The rate constant for the decomposition of a hydrocarbon 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?

Question

The rate constant for the decomposition of a hydrocarbon 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? (A) 2.9 × 10-12 s-1  (B) 4.5 × l0-5s-1m (C) 3.2 × 10-4 s-1  (D) 3.9 × 1012 s-1

Tardigrade Admin · Accepted Answer

According to Arrhenius equation log A=log k +( Ea/2.303 RT) log (2.418×10-5) +(179.9×103/2.303×8.314×546) (-5+0.3834)+ 17.2081=12.5924s-1 or A =Antilog(12.5924)s-1=3.912 ×1012 s-1

Q. The rate constant for the decomposition of a hydrocarbon is $2.418 x 1 0^{- 5} s^{- 1}$ 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?

$2.9 \times 1 0^{- 12} s^{- 1}$

$4.5 \times l 0^{- 5} s^{- 1} m$

$3.2 \times 1 0^{- 4} s^{- 1}$

$3.9 \times 1 0^{12} s^{- 1}$

According to Arrhenius equation
$l o g A = l o g k + \frac{E _{a}}{2.303 RT}$
log $(2.418 \times 1 0^{- 5}) + \frac{179.9 \times 1 0 ^{3}}{2.303 \times 8.314 \times 546}$
$(- 5 + 0.3834) + 17.2081 = 12.5924 s^{- 1}$
or $A = A n t i l o g (12.5924) s^{- 1} = 3.912 \times 1 0^{12} s^{- 1}$

Q. The rate constant for the decomposition of a hydrocarbon is 2.418x10−5s−1 at 546K. If the energy of activation is 179.9kJ/mol, what will be the value of pre exponential factor?

2.9×10−12s−1

4.5×l0−5s−1m

3.2×10−4s−1

3.9×1012s−1