Question

Solve

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

Answer 1

Given:

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

Temperature (T) = 300 K

To find:

Fraction of molecule (f) with energy equal to E_a

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.