MCQ
For a first order reaction $\mathrm{A} \rightarrow \mathrm{P}$, the temperature ( $\mathrm{T}$ ) dependent rate constant ( $k$ ) was found to follow the equation $\log k=-(2000) \frac{1}{\mathrm{~T}}+6.0$. The pre-exponential factor $\mathrm{A}$ and the activation energy $\mathrm{E}_{\mathrm{a}}$, respectively, are
  • A
    $1.0 \times 10^6 \mathrm{~s}^{-1}$ and $9.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $6.0 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C
    $1.0 \times 10^6 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • $1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

Correct option: D.
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
d
The correct option is $D$ $1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$k= \mathrm{Ae}^{-Ea/R}$

or log $K=log A -  \frac{Ea} {2.303R T}$

$\log k=-(2000) \frac{1}{~T}+6.0$

On comparison, we get

log $A = 6$

or $A=10^6 \mathrm{~s}^{-1}$

$ \frac{Ea}{2.303R}=2000$

or$ \mathrm{E}^{a}= 2.303 \times 80314 \times 2000 $ or $E$

$=38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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