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[Solved] The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol

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29/01/2022 6:28 pm

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The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Shilpi98

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class 12 cbse ncert chemical kinetics

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29/01/2022 6:51 pm

k= 2.418 × 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 × 10³ J mol^-1

According to the Arrhenius equation,

k = Ae^-E_a/RT

⇒ In k = In A - \(\frac{E_a}{RT}\)

⇒ In k = log A - \(\frac{E_a}{2.303RT}\)

⇒ log A = log k + \(\frac{E_a}{RT}\)

= log (2.418 x 10^-5 s^-1) + \(\frac{179.9 \times 10^3\ J \ mol^{-1}}{2.303 \times 8.314 \ Jk^{-1} mol^{-1} \times 546 K}\)

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s^-1 (approximately)

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[Solved] The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol

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