The rate constant for the decomposition of hydrocarbons is 2-418 - 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-

K = 2.418 × 10−5 s−1 T = 546 K Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1 According to the Arrhenius equation, = (0.3835 − 5) + 17.2082 = 12.5917 Therefore, A ...

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Standard XII

Chemistry

Stability of Conformations

The rate cons...

Question

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

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Solution

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,

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s⁻¹ (approximately)

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The rate constant for the decomposition of hydrocarbons is 2.418 × 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.

k = 2.418 × 10−5 s−1 T = 546 K Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1 According to the Arrhenius equation, = (0.3835 − 5) + 17.2082 = 12.5917 Therefore, A = antilog (12.5917) = 3.9 × 1012 s−1 (approximately)

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

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,

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s⁻¹ (approximately)