ix. What fraction of molecules in a gas at 300 K collide with an energy

chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution

ix. What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol ? (2 × 10-9)

Answer:-

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 values:

f = e^{((-50 × 10³ J mol^-1)/(8.314 J K^-1 mol^{-1> × 300 K))}}

log₁₀f = ((-50 × 10³)/(8.314 × 300 × 2.303))

log₁₀f = -8.70

f = antilog (-8.70)

f ≈ 2.0 × 10^-9

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