A certain reaction has an activation energy of 40.02 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 331 K?

Question

I've tried this question so many times and keep getting 347 K but the website says that's the wrong answer. Can someone help me please?

Answer 1

You can use the Arrhenius equation and set k1 = 1 and k2 = 6.50 because rate = k[A] so as long as concentration of reactants doesn't vary, rate is proportional to k.

ln (k2/k1) = Ea/R(1/T1 - 1/T2)

k2 = 6.50

k1 = 1

Ea = 40,020 J/mole (note: change units to Joules so is consistent with units of R)

R = 8.314 J/mol-K

T1 = 331

T2 = ?

ln(6.50/1) = 40,020/8.314 (1/331 - 1/T2)

1.87 = 4814 (1/331 - 1/T2)

1.87 = 14.5 - 4814/T2

-12.7 = - 4814/T2

T2 = 379K

You might want to check my math, but this is the way the problem should be approached. Hope this helps.