Interpret the enthalpy profile

The reaction is: $A+B \rightarrow C+D$

Such a diagram typically shows two possible product channels:

• one product formed via a lower activation enthalpy but ending at higher enthalpy → kinetic product
• another product formed via a higher activation enthalpy but ending at lower enthalpy → thermodynamic product

From the statements given, we infer:

• $C$ is the thermodynamically stable product, so $H_C < H_D$
• $D$ is the kinetically stable/product formed faster, so formation of $D$ has lower activation enthalpy than formation of $C$

Check each option

Option A

Formation of $A$ and $B$ from $C$ has highest enthalpy of activation.

Since $C$ is the lowest-enthalpy (most stable) species, going from $C$ back to the transition state requires the largest energy rise.

So this statement is correct.

Option B

$D$ is kinetically stable product.

The kinetic product is the one formed through the lower activation barrier. From the profile, $D$ corresponds to that product.

So this statement is correct.

Option C

$C$ is the thermodynamically stable product.

Thermodynamic stability means lower final enthalpy. From the profile, $C$ lies lower than $D$.

So this statement is correct.

Option D

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

This contradicts the usual kinetic/thermodynamic interpretation above: if $D$ is the kinetic product, then $\Delta H^\ddagger(\text{to form }D) < \Delta H^\ddagger(\text{to form }C)$ not the reverse.

Therefore this statement is incorrect.

Conclusion

The incorrect statement is: $\boxed{\text{D}}$

Comparison with stored answer

Stored correct answer: D

My derived answer also is D, so they agree.