Question

The rate of the chemical reaction doubles for an increase of $ 10\,K $ in absolute temperature from $ 298\,K $ . Calculate $ E_{a} $

• A. 52.988 kJ
• B. 42.024 kJ
• C. 52.898 kJ
• D. 51.898 kJ

Answer 1

Answer: (C)

$\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]$
$\frac{k_{2}}{k_{1}}=2, T_{1}=298\, K, T_{2}=308\, K$
$R=8.314 \,J \,K^{-1}\, mol^{-1}$
$\log 2=\frac{E_{a}}{2.303 \times 8.314} \times\left[\frac{308-298}{308 \times 298}\right]$
$0.3010=\frac{E_{a}}{2.303 \times 8.314} \times \frac{10}{298 \times 308}$
$\therefore E_{a}=\frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$
$=52.898\, k J$