For the reaction $A + B \to C + D Δ H = - 20 k J m o l^{- 1}$ . The activation energy of the forward reaction is $85 k J m o l^{- 1}$ . The activation energy for backward reaction is

$65 k J m o l^{- 1}$

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$85 k J m o l^{- 1}$

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$40 k J m o l^{- 1}$

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$105 k J m o l^{- 1}$

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Solution

The correct option is D
$105 k J m o l^{- 1}$

$Δ$ H = $E_{a (f o r w a r d)} - E_{a (b a c k w a r d)}$

$- 20 = (85) - E_{a (b a c k w a r d)}$

$E_{a (b a c k w a r d)} = 105 K J . m o l^{- 1}$

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For the reaction $A + B \to C + D Δ H = - 20 k J m o l^{- 1}$ . The activation energy of the forward reaction is $85 k J m o l^{- 1}$ . The activation energy for backward reaction is

Q. For a reaction,

A \to B; Δ H = 20 k J . {m o l}^{- 1}

, the activation energy of the forward reaction is

85 k J / m o l

. The activation energy of the backward reaction will be:

Q. The activation energy for the forward reaction X Y is

60

m o l^{- 1}

and

Δ H

- 20

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. The activation energy for the backward reaction Y X is:

For the reaction $A + B \to C + D Δ H = - 20 k J m o l^{- 1}$ . The activation energy of the forward reaction is $85 k J m o l^{- 1}$ . The activation energy for backward reaction is

Q. For a reaction,

A + B \to C + D; Δ H = - 20 K J m o l e^{- 1}

The activation energy for the forward reaction is 85

k J

. Then the activation energy for the backward reaction is:

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For the reaction A+B→C+D ΔH=−20 kJ mol−1 . The activation energy of the forward reaction is 85kJ mol–1. The activation energy for backward reaction is

The correct option is D 105 kJ mol−1 ΔH = Ea(forward)−Ea(backward) −20=(85)−Ea(backward) Ea(backward)=105KJ.mol−1

For the reaction $A + B \to C + D Δ H = - 20 k J m o l^{- 1}$ . The activation energy of the forward reaction is $85 k J m o l^{- 1}$ . The activation energy for backward reaction is

The correct option is D
$105 k J m o l^{- 1}$

$Δ$ H = $E_{a (f o r w a r d)} - E_{a (b a c k w a r d)}$

$- 20 = (85) - E_{a (b a c k w a r d)}$

$E_{a (b a c k w a r d)} = 105 K J . m o l^{- 1}$