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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

3 answers

Never mind! I didn't change my R to kJ
Would you show how you arrived at the right answer?
f=e^(-Ea/RT)
f=e^(-151.0/(8.314e-3 x 292)
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