Question

The activation energy of a certain reaction is 40.1 kj/mol . at 26 ∘c , the rate constant is 0.0160s−1. at what temperature in degrees celsius would this reaction go twice as fast? express your answer with the appropriate units. view available hint(s)

Answer 1

Answer:

$\boxed{\text{39 }^{\circ}\text{C}}$

Explanation:

We can use the Arrhenius equation:

$\ln(\frac{k_2 }{k_1}) = \frac{E_{a} }{R}(\frac{1}{T_1} - \frac{1 }{T_2 })$

Data:

Eₐ = 40.1 kJ·mol⁻¹

k₁ = 0.0160 s⁻¹;              k₂ = 0.0320 s⁻¹

T₁ = 26 °C = 299.15 K; T₂ = ?  

Calculations:

$\ln(\frac{0.0320}{0.0160}) = \frac{40 100 }{8.314}(\frac{ 1}{299.15}} - \frac{1 }{T_{2} })\\\\\ln2 = 4823(\frac{ 1}{299.15}} - \frac{1}{T_{2} }) \\\\\ln2 - 16.12 = -\frac{4823}{T_{2} }\\\\-15.43 = -\frac{4823}{T_{2} }\\\\T_{2} = \frac{4823}{15.43} =\textbf{312.6 K}$

T₂ = 312.6 K = 39 °C

The reaction will go twice as fast at $\boxed{\text{39 }^{\circ}\text{C}}$.