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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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 $179.9 k J / m o l$ , what will be the value of pre-exponential factor?

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Solution

The expression for the preexponentail factor can be obtained from Arrhenius equation.
$l o g k = l o g A - \frac{E_{a}}{2.303 R T}$

$l o g A = l o g k + \frac{E_{a}}{2.303 R T}$

$l o g A == l o g 2.418 \times 10^{- 5} + \frac{179.9 \times 10^{3}}{2.303 \times 8.314 \times 546}$

$l o g A = 12.592$

The pre-exponential factor, $A = 3.90 \times 10^{12} / s$ .

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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?

The expression for the preexponentail factor can be obtained from Arrhenius equation.logk=logA−Ea2.303RTlogA=logk+Ea2.303RTlogA==log2.418×10−5+179.9×1032.303×8.314×546logA=12.592The pre-exponential factor, A=3.90×1012/s.

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 $179.9 k J / m o l$ , what will be the value of pre-exponential factor?