Question

A certain reaction has an activation energy 52.93 kJ/mol. At what kelvin temperature will the reaction proceed 8.00 times faster than it did at 315K?

Ea = 52.93 kJ/mol = 52930J
R (gas constant) = 8.314
T1 = 315K
k1 = rate constant at T1
T2 = ?
k2 = rate constant at k2 = 8.00

Equation:
Ea = -Rln (k2/k1) / (1/T2 - 1/T1)

Answer 1

Since the equation is already given, we just plug in the values.
It was said in the problem that the reaction should proceed 8 times faster than the original at 315 K. Thus the ratio k2/k1 = 8.
R is the universal gas constant = 8.314 J/mol-K
All temperatures must be absolute (in Kelvin units).
Substituting,
Ea = -R * ln (k2/k1) / (1/T2 - 1/T1)
52.93 kJ/mol * (1000 J / 1 kJ) = -8.314 J/mol-K * ln(8) / (1/T2 - 1/315 K)
Solve for T2. Units in K. Note that the answer should be more than 315 K.

hope this helps~ `u`