product and C is not, would you rather run the reaction at low or high temperatures

Explanation / Answer

Temperature = 144 deg.F = (144-32)/1.8= 62.2 deg.c = 62.2+273= 335.2 K

From Arhenius equaton, ln K= lnKo-Ea/RT

for the reaction A--->B, lnK= ln(1010)- Ea/(R*335.2)

ln (0.001)= 6.92- Ea/(R*335.2)

Ea/(R*335.2)= 13.83

Ea= 13.83*8.314* 335.2 J/mole -38542 J/mole

for the reaction A--->C, lnK= ln (1013)- Ea/(R*335.2)

since lnK= rate constants are same, ln(0.001) = ln (1010)- Ea/(R*335.2), Ea= 38530 J/mole

let T = temperature at which the reactin of B is twice as that of C

ln(2K)= ln(1013)- 38542/RT, ln K= ln(1010)- 38530/ RT

hence subtracting the equations gives ln 2= ln(1013/1010)-1/RT*(38542-38530)=0.003- (1/RT)*12

0.693 = 0.003- (1/RT)*12

(1/RT)*12= -0.69

T = -12/(R*0.69)=-2.1 K

=-2.1-273= -275.1 deg.c

3. for this case       ln(K)= ln(1013)- 38542/RT, ln 2K= ln(1010)- 38530/ RT

subtracting ln (1/2)= 0.003- 12/RT

12/RT= 0.003+0.693 = 0.696

T= 12/(8.314*0.696)= 2.073 K

when reaction is run at high temperature, since the activation energy of formation of C is lower than B, more C is formed. At lower temperature, the rate of B formation is twice that of C. So low temperature is prefered,

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