A certain reaction has an activation energy of 31.16 kJ/mol.

At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 321 K?

Use the following equation to solve :

Ea = 2.303xRx{T1xT2/(T2-T1)}x log k2/k1
Where Ea = Energy of activation = 31.16kJ/mol = 31160J/mol
R = molar gas constant = 8.314J/degree/mole
T1 = 321 K, k1 = rate constant at T1
T2 = ? , k2 = rate constant at T2 = 3.50 k1,
Substituting all these values in the equation
31160 = 2.303x8.314x321xT2/(T2-321) x log 3.50k1/k1
31160 = 2.303x8.314x321xT2/(T2-321) x log 3.50
31160 = 6146.2 T2/(T2-321) x 0.5441 = 3344.15 T2/(T2-321)
or 31160x(T2-321) = 3344.15 T2
or 31160 T2 - 10002360 = 3344.15 T2
or (31160 - 3344) T2 = 10002360
or 27816 T2 = 10002360
or T2 = 10002360/27816 = 359.6K