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-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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?

Open in App

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, $A = 3.90 \times 10^{12} / s$ .

Suggest Corrections

Similar questions

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?

Q. The rate constant for the decomposition of a hydrocarbon is

2.418 \times 10^{- 5} s e c^{- 1}

546

K. If the energy of activation is

179.9

kJ/mol, what will be the value of pre-exponential factor?

Q. The decomposition of methyl iodide,

2 C H_{3} I (g) \to C_{2} H_{6} (g) + I_{2} (g)

, at

273^{o} C

has a rate constant of

2.418 \times 10^{- 5} s^{- 1}

. If activation energy for the reaction is

+ 179.9

k J

{m o l}^{- 1}

, what is the value of collision factor '

A

' at

273^{o} C

The first order rate constant for the decomposition of ethyl iodide by the reaction:

$C_{2} H_{5} I (g) \to C_{2} H_{4} (g) + H I (g)$

at $600 K$ is $1.60 \times 10^{- 5} s^{- 1}$ . Its energy of activation is $209 k J m o l^{- 1}$ . The rate constant of the reaction at $700 K$ will be:

Q. The decomposition rate constant of hydrocarbon is

2.428 \times 10^{- 5} {sec}^{- 1}

550 K

and activation energy is

197.7 k

joule

{mol}^{- 1}

, the calculate. Arrhenius constant.

Join BYJU'S Learning Program

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?