Question

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

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

Answer 1

The correct option is D $1.0\times {10}^6{s}^{-1}$ and $38.3kJmo{l}^{-1}$

$logk=\frac{-2000}{T}+6.0$

Multiply by $2.303$

$ln\frac{k}{1\times {10}^6}=\frac{-4606}{T}$

$k=1\times {10}^6{e}^{\frac{-38.3\times {10}^3}{RT}}$

$A=1\times {10}^6{s}^{-1}$ and

$E_a=38.3kJ{mol}^{-1}$