Answer:
2.2 × 10⁻² min⁻¹
Step-by-step explanation:
We can 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 = 9.8 × 10⁻³ min⁻¹; k₂ = ?
Eₐ = 126 kJ·mol⁻¹
T₁ = 30 °C; T₂ = 35°C
Calculations:
(a) Convert temperatures to kelvins.
T₁ = (30 + 273.15) K = 303.15 K
T₂ = (35 + 273.15) K = 308.15 K
(b) Activation energy
[tex]\ln(\frac{k_{2} }{9.8 \times 10^{-3}}) = (\frac{126 000 }{8.314})(\frac{ 1}{303.15} - \frac{1 }{308.15 })[/tex]
[tex]\ln(\frac{k_{2} }{9.8\times10^{-3}}) = 15160\times 5.35 \times10^{-5}[/tex]
[tex]\ln(\frac{k_{2} }{9.8 \times 10^{-3}}) = 0.811[/tex]
[tex]\frac{k_{2} }{9.8\times 10^{-3}} = \text{e}^{0.811}[/tex]
[tex]\frac{k_{2} }{9.8 \times 10^{-3}} = 2.25[/tex]
k₂ = 9.8 × 10⁻³× 2.25
= 0.022 min⁻¹
= 2.2 × 10⁻² min⁻¹
The rate constant of the reation at 35 degree is 2.1 × 10^-2.
Using the Arrhnius equation at two temperatures, we have;
ln(k2/k1) = -Ea/R(1/T2 - 1/T1)
Where;
k2 = ?
k1 = 9.8 × 10-3 min-1
T2 = 35 + 273 = 308 K
T1 = 30 + 273 = 303 K
R = 8.314 J/K/mol
Ea = 126 * 10^3 J/mol
Substituting the values;
ln(k2/ 9.8 × 10-3) = -( 126 * 10^3)/ 8.314 (1/308 - 1/303)
ln(k2/ 9.8 × 10-3) = 0.758
e^ln(k2/ 9.8 × 10-3) = e^0.758
k2/ 9.8 × 10-3 = 2.134
k2 = 2.134 * 9.8 × 10-3
k2 = 2.1 × 10^-2
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