Observe the following reaction: $ 2A_2(g) + B_2(g) \xrightarrow{T(K)} 2A_2B(g) + 600 \text{ kJ} $. The standard enthalpy of formation $ (\Delta_f H^\circ) $ of $ A_2B(g) $ is:
The isotherms of an ideal gas at $ T_1, T_2, T_3 $ along with their slopes (m) (in the brackets) are shown here. If $ T_1 > T_2 > T_3 $, then the correct order of slopes of these isotherms is: