JEE Mains · Chemistry · STD 12 - 3. 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 \(D\)
Answer & Solution
Correct Answer
(D) Activation enthalpy to form \(C\) is \(5\,kJ\, mol^{-1}\) less than that to form \(D\)
Step-by-step Solution
Detailed explanation
\({E_a} = (D \to C)\) \( = 15 - 0 = 15\,kJ\,mo{l^{ - 1}}\) \({E_a} = (A + B) \to C = 15\,kJ\,mo{l^{ - 1}}\) \({E_a} = (A + B) \to D = 10\,kJ\,mo{l^{ - 1}}\) \({E_a} = C \to (A + B) = 20\,kJ\,mo{l^{ - 1}}\) (high activation enthalpy in the reaction) \(\therefore \) Activation…
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