Question

The rate of chemical reaction doubles for an increase of 10k from 298 k. Calculate ea

Answer 1

Hey !!

log k₂/k₁ = Ea / 2.303 R [T₂ - T₁] / T₂T₁

 Ea = 2.303 R log k₂/k₁ [T₁T₂ / T₂ - T₁]

= (2.303) (8.314 J K⁻¹ mol⁻¹) (log 2/1) × ( 298 K × 308 K ) /( 308 - 298 K )

= 52898 J mol⁻¹ = 52.9 kJ mol⁻¹

Hope it helps you !!

Answer 2

Answer:

• 52.9 KJ mol$^{-1}$

Explanation:

Given

• Initial temperature (T1) = 298 K
• Final temperature (T2) = (298 + 10)K = 308 K
• Reaction doubles when temperature is increased by 10°

Let's assume

• Value of k1 = k
• k2 = 2k
• R = 8.314JK$^{-1}$mol$^{-1}$

Formula to be used:

$\boxed{ln\frac{k_{2}}{k_{1}}=\frac{E}{R}\bigg[\frac{T_{2}-T_{1}}{T_{1}T_{2}}\bigg]}$

$\implies\:\:log\frac{k_{2} }{k_{1}}= \frac{E}{2.303R} \bigg[\frac{T_{2}-T_{1}}{T_{1}T_{2}}\bigg]$

$\implies\:\:log\frac{2k}{k}\frac{E}{2.303R}\bigg[\frac{308-298}{298\times308}\bigg]$

$\implies\:\:log\frac{2k}{k}\frac{E_{a}}{2.303 \times 8.314}\bigg[ \frac{10}{298\times308}\bigg]$

$\implies\:\:log2= \frac{E_{a}}{2.303 \times 8.314}\bigg[ \frac{10}{298\times308}\bigg]$

$\implies\:\:E_{a} = \frac{ 2.303 \times 8.314 \times298 \times 308 \times log 2}{10}$

$\implies\:\:52897.78 J mol^{-1}$

$\implies\:\:52.9 \:KJ\: mol</p><p>^{-1}$