Chemistry

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

asked by Caroline
  1. Never mind! I didn't change my R to kJ

    posted by Caroline
  2. Would you show how you arrived at the right answer?

    posted by Anonymous
  3. f=e^(-Ea/RT)
    f=e^(-151.0/(8.314e-3 x 292)

    posted by Caroline

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