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.

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#### Solution

*k* = 2.418 × 10^{−5} s^{−1}

*T* = 546 K

*E*_{a} = 179.9 kJ mol^{−1} = 179.9 × 10^{3} J mol^{−1}

According to the Arrhenius equation,

`k = Ae^((-E_a)/"RT")`

`=>In k = In A - E_a/"RT"`

`=>log k = log A - E_a/(2.303 RT)`

`=> logA = log k + E_a/(2.303 RT)`

`=log(2.418 xx10^(-5) s^(-1))+ (179.9xx10^3 "J mol"^-1)/(2.303xx8.314 " JK"^(-1) mol^(-1)xx546K)`

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10^{12} s^{−1} (approximately)

Is there an error in this question or solution?

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