Question

The rate constant for the first order decomposition of H₂O₂ is given by the following equation:

log k = 14.34 − 1.25 × 10⁴ K/T

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

Answer 1

According to the Arrhenius equation,

k = `Ae^(-E_a//RT)`

or, log_e k = `log_e A - E_a/(RT)`

or, log₁₀ k = `log_10 A - E_a/(2.303 RT)`    ...(i)

The given equation is:

log₁₀ k = `14.34 - (1.25 xx 10^4 (K))/T`    ...(ii)

Comparing equations (i) and (ii), we have,

`E_a/(2.303 RT) = (1.25 xx 10^4 (K))/T`

or, E_a = 2.303 × R × 1.25 × 10⁴ (K)

= 2.303 × (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × 1.25 × 10⁴ (K)

= 239.34 kJ mol⁻¹

When t_1/2 = 256 min, i.e., 256 × 60 s, we have 

k = `0.693/t_(1//2)`

= `0.693/(256 xx 60)`

= 4.51 × 10⁻⁵ s⁻¹

Substituting this value in equation (ii), we have,

log₁₀ (4.51 × 10⁻⁵) = `14.34 - (1.25 xx 10^4 (K))/T`

⇒ −4.346 = `14.34 - (1.25 xx 10^4 (K))/T`

⇒ `(1.25 xx 10^4 (K))/T` = 14.34 + 4.346

⇒ `(1.25 xx 10^4 (K))/T` = 18.686

T = `(1.25 xx 10^4 (K))/18.686`

= 668.95 K