In: Chemistry
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
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,