The decomposition rate constant of hydrocarbon is 2-428 - 10-5-1 at 550 K and activation energy is 197-7 k joule mol-1- the calculate- Arrhenius constant-

Arrenhius equation is given by, k=Ae^ E_aRT logA=logk+E_a2.303RT logA=1.877 Taking antilog of both sides, A=75.16(sec)^ 1 ...

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

Byju's Answer

Standard XII

Chemistry

Rate Constant

The decomposi...

Question

The decomposition rate constant of hydrocarbon is $2.428 \times 10^{- 5} {sec}^{- 1}$ at $550 K$ and activation energy is $197.7 k$ joule ${mol}^{- 1}$ , the calculate. Arrhenius constant.

Open in App

Solution

Arrenhius equation is given by,
$k = A e^{\frac{- E_{a}}{R T}}$
$l o g A = l o g k + \frac{E_{a}}{2.303 R T}$

$l o g A = 1.877$
Taking antilog of both sides,
$A = 75.16 (s e c)^{- 1}$

Suggest Corrections

Similar questions

Q. At

25^{o} C

, the values of rate constant, activation energy and Arrhenius constant of a reaction are

3 \times 10^{- 4} {s e c}^{- 1}

129 k J / m o l

and

2 \times 10^{15} {s e c}^{- 1}

respectively.
The value of rate constant (in

s e c^{- 1}

) as

T \to \infty

is:

Q. The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at

25^{0} C

are

3 \times 10^{- 4} s^{- 1}

, 104.4KJ / mol and

6 \times 10^{- 4} s^{- 1}

respectively. The value of the rate constant at

T \to \infty

is:

Q. The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at

25^{o} C

are

3.0 \times 10^{- 4} s^{- 1}

104.4

k J

{m o l}^{- 1}

and

6.0 \times 10^{14} s^{- 1}

respectively. The value of rate constant at

T ⟶ \infty

is:

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, the activation energy and the Arrhenius parameter (A) of a chemical reaction at

25^{\circ} C

are

3.0 \times 10^{- 22} s^{- 1},

104.4 kJ

m o l^{- 1}

6.0 \times 10^{14} s^{- 1}

and respectively. The value of the rate constant at

T \to \infty

is:

Join BYJU'S Learning Program

The decomposition rate constant of hydrocarbon is 2.428×10−5sec−1 at 550K and activation energy is 197.7 k joule mol−1, the calculate. Arrhenius constant.

Arrenhius equation is given by,k=Ae−EaRTlogA=logk+Ea2.303RTlogA=1.877Taking antilog of both sides,A=75.16(sec)−1

The decomposition rate constant of hydrocarbon is $2.428 \times 10^{- 5} {sec}^{- 1}$ at $550 K$ and activation energy is $197.7 k$ joule ${mol}^{- 1}$ , the calculate. Arrhenius constant.

Arrenhius equation is given by,
$k = A e^{\frac{- E_{a}}{R T}}$
$l o g A = l o g k + \frac{E_{a}}{2.303 R T}$

$l o g A = 1.877$
Taking antilog of both sides,
$A = 75.16 (s e c)^{- 1}$