Answer:
0.314 s⁻¹
Step-by-step explanation:
Use the Arrhenius equation.
[tex]\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\[/tex]
Data:
k₁ = 0.0275 s⁻¹; k₂ = ?
Eₐ = 75.5 kJ·mol⁻¹
T₁ = 20 °C = 293.15 K; T₂ = 45°C = 318.15 K
Calculation:
[tex]\ln(\frac{k_{2} }{0.0275}) = (\frac{75 500 }{8.314})(\frac{ 1}{293.15} - \frac{1 }{318.15 })[/tex]
[tex]\ln(\frac{k_{2} }{0.0275}) = 9081\times 2.681 \times10^{-4}[/tex]
[tex]\ln(\frac{k_{2} }{0.0275}) = 2.434[/tex]
[tex]\frac{k_{2} }{0.0275} = \text{e}^{2.434}[/tex]
[tex]\frac{k_{2} }{0.0275} = 11.41[/tex]
k₂ = 0.0275 × 11.41
k₂ = 0.314 s⁻¹
The value of [tex]k[/tex] (rate constant) will be 0.314 s⁻¹ at 45°C if activation energy is 75.5 kJ/mol.
It describes the effect of temperature change on the rate of reaction.
[tex]{\rm ln}\dfrac{k_2}{k_1} = (\dfrac {E_a}{R}){\dfrac 1{T_1}} - \dfrac 1{T_2}}[/tex]
Where,
k₁ - rate constant = 0.0275 s⁻¹
k₂ = ?
Eₐ - Activation energy = 75.5 kJ·mol⁻¹
T₁- Initial temperature = 20 °C = 293.15 K
T₂- Final temperature = 45°C = 318.15 K
Put the values in the formula,
[tex]{\rm ln}\dfrac{k_2}{0.0275} = (\dfrac {75.5}{8.314}){\dfrac 1{293.15 }} - \dfrac 1{318.15}}\\\\K_2 = 0.314[/tex]
Therefore, the value of [tex]k[/tex] (rate constant) will be 0.314 s⁻¹ at 45°C if activation energy is 75.5 kJ/mol.
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