Question

For the reaction of H₂ with I₂, the rate constant is 2.5 × 10⁻⁴ dm³ mol⁻¹ s⁻¹ at 327°C and 1.0 dm³ mol⁻¹ at 527°C. The activation energy for the reaction, in kJ mol⁻¹ is:
(R = 8.314 JK⁻¹ mol⁻¹)

पर्याय

• 59

• 166

• 72

• 150

Answer 1

166

Explanation:

Given: k_1​ = 2.5 × 10⁻⁴ dm³ mol⁻¹ s⁻¹ at T₁ = 327°C = 600 K

k₂ = 1.0 dm³ mol⁻¹ s⁻¹ at T₂ = 527°C = 800 K

R = 8.314 J K⁻¹ mol⁻¹

To find the activation energy E_a​, we use the Arrhenius equation in two-point form:

`ln(k_2/k_1) = E_a/R (1/T_1 - 1/T_2)`

⇒ `ln(1.0/(2.5 xx 10^-4)) = E_a/8.314 (1/600 - 1/800)`

ln(4000) = `E_a/8.314 (1/600 - 1/800)`    ...(i)

ln(4000) ≈ ln(4 × 10³) = ln(4) + ln(1000) ≈ 1.386 + 6.908 = 8.294

`(1/600 - 1/400) = ((1 xx 4)/(600 xx 4) - (1 xx 3)/(800 xx 3)) = (4 - 3)/2400 = 1/2400`

By putting the above-calculated value in equation (i), we get,

8.294 = `E_a/8.314 * 1/2400`

E_a ​= 8.294 × 8.314 × 2400

E_a ​= 165857.6 J mol⁻¹

= 165.9 kJ mol⁻¹

= 166 kJ mol⁻¹