The activation energy of the backward reaction exceed, that of the forward reaction by 2RT $(i n J m o l^{- 1})$ . If the Pre- Exponential factor of the forward reaction is 4 times that of the reverse reaction. The absolute value of $Δ G^{0} (i n J m o l^{- 1})$ for the reaction at 300 k is_______
(given : ln(2)=0.7 RT = $2500 J m o l^{- 1}$ at 300k and G is the Gibb’s energy)

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

answer is 8500.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \Rightarrow E_{b} = E_{f} + 2 R T \\ (E_{f} - E_{b}) = - 2 R T \to (1) \\ A_{f} = 4 A b \\ K_{f} = A_{F} \cdot e^{- E f / R T} \\ K_{b} = A b \cdot e^{- E b / R T} \end{array}$
$\begin{array}{l} K_{e q m} = \frac{K_{f}}{K} \\ K_{e q m} = \frac{4 A b e^{- E f / R T}}{A b \cdot e^{- E b / R T}} \\ K_{e q m} = \frac{K_{f}}{K_{b}} = 4 e^{- (\frac{E_{f} - E_{b}}{R T})} \\ K_{eqm} = 4 r^{+ 2} \\ Δ G = - R T \ln K e q m \\ Δ G^{0} = - 2500 \ln (4 . e^{+ 2}) \\ Δ G^{0} = - 2500 \ln (2 + \ln 4) \\ Δ G^{0} = - 2500 \ln (3.4) \\ Δ G^{0} = - 8500 J / mol \end{array}$