Question

A certain reaction has an activation energy of 50.05 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 307 K?

Answer 1

Arrhenius equation

K=Aexp(-Ea/RT)

Where K=rate constant

A=frequency factor

Ea=activation energy

R=gas constant=8.314J/K.mol

T=temperature

Higher the Rate constant ,faster is the reaction

Or, ln K=lnA-Ea/RT

Case 1)

ln k1=lnA-(50.05*1000 J/mol/8.314 J/K.mol*307K)

or,ln K1=lnA-19.609 K^-1…………….(1)

case 2)

ln k2=lnA-(50.05*1000 J/mol/8.314 J/K.mol*T2)

Also k2=6.5 K1

ln 6.5K1= lnA-(50.05*1000 J/mol/8.314 J/K.mol*T2)

ln 6.5K1=lnA-6019.966/T2…………..(2)

equation (1)-(2)

ln K1/6.5K1=-19.609 K-1+6019.966/T2

ln 1/6.5=-19.609 K-1+6019.966/T2

ln 0.154=-19.609 K-1+6019.966/T2

-1.871=-19.609 K-1+6019.966/T2

17.738=6019.966/T2

T2=339.38K (answer)