Question

The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.

Answer 1

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

$ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})$

And as Half-life in a first order reaction is:

$t_{1/2}=\frac{ln2}{K}$

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

$58.0min=\frac{ln2}{K}$

K = 0.01195min⁻¹ = K₁

$ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})$

$ln\frac{0.01195min^{-1}}{K_2} =1.47$

$\frac{0.01195min^{-1}}{K_2} =4.35$

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

$t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}$

Half-life at 629K = 252.4min