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Q. For a reaction $\text{A} \rightarrow \text{B,} \, \text{E}_{\text{a}} = 10 \text{ kJ/mol,} \, \text{ΔH} = 5 \, \text{kJ/mol} \text{.}$ Thus potential energy profile for this reaction is

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

$\Delta H =\left( E _{ a }\right)_{ f }-\left( E _{ a }\right)_{ b }$

$5=10-\left( E _{ a }\right)_{ b }$

$\left( E _{ a }\right)_{ b }=5 kJ / mol$

$\left( E _{ a }\right)_{ f }>\left( E _{ a }\right)_{ b }$ hence option $( B )$ is correct