# Solve What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? - Chemistry

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What fraction of molecules in a gas at 300 K collide with an energy equal to the activation energy of 50 kJ/mol?

#### Solution

Given:

Activation energy (Ea) = 50 kJ mol-1 = 50 × 103 J mol-1

Temperature (T) = 300 K

To find:

Fraction of molecule (f) with energy equal to Ea

Formula:

f = "e"^((-"E"_"a")/("RT"))

Calculation:

Substituting the given value above

f = "e"^(((-50 xx 10^3 "J"  "mol"^-1)/(8.314  "J" "K"^-1 "mol"^-1 xx  300  "K")))

log10f = ((-50 xx 10^3)/(8.314 xx 300 xx 2.303))

log10f = -8.70

f = antilog (-8.70)

f = 1.99 × 10-9 ≈ 2.0 × 10-9

Fraction of molecules having energy equal to activation energy is 2.0 × 10-9.

Concept: Temperature Dependence of Reaction Rates
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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercises | Q 4.9 | Page 137