Consider an endothermic reaction $$X \to Y$$ with the activation energies $${E_b}$$ and $${E_f}$$ for the backward and forward reactions, respectively. In general
A.
there is no definite relation between $${E_b}$$ and $${E_f}$$
B.
$${E_b} = {E_f}$$
C.
$${E_b} > {E_f}$$
D.
$${E_b} < {E_f}$$
Answer :
$${E_b} < {E_f}$$
Solution :
Enthalpy of reaction $$\left( {\Delta H} \right) = {E_{{a_{\left( f \right)}}}} - {E_{{a_{\left( b \right)}}}}$$
for an endothermic reaction $$\Delta H = + ve$$ hence for $$\Delta H$$ to be negative
$${E_{{a_{\left( b \right)}}}} < {E_{{a_{\left( f \right)}}}}$$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$