Consider the following reaction with given energy profile diagram
$A ⇌_{K_{- 1}}^{K_{1}} B ⇌_{K_{- 2}}^{K_{2}} C ⇌_{K_{- 3}}^{K_{3}} D$

Here assume that energy corresponds to enthalpy (H)
If all the reactions in any direction have same pre-exponential factor, then select the CORRECT statement.

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change produced in $K_{- 3}$ is more than any other rate constants per degree change in temperature at lower temperature range.

Enthalpy change for the reaction, A $\to$ C, is 20.8 K cal

C $\to$ D is the fastest step2 among all when equilibrium is attained

rate-determining step in backward direction is D $\to$ C

answer is A, B, D.

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Detailed Solution

$K = A e^{- E_{a} / R T}$
A) More is the value of $E_{a},$ more change will be produced per degree change in temperature at lower temperature range
$d K = A e^{- E a / R T} (- \frac{1}{T^{2}}) d T \times \frac{- E_{a}}{K}$
$\frac{d K}{d T} = \frac{A}{T^{2}} e^{- E a / R T} (\frac{E_{a}}{R})$
As $E_{a}$ is large at lower temperature more change is produced in K with temperature
C) For step $C \to D, E_{a} = 4.3 K . C a l / m o l$
It is the smallest value among all energy of activation