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Standard XII

Chemistry

Activation Energy

In a reversib...

Question

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively $- x k J m o l^{- 1}$ and $y k J m o l^{- 1}$ . Therefore, the energy of activation in the backward direction, in $k J m o l^{- 1}$ is:

(y - x)

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(x + y)

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(x - y)

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- (x + y)

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Solution

The correct option is A $(y - x)$
In case of exothermic reaction,

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

$= y - x$ .

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The energies of activation for forward and reverse reactions for $A_{2} + B_{2} ⇌ 2 A B$ are $180 k J m o l^{- 1}$ and $200 k J m o l^{- 1}$ respectively. The presence of a catalyst lowers activation energy of both reactions by $100 k J m o l^{- 1}$ . The enthalpy change of the reaction $(A_{2} + B_{2} \to 2 A B)$ in the presence of a catalyst will be (in kJ $m o l^{- 1}$ )

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In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively −x kJ mol−1 and y kJ mol−1. Therefore, the energy of activation in the backward direction, in kJ mol−1 is:

The correct option is A (y−x)In case of exothermic reaction,Ea(back)=Ea(forward)+△H=y−x.

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively $- x k J m o l^{- 1}$ and $y k J m o l^{- 1}$ . Therefore, the energy of activation in the backward direction, in $k J m o l^{- 1}$ is:

The correct option is A $(y - x)$
In case of exothermic reaction,

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

$= y - x$ .