Question

or 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 (ln 2.06 = 0.727, ln 10 = 2.303) -
(1)
16 kJ/mole
(2)
16 kcal/mole
(3)
8 kcal/mole
(4)
80 kcal/mole

Answer 1

The activation energy for reaction is 27514.34 J/mol

Explanation:

The activation energy is given by the formula:

$\mathbf{A}=\mathbf{e}^{-\frac{E_{a}}{\mathbf{R}{T}}}$

Where,

A = Fraction of molecules that have enough energy to form product = 2.06 × 10⁻⁹

Ea = Activation energy = ?

T = Temperature in Kelvin = 127 + 273 = 400 K

R = Gas constant = 8.3145 J/(mol.K)

On substituting the values, we get,

$2.06 \times 10^{-9}=e^{-\frac{E_{a}}{(8.3145 \times 400)}}$

$2.06 \times 10^{-9}=e^{-\frac{E_{a}}{3325.8}}$

On taking log on both sides, we get,

$\log \left(2.06 \times 10^{-9}\right)=-\frac{E_{a}}{3325.8}$

$\log (2.06)+\log \left(10^{-9}\right)=-\frac{E_{a}}{3325.8}$

$0.727+(-9) \log (10)=-\frac{E_{a}}{3325.8}$

$0.727-9=-\frac{E_{a}}{3325.8}$

$-8.273=-\frac{E_{a}}{3325.8}$

$E_{a}=8.273 \times 3325.8$

$\therefore E_{a}=27514.34 \ \mathrm{J} / \mathrm{mol}$