3 years ago

Please Help!!!!! Need by 11:30

Chemistry

1 answer:

Hitman42 [59]3 years ago

7 0

Answer:

D i think

Explanation:

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A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol

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Answer:

0.314 s⁻¹

Step-by-step explanation:

Use the Arrhenius equation.

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

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:

$\ln(\frac{k_{2} }{0.0275}) = (\frac{75 500 }{8.314})(\frac{ 1}{293.15} - \frac{1 }{318.15 })$

$\ln(\frac{k_{2} }{0.0275}) = 9081\times 2.681 \times10^{-4}$

$\ln(\frac{k_{2} }{0.0275}) = 2.434$

$\frac{k_{2} }{0.0275} = \text{e}^{2.434}$

$\frac{k_{2} }{0.0275} = 11.41$

k₂ = 0.0275 × 11.41

k₂ = 0.314 s⁻¹

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