$$\ce{2 NH4OH(aq) + H2SO4(aq) -> (NH4)2SO4(aq) + 2 H2O(l)}$$
This is an acid base reaction. $\ce{NH4}$ compounds are always water-soluble, so no precipitate, and no color change will be observed. Neutralization reactions like the one above are generally exothermic (they release heat).
This can be proved by observing their heats of formation:
- $\Delta H_{f(\ce{(NH4)2SO4})} = \pu{-1181 kJ/mol * 1 mol = -1181 kJ}$
- $\Delta H_{f(\ce{H2O(l))}} = \pu{-285.8 kJ/mol * 2 mol = -571.6 kJ}$
- $\Delta H_{f(H2SO4)} = \pu{-909 kJ/mol * 1 mol = -909 kJ}$
- $\Delta H_{f(NH4OH)} = \pu{-80.8 kJ/mol * 2 mol = -161.6 kJ}$
Thus the $\Delta H_{rxn}=\pu{(-1181 kJ + -571.6 kJ) - ( -909 kJ + -161.6 kJ) = -682 kJ}$ is negative, hence the reaction is exothermic.
