The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10¹⁰s^–1. Calculate k at 318K and E_a.

Question

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10 10 s –1 . Calculate k at 318K and E a .

Philoid · Accepted Answer

We know, time t = (2.303/k) × log([R]₀/[R])

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]₀/0.9[R]₀)

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]₀/0.75[R]₀)

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k₂/k₁ = (E_a / 2.303 R) × (T₂ - T₁) / T₁T₂

Substituting the values,

⇒

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1

We know, log k = log A –E_a/RT

Log k = log(4 × 10¹⁰)-(76.64kJ mol^-1/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^-2 s ^-1

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s–1. Calculate k at 318K and Ea.

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10¹⁰s^–1. Calculate k at 318K and E_a.