For a reaction, given below is the graph of ln k vs (1/ T ). The activation energy for the reaction is equal to cal mol -1. (Nearest integer). (Given: R =2 cal K -1 mol -1 )

Question

For a reaction, given below is the graph of ln k vs (1/ T ). The activation energy for the reaction is equal to cal mol -1. (Nearest integer). (Given: R =2 cal K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K = Ae - Ea / RT ln k =(- Ea / RT )+ ln A text Slope =( Ea / R )=(20/5) E a =4 R =8 Cal / mol

Q. For a reaction, given below is the graph of $ln k$ vs $\frac{1}{T}$ . The activation energy for the reaction is equal to ___ cal $m o l^{- 1}$ . (Nearest integer).
(Given : $R = 2 c a l K^{- 1} m o l^{- 1}$ )

$K = A e^{- E a / RT}$
$ln k = \frac{- E a}{RT} + ln A$
$Slope = \frac{E a}{R} = \frac{20}{5}$
$E_{a} = 4 R = 8 C a l / m o l$

Q. For a reaction, given below is the graph of lnk vs T1​. The activation energy for the reaction is equal to ___ cal mol−1. (Nearest integer). (Given : R=2calK−1mol−1 )

K=Ae−Ea/RT lnk=RT−Ea​+lnA Slope =REa​=520​ Ea​=4R=8Cal/mol

Q. For a reaction, given below is the graph of $ln k$ vs $\frac{1}{T}$ . The activation energy for the reaction is equal to ___ cal $m o l^{- 1}$ . (Nearest integer).
(Given : $R = 2 c a l K^{- 1} m o l^{- 1}$ )

$K = A e^{- E a / RT}$
$ln k = \frac{- E a}{RT} + ln A$
$Slope = \frac{E a}{R} = \frac{20}{5}$
$E_{a} = 4 R = 8 C a l / m o l$