The correct expression for Arrhenius equation showing the effect of temperature on rate constant is $[T_{2} > T_{1}]$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{R}{2.303 E_{a}} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{1} T_{2}}{T_{2} - T_{1}}]$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

For a reaction with the effect of temperature on rate constant at two different temperature $T_{1} and T_{2}$ Arrhenius equation is as below.

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

The modified Arrhenius equation is

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE