Rate constant k of a reaction varies with temperature according to the equation log k=constant-(Ea/2.303 RT)⋅ When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

Question

Rate constant k of a reaction varies with temperature according to the equation log k=constant-(Ea/2.303 RT)⋅ When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is (A) 127.67 KJ mol-1 (B) 107.84 KJ mol-1 (C) 86 KJ mol-1 (D) 246.8 KJ mol-1.

Tardigrade Admin · Accepted Answer

log k=constant -(Ea/2.303 RT) (Ea/2.303R)=-5632 Ea=5632×2.303×8.314=107836 Ea ≈107.836 KJ mol-1.

Q. Rate constant k of a reaction varies with temperature according to the equation
$l o g k = co n s t an t - \frac{E _{a}}{2.303 RT} \cdot$
When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

$127.67 K J m o l^{- 1}$

$107.84 K J m o l^{- 1}$

$86 K J m o l^{- 1}$

$246.8 K J m o l^{- 1} .$

log k=constant $- \frac{E _{a}}{2.303 RT}$
$\frac{E _{a}}{2.303 R} = - 5632$
$E_{a} = 5632 \times 2.303 \times 8.314 = 107836$
$E_{a} \approx 107.836 K J m o l^{- 1} .$

Q. Rate constant k of a reaction varies with temperature according to the equation logk=constant−2.303RTEa​​⋅ When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

127.67KJmol−1

107.84KJmol−1

86KJmol−1

246.8KJmol−1.