Question

For a first order reaction $A \rightarrow P$, the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation
$\log k =-(2000) \frac{1}{ T }+6.0$
The pre-exponential factor $A$ and the activation energy $E_{a}$ respectively, are

• A. $1.0 \times 10^{6} s ^{-1}$ and $9.2\, kJ\, mol ^{-1}$
• B. $6.0 \times s ^{-1}$ and $16.6\, kJ\, mol ^{-1}$
• C. $1.0 \times 10^{6} s ^{-1}$ and $16.6\, kJ\, mol ^{-1}$
• D. $1.0 \times 10^{6} s ^{-1}$ and $38.3\, kJ\, mol ^{-1}$

Answer 1

Answer: (D)

Given, $\log K =6-\frac{2000}{ T }$
Since, $\log K =\log A -\frac{ Ea }{2.303 RT }$
$\Rightarrow \log A =6$ and $\frac{- E _{ a }}{2.303 R }=-2000$
So, $A=10^{6} \sec ^{-1}$ and $E a=38.3\, kJ / mole$