The specific rate constant of decomposition of a compound is represented by ln k-5-0-12000T- The activation energy in Kcal-mol of decomposition for this compound at 300 K is-

According Arrhenius equation, we have nbsp; k= A× e^ E_a/RT ⇒ln k = ln A EaRT .....(1) In the question, it is given that ln k =5.0 12000T .....(2) Now ...

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Standard XII

Chemistry

Derivation of Kp and Kc

The specific ...

Question

The specific rate constant of decomposition of a compound is represented by $ln k = 5.0 - \frac{12000}{T}$ . The activation energy (in $Kcal/mol$ ) of decomposition for this compound at $300 K$ is:

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Solution

According Arrhenius equation, we have
$k = A \times e^{- E_{a} / R T}$
$\Rightarrow ln k = ln A - \frac{E a}{R T} . . . . . (1)$
In the question, it is given that
$ln k = 5.0 - \frac{12000}{T} . . . . . (2)$
Now comparing the equation $1$ and $2$ , we get
$\frac{E_{a}}{R} = 12000 \Rightarrow E_{a} = 12000 \times 2 = 24000 cal/mol = 24 kcal/mol$

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The specific rate constant of decomposition of a compound is represented by lnk=5.0−12000T. The activation energy (in Kcal/mol) of decomposition for this compound at 300 K is:

According Arrhenius equation, we have k=A×e−Ea/RT ⇒lnk=lnA−EaRT .....(1) In the question, it is given that lnk=5.0−12000T .....(2) Now comparing the equation 1 and 2, we get EaR=12000⇒Ea=12000×2=24000 cal/mol=24 kcal/mol

The specific rate constant of decomposition of a compound is represented by $ln k = 5.0 - \frac{12000}{T}$ . The activation energy (in $Kcal/mol$ ) of decomposition for this compound at $300 K$ is: