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 preexponential factor-

According to Arrhenius equation, log k = log A E_a/2.303 RT k = 2.418 × 10^ 5 s^ 1; E_Q = 179.9 kJ mol^ 1 or 179900 J mol^ 1 R = 8.314 JK^ 1 mol^ 1 T = 546 ...

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Standard XII

Chemistry

Arrhenius Equation

The rate cons...

Question

The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{- 5} s^{- 1}$ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of preexponential factor?

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Solution

According to Arrhenius equation,
$l o g k = l o g A - \frac{E_{a}}{2.303 R T}$
$k = 2.418 \times 10^{- 5} s^{- 1}; E_{Q} = 179.9 k J m o l^{- 1} o r 179900 J m o l^{- 1} R = 8.314 J K^{- 1} m o l^{- 1} T = 546 K l o g A = l o g k + \frac{E_{a}}{2.303 R T} = l o g (2.418 \times 10^{- 5} s^{- 1}) + \frac{179900 J m o l^{- 1}}{2.303 \times (8.314 J k^{- 1} m o l^{- 1}) \times 546 K} = - 4.6184 + 17.21 = 12.5916 A = A n t i l o g 12.5916 = 3.9 \times 10^{12} s^{- 1} A = 3.9 \times 10^{12} s^{- 1}$

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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 preexponential factor?

The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{- 5} s^{- 1}$ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of preexponential factor?