From the following data at 25°C, which of the following statement(s) is/are correct?<br>1/2 (g) + 1/ — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Let's analyze each statement:<br>(a) Reaction: $H_2O$(g) → 2H(g) + O(g).<br>This represents the complete atomization of $H_2O$(g). By Hess's Law:<br>ΔH = 2 × ΔfH°(H, g) + ΔfH°(O, g) - ΔfH°($H_2O$, g).<br>From the data:<br>- ΔfH°(H, g) = 1/2 × ΔH_diss($H_2$) = 436 / 2 = +218 kJ/mol.<br>- ΔfH°(O, g) = 1/2 × ΔH_diss($O_2$) = 495 / 2 = +247.5 kJ/mol.<br>- ΔfH°($H_2O$, g) = -242 kJ/mol.<br>Thus:<br>ΔH = 2 × (218) + 247.5 - (-242) = 436 + 247.5 + 242 = 925.5 kJ. Thus, Statement A is correct.<br><br>(b) Reaction: OH(g) → H(g) + O(g).<br>This is the bond dissociation of OH(g):<br>ΔH = ΔfH°(H, g) + ΔfH°(O, g) - ΔfH°(OH, g).<br>Given ΔfH°(OH, g) = 42 kJ/mol:<br>ΔH = 218 + 247.5 - 42 = 465.5 - 42 = 423.5 kJ.<br>Since the statement says 502 kJ, Statement B is incorrect.<br><br>(c) Enthalpy of formation of H(g): 1/2 $H_2$(g) → H(g) has ΔH = +218 kJ/mol. (Endothermic, so positive). Statement C says -218 kJ/mol, so it is incorrect.<br><br>(d) Enthalpy of formation of OH(g): The first reaction 1/2 $H_2$(g) + 1/2 $O_2$(g) → OH(g) has ΔH° = 42 kJ/mol, which is by definition the standard enthalpy of formation of OH(g). Thus, Statement D is correct.<br><br>Thus, the correct statements are A and D.