For a first order reaction, Aarrow P, the temperature (T) dependent rate constant (k) was found to follow the equation: log k= (-2000) (1/T)+6.0. The pre-exponential factor A and the activation energy Ea, respectively, are

Question

For a first order reaction, Aarrow P, the temperature (T) dependent rate constant (k) was found to follow the equation: log k= (-2000) (1/T)+6.0. The pre-exponential factor A and the activation energy Ea, respectively, are (A) 1.0×106 s-1 and 9.2 kJ mol-1 (B) 6.0 s-1 and 16.6 kJ mol-1 (C) 1.0×106 s-1 and 16.6 kJ mol-1 (D) 1.0×106 s-1 and 38.3 kJ mol-1

Tardigrade Admin · Accepted Answer

The logarithmic form of Arrhenius equation is

lo g k = lo g A - \frac{E _{a}}{2.303 RT}

Given,

lo g k = 6 - \frac{2000}{T}

Comparing the above two equations :

lo g A = 6 \Rightarrow A = 1 0^{6}

and

\frac{E _{a}}{2.303 R} = 2000

\Rightarrow E_{a} = 2000 \times 2.303 \times 8.314 J

= 38.3 k J m o l^{- 1}

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

$1.0 \times 1 0^{6} s^{- 1}$ and $9.2 k J m o l^{- 1}$

$6.0 s^{- 1}$ and $16.6 k J m o l^{- 1}$

$1.0 \times 1 0^{6} s^{- 1}$ and $16.6 k J m o l^{- 1}$

$1.0 \times 1 0^{6} s^{- 1}$ and $38.3 k J m o l^{- 1}$

Q. For a first order reaction, A→P, the temperature (T) dependent rate constant (k) was found to follow the equation: logk=(−2000)T1​+6.0. The pre-exponential factor A and the activation energy Ea​, respectively, are

1.0×106s−1 and 9.2kJmol−1

6.0s−1 and 16.6kJmol−1

1.0×106s−1 and 16.6kJmol−1

1.0×106s−1 and 38.3kJmol−1

The logarithmic form of Arrhenius equation is logk=logA−2.303RTEa​​ Given, logk=6−T2000​ Comparing the above two equations : logA=6⇒A=106 and 2.303REa​​=2000 ⇒Ea​=2000×2.303×8.314J =38.3kJmol−1

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

$1.0 \times 1 0^{6} s^{- 1}$ and $9.2 k J m o l^{- 1}$

$6.0 s^{- 1}$ and $16.6 k J m o l^{- 1}$

$1.0 \times 1 0^{6} s^{- 1}$ and $16.6 k J m o l^{- 1}$

$1.0 \times 1 0^{6} s^{- 1}$ and $38.3 k J m o l^{- 1}$