The slope of Arrhenius Plot ln k v/s1T of first oder reaction is -5 x 103 K . The value of Ea of the reaction is. Choose the correct option for your answer [Given R = 8.314 JK-1mol-1]

K=Ae−EaRT In K=ln⁡A−EaR⋅1Ty C m x Slope =−EaR−EaR=−5×103KEa=5×8.314JK−1mole−1×103K

The slope of Arrhenius Plot ln k v/s1T of first oder reaction is -5 x 103 K . The value of Ea of the reaction is. Choose the correct option for your answer [Given R = 8.314 JK-1mol-1]

K=Ae−EaRT In K=ln⁡A−EaR⋅1Ty C m x Slope =−EaR−EaR=−5×103KEa=5×8.314JK−1mole−1×103K

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The slope of Arrhenius Plot $(\ln k v / s \frac{1}{T})$ of first oder reaction is -5 x 10³ K . The value of E_a of the reaction is. Choose the correct option for your answer
[Given R = 8.314 JK^-1mol^-1]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

-83 kJ mol^-1

41.5 kJ mol^-1

83.0 kJ mol^-1

166 kJ mol^-1

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} K = {Ae}^{- \frac{E_{a}}{RT}} \\ In K = \ln A - \frac{E_{a}}{R} \cdot \frac{1}{T} \\ y C m x \\ Slope = - \frac{E_{a}}{R} \\ - \frac{E_{a}}{R} = - 5 \times 10^{3} K \\ E_{a} = 5 \times 8 .314 {JK}^{- 1} {mole}^{- 1} \times 10^{3} K \end{array}$