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-y - x-x - y-x - y-x - y-

Question

Toppr Admin · Accepted Answer

The correct option is A -y-x2212-x-In case of exothermic reaction-Ea-back-Ea-forward-x25B3-H-y-x2212-x-xA0-

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:
$(y - x)$
$(x + y)$
$(x - y)$
$- (x + y)$

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

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

$= y - x$ .

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:(y−x)(x+y)(x−y)−(x+y)

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:
$(y - x)$
$(x + y)$
$(x - y)$
$- (x + y)$

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

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

$= y - x$ .