Rate constant k of a reaction varies with temperature according to the equation log k= constant -(Ea/2.303 R) × (1/T); where, Ea is the energy of activation for the reaction. When a graph is plotted for log k v s (1/T) a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R=8.314 JK -1 mol -1).

Question

Rate constant k of a reaction varies with temperature according to the equation log k= constant -(Ea/2.303 R) × (1/T); where, Ea is the energy of activation for the reaction. When a graph is plotted for log k v s (1/T) a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R=8.314 JK -1 mol -1). (A) 122.65 kJ mol-1 (B) 127.71 kJ mol-1 (C) 142.34 kJ mol-1 (D) 150.00 kJ mol-1

Tardigrade Admin · Accepted Answer

Slope of the line =-(Ea/2303 R)=-6670 K Ea =2.303 × 8.314( JK -1 mol -1) × 6670 K =127711.4 J mol -1=127.71 kJ mol -1

$122.65 k J m o l^{- 1}$

$127.71 k J m o l^{- 1}$

$142.34 k J m o l^{- 1}$

$150.00 k J m o l^{- 1}$

Slope of the line $= - \frac{E _{a}}{2303 R} = - 6670 K$
$E_{a} = 2.303 \times 8.314 (J K^{- 1} m o l^{- 1}) \times 6670 K$
$= 127711.4 J m o l^{- 1} = 127.71 k J m o l^{- 1}$

122.65kJmol−1

127.71kJmol−1

142.34kJmol−1

150.00kJmol−1

Slope of the line =−2303REa​​=−6670K Ea​=2.303×8.314(JK−1mol−1)×6670K =127711.4Jmol−1=127.71kJmol−1

$122.65 k J m o l^{- 1}$

$127.71 k J m o l^{- 1}$

$142.34 k J m o l^{- 1}$

$150.00 k J m o l^{- 1}$

Slope of the line $= - \frac{E _{a}}{2303 R} = - 6670 K$
$E_{a} = 2.303 \times 8.314 (J K^{- 1} m o l^{- 1}) \times 6670 K$
$= 127711.4 J m o l^{- 1} = 127.71 k J m o l^{- 1}$