The energies of activation for forward and reverse reactions for $${A_2} + {B_2} \rightleftharpoons 2AB$$ are $$180\,kJ\,mo{l^{ - 1}}$$ and $$200\,kJ\,mo{l^{ - 1}}$$ respectively, The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by $$100\,kJ\,mo{l^{ - 1}}$$ . The enthalpy change of the reaction $$\left( {{A_2} + {B_2} \to 2AB} \right)$$ in the presence of a catalyst will be ( in $$kJ\,mo{l^{ - 1}}$$ )
A.
20
B.
300
C.
120
D.
280
Answer :
20
Solution :
$$\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol$$
The nearest correct answer given in choices may be obtained by neglecting sign.
Releted MCQ Question on Physical Chemistry >> Chemical Kinetics
Releted Question 1
If uranium (mass number 238 and atomic number 92) emits an $$\alpha $$ -particle, the product has mass no. and atomic no.