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 ...

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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.

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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,
$A = 75.16 (s e c)^{- 1}$

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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,
$A = 75.16 (s e c)^{- 1}$