JEE Main · 2021 · Shift-ImediumCK-126

Consider the given plot of enthalpy of the following reaction between A and B: {A + B C + D} Identify the incorrect…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

Consider the given plot of enthalpy of the following reaction between A and B: A+BC+D\mathrm{A + B \rightarrow C + D} image Identify the incorrect statement.

Options
  1. a

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

  2. b

    D is kinetically stable product.

  3. c

    C is the thermodynamically stable product.

  4. d

    Activation enthalpy to form C is 5kJmol15\,\mathrm{kJ\,mol^{-1}} less than that to form D.

Correct Answerd

Activation enthalpy to form C is 5kJmol15\,\mathrm{kJ\,mol^{-1}} less than that to form D.

Detailed Solution

From the enthalpy profile for A+BC+D\mathrm{A + B \rightarrow C + D} (two competing pathways):

  • C is the thermodynamically stable product (lower energy/enthalpy).
  • D is the kinetically stable product (lower activation energy to form D, so it forms faster).
  • The reverse reaction (C or D → A + B) has the highest activation energy for the C pathway.

Evaluating statements:

(1) TRUE — Formation of A and B from C requires overcoming the highest energy barrier (reverse of the C pathway). ✓

(2) TRUE — D is kinetically stable (formed faster due to lower EaE_a). ✓

(3) TRUE — C is thermodynamically stable (lower energy product). ✓

(4) FALSE — From the graph, the activation enthalpy to form C is higher than to form D (D is kinetically favoured). The statement says C's EaE_a is 5 kJ/mol less than D's, which contradicts the graph. ✗

Answer: Option (4)

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