Question:

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

Show Hint

Read the energy of A+B, D, C and the two transition-state peaks from the graph, then compare the barriers A+B to D and A+B to C.
Updated On: Nov 5, 2026
  • Formation of A and B from C has highest enthalpy of activation.
  • D is kinetically stable product.
  • C is the thermodynamically stable product
  • Activation enthalpy to form C is 5 kJ mol\(^{-1}\) less than that to form D.
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The Correct Option is D

Solution and Explanation

From the plot: A+B is at 5 kJ/mol, the peak leading to D is at 15, D is at 10, the peak leading to C is at 20, and C is the lowest at about 0.
Activation enthalpy for A+B to D = 15 - 5 = 10 kJ/mol.
Activation enthalpy for A+B to C = 20 - 5 = 15 kJ/mol.
Activation enthalpy for C to A+B (reverse) = 20 - 0 = 20 kJ/mol, the highest barrier in the diagram, so statement (1) is correct.
D has the smaller forward barrier, so it forms faster (kinetic product); statement (2) is acceptable.
C has the lowest enthalpy, so it is the thermodynamically stable product; statement (3) is correct.
Barrier to form C (15) is 5 kJ/mol MORE than the barrier to form D (10), not less, so statement (4) is incorrect.
Answer: (4) Activation enthalpy to form C is 5 kJ mol\(^{-1}\) less than that to form D.
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