JEE MAIN - Chemistry (2007 - No. 26)
The energies of activation for forward and reverse reactions for A2 + B2 $$\leftrightharpoons$$ 2AB are 180 kJ mol−1
and 200 kJ mol−1 respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol−1. The enthalpy change of the reaction ( A2 + B2 $$\to$$ 2AB) in the presence of catalyst will be (in kJ mol−1)
300
120
200
20
स्पष्टीकरण
$$\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol$$
The nearest correct answer given in choices may be obtained by neglecting sign.
The nearest correct answer given in choices may be obtained by neglecting sign.