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(D) From the graph: Enthalpy of $(A+B) = 5\,kJ\,mol^{-1}$. Enthalpy of $D = 10\,kJ\,mol^{-1}$. Enthalpy of $C = 0\,kJ\,mol^{-1}$. Activation energy fo

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Activation enthalpy to form $C$ is $5\,kJ\,mol^{-1}$ less than that to form $D$.

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(D) From the graph: Enthalpy of $(A+B) = 5\,kJ\,mol^{-1}$. Enthalpy of $D = 10\,kJ\,mol^{-1}$. Enthalpy of $C = 0\,kJ\,mol^{-1}$. Activation energy for $(A+B 	o D) = 15 - 5 = 10\,kJ\,mol^{-1}$. Activation energy for $(A+B 	o C) = 20 - 5 = 15\,kJ\,mol^{-1}$. Activation energy for $(C 	o A+B) = 20 - 0 = 20\,kJ\,mol^{-1}$. Activation energy for $(D 	o A+B) = 15 - 10 = 5\,kJ\,mol^{-1}$. Comparing the activation energies,the formation of $(A+B)$ from $C$ has the highest activation energy $(20\,kJ\,mol^{-1})$. $C$ is the most stable product (lowest enthalpy). $D$ is formed faster (lower activation energy),so it is the kinetically stable product. Activation enthalpy to form $C$ $(15\,kJ\,mol^{-1})$ is $5\,kJ\,mol^{-1}$ more than that to form $D$ $(10\,kJ\,mol^{-1})$. Therefore,statement $D$ is incorrect.

Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Consider the given plot of enthalpy of the following reaction between $A$ and $B$: $A + B \to C + D$. Identify the incorrect statement.

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