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

`\boxed{\text{39 }^(\circ)\text{C}}`

Step-by-step explanation:

We can use the Arrhenius equation:

`\ln((k_2 )/(k_1)) = (E_(a) )/(R)((1)/(T_1) - (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((0.0320)/(0.0160)) = (40 100 )/(8.314)(( 1)/(299.15)} - (1 )/(T_(2) ))\\\\\ln2 = 4823(( 1)/(299.15)} - (1)/(T_(2) )) \\\\\ln2 - 16.12 = -(4823)/(T_(2) )\\\\-15.43 = -(4823)/(T_(2) )\\\\T_(2) = (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}}`.