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

(D) Activation enthalpy to form C is $5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ less than that to form D

(A) C is the thermodynamically stable product

(B) Formation of A and B from C has highest enthalpy of activation

(C) D is kinetically stable product

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Chemical Kinetics

Consider the given plot of enthalpy of the following reaction between A and B .

A + B \to C + D

Identify the incorrect statement

(A)C is the thermodynamically stable product
(B)Formation of A and B from C has highest enthalpy of activation
(C)D is kinetically stable product
(D)Activation enthalpy to form C is $5 kJ mol^{- 1}$ less than that to form D✓ Correct

Answer: (D)

Solution

Reading the enthalpy profile, forming C needs an activation enthalpy of

15 kJ mol^{- 1}

while forming D needs

10 kJ mol^{- 1}

, so the barrier to C is actually

5 kJ mol^{- 1}

higher than that to D, not lower, which makes that the incorrect statement. The rest match the diagram: C lies lowest in enthalpy and is the thermodynamically stable product, D forms over the smaller barrier and is the kinetically favoured product, and the reverse step turning C back into A and B must climb the largest barrier of all.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Chemical Kinetics →

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Chemical Kinetics

Solution

More JEE Main questions on Chemical Kinetics