For the bimolecular gaseous reaction: 2R(g) arrow Product, the fraction of molecules having sufficient energy for effective collision is 2.06× 10- 9 at 127° C . The activation energy for reaction is about (ln 2 . 06 = 0 . 727 , ln 10 = 2 . 303)

Question

For the bimolecular gaseous reaction: 2R(g) arrow Product, the fraction of molecules having sufficient energy for effective collision is 2.06× 10- 9 at 127° C . The activation energy for reaction is about (ln 2 . 06 = 0 . 727 , ln 10 = 2 . 303) (A) 16kJ/mole (B) 16kcal/mole (C) 8kcal/mole (D) 80kcal/mole

Tardigrade Admin · Accepted Answer

The fraction of effective collisions =e(- (Ea/RT)) 2.06× 10- 9=e(- (Ea/1 . 98 × 400)) ln (2.06 × 10-9 )=-( Ea /1.98 × 400) Ea≈16kcal/mole

Q. For the bimolecular gaseous reaction: $2 R (g) \to$ Product, the fraction of molecules having sufficient energy for effective collision is $2.06 \times 1 0^{- 9}$ at $12 7^{\circ} C$ . The activation energy for reaction is about $(l n 2.06 = 0.727, l n 10 = 2.303)$

$16 k J / m o l e$

$16 k c a l / m o l e$

$8 k c a l / m o l e$

$80 k c a l / m o l e$

The fraction of effective collisions $= e^{(- \frac{E a}{RT})}$
$2.06 \times 1 0^{- 9} = e^{(- \frac{E a}{1.98 \times 400})}$
$ln (2.06 \times 1 0^{- 9}) = - \frac{E a}{1.98 \times 400}$
$E a \approx 16 k c a l / m o l e$

Q. For the bimolecular gaseous reaction: 2R(g)→ Product, the fraction of molecules having sufficient energy for effective collision is 2.06×10−9 at 127∘C . The activation energy for reaction is about (ln2.06=0.727,ln10=2.303)

16kJ/mole

16kcal/mole

8kcal/mole

80kcal/mole