For a first order reaction A arrow 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 A arrow 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

Given, log K =6-(2000/ T ) Since, log K = log A -( Ea /2.303 RT ) ⇒ log A =6 and (- E a /2.303 R )=-2000 So, A=106 sec -1 and E a=38.3 kJ / mole

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 $9.2 k J m o l^{- 1}$

$6.0 \times s^{- 1}$ 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 $38.3 k J m o l^{- 1}$

Given, $lo g K = 6 - \frac{2000}{T}$
Since, $lo g K = lo g A - \frac{E a}{2.303 RT}$
$\Rightarrow lo g A = 6$ and $\frac{- E _{a}}{2.303 R} = - 2000$
So, $A = 1 0^{6} sec^{- 1}$ and $E a = 38.3 k J / m o l e$

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

6.0×s−1 and 16.6kJmol−1

1.0×106s−1 and 16.6kJmol−1

1.0×106s−1 and 38.3kJmol−1