Question

Two reactions, (i) \[\ce{A -> Products}\], and (ii) \[\ce{B -> Products}\], follow first order kinetics. The rate of reaction (i) is doubled when the temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minutes. At the same temperature, B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that of reaction (i) calculate the rate constant of the reaction (ii) at 300 K.

Answer 1

The given reactions are

(i) \[\ce{A -> Products}\], and

(ii) \[\ce{B -> Products}\]

Both the reactions are first order reactions.

According to the Arrhenius equation

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

If T₁ = 300 K and T₂ = 310 K, we have

For reaction (i),

`log_10  (k_((i))^(310))/(k_((i))^(300)) = (E_a (i))/(2.303 R) [1/300 - 1/310]`    ...(i)

For reaction (ii),

`log_10  (k_((ii))^(310))/(k_((ii))^(300)) = (E_a (ii))/(2.303 R) [1/300 - 1/310]`    ...(ii)

Dividing eq. (i) by eq. (ii), we have

`(log_10  (k_((i))^(310))/(k_((i))^(300)))/(log_10  (k_((ii))^(310))/(k_((ii))^(300))) = (E_a (i))/(E_a (ii))`    ...(iii)

Given that

1. `k_((i))^310 = 2 xx k_((i))^300`    ...[∵ The rate of reaction (i) gets doubled on increasing the temperature from 300 K to 310 K]
2. `k_((i))^310 = 0.693/30` = 0.231 min⁻¹    ...[∵ The half-life of reaction (i) at 310 K is 30 minutes]
3. `k_((ii))^310 = 2 xx k_((i))^300`    ...[∵ At 310 K, B decomposes twice as fast as A]
4. E_a (ii) = `1/2 E_a (i)`    ...[∵ The energy of activation for reaction (ii) is half that of (i)]

From eq. (iii) and relations (a), (b), (c) and (d), we have

`(log_10  (2 xx k_((i))^300)/(k_((i))^300))/(log_10  (2 xx 0.0231)/(k_((ii))^300)) = (E_a (i))/(1/2 xx E_a (i))`

or, `(log_10 2)/(log_10  (2 xx 0.0231)/(k_((ii))^300)) = 2`

or, `log_10  (2 xx 0.0231)/(k_((ii))^300) = (log_10 2)/2 = 0.1505`

or, `(2 xx 0.0231)/(k_((ii))^300)` antilog₁₀ 0.1505 = 1.4142

or, `k_((ii))^300 = (2 xx 0.0231)/1.4142 = 0.0327`

= 3.27 × 10⁻² min⁻¹

Hence, the rate constant of the reaction (ii) at 300 K is 3.27 × 10⁻² min⁻¹.