Question

At 19.0°C, what is the fraction of collisions with energy equal to or greater than an activation energy of 151.0 kJ/mol?

I got 1.06 which wasn't right

Answer 1

Never mind! I didn't change my R to kJ

Answer 2

Would you show how you arrived at the right answer?

Answer 3

f=e^(-Ea/RT)
f=e^(-151.0/(8.314e-3 x 292)