From the above figure, the activation energy for the reverse reaction would be

Question

From the above figure, the activation energy for the reverse reaction would be (A) -120 KJ mol-1 (B) +152 KJ mol-1  (C) +120 KJ mol-1  (D) +1760 KJ mol-1

Tardigrade Admin · Accepted Answer

Given Δ H=-120 KJ mol-1; Ea=1640 KJ mol-1 For reverse reaction, Δ H=+120 KJ mol-1 ∴ Activation energy for the reverse reaction =1640+120=+1760 KJ mol-1

Q. From the above figure, the activation energy for the reverse reaction would be

$- 120 K J m o l^{- 1}$

$+ 152 K J m o l^{- 1}$

$+ 120 K J m o l^{- 1}$

$+ 1760 K J m o l^{- 1}$

Given $Δ H = - 120 K J m o l^{- 1}; E_{a} = 1640 K J m o l^{- 1}$
For reverse reaction, $Δ H = + 120 K J m o l^{- 1}$
$∴$ Activation energy for the reverse reaction
$= 1640 + 120 = + 1760 K J m o l^{- 1}$

Q. From the above figure, the activation energy for the reverse reaction would be

−120KJmol−1

+152KJmol−1

+120KJmol−1

+1760KJmol−1

Given ΔH=−120KJmol−1;Ea​=1640KJmol−1 For reverse reaction, ΔH=+120KJmol−1 ∴ Activation energy for the reverse reaction =1640+120=+1760KJmol−1

$- 120 K J m o l^{- 1}$

$+ 152 K J m o l^{- 1}$

$+ 120 K J m o l^{- 1}$

$+ 1760 K J m o l^{- 1}$

Given $Δ H = - 120 K J m o l^{- 1}; E_{a} = 1640 K J m o l^{- 1}$
For reverse reaction, $Δ H = + 120 K J m o l^{- 1}$
$∴$ Activation energy for the reverse reaction
$= 1640 + 120 = + 1760 K J m o l^{- 1}$