A volume of 100 ml of 0.5 N- solution is neutralized with 200 ml of 0.2 M- in a constant pressure ca — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Let's verify each statement step-by-step:<br>1. Heat released in the experiment: q = C × ΔT = 1.5 kJ/°C × 1.4°C = 2.1 kJ.<br>Let's check the moles of H+ and $NH_4OH$:<br>- $H_2SO_4$: 100 mL of 0.5 N → equivalents of H+ = 0.1 × 0.5 = 0.05 eq (or 0.05 mol of H+).<br>- $NH_4OH$: 200 mL of 0.2 M → moles of $NH_4OH$ = 0.2 × 0.2 = 0.04 mol.<br>Since $NH_4OH$ is limiting, 0.04 moles of $NH_4OH$ are neutralized, releasing 2.1 kJ of heat.<br>Therefore, the enthalpy of neutralization of $NH_4OH$ (a weak base) by $H_2SO_4$ (a strong acid) is:<br>ΔH_neutralization = -2.1 kJ / 0.04 mol = -52.5 kJ/mol.<br>Since $H_2SO_4$ is a strong acid, this is equal to the neutralization of $HCl$ vs. $NH_4OH$. Thus, Statement A is correct.<br><br>2. Enthalpy of dissociation of $NH_4OH$:<br>We know that for strong acid + strong base, ΔH_neutralization = -57 kJ/mol.<br>For strong acid + weak base ($NH_4OH$): ΔH_neutralization = -52.5 kJ/mol.<br>The difference is due to the ionization of the weak base:<br>ΔH_diss($NH_4OH$) = -52.5 - (-57.0) = +4.5 kJ/mol. Thus, Statement B is correct.<br><br>3. Enthalpy of dissociation of CH3COOH:<br>For weak acid (CH3COOH) + weak base ($NH_4OH$), the given heat of neutralization is -48.1 kJ/mol.<br>We know:<br>ΔH_neutralization = ΔH_neut(SA/SB) + ΔH_diss(CH3COOH) + ΔH_diss($NH_4OH$)<br>-48.1 = -57.0 + ΔH_diss(CH3COOH) + 4.5<br>-48.1 = -52.5 + ΔH_diss(CH3COOH) → ΔH_diss(CH3COOH) = -48.1 - (-52.5) = +4.6 kJ/mol. Thus, Statement C is correct.<br><br>4. For H+(aq) + OH-(aq) → $H_2O$(l), ΔH° = -57 kJ/mol.<br>Therefore, for the reverse process: $H_2O$(l) → H+(aq) + OH-(aq), ΔH° = +57 kJ/mol.<br>For 2 moles of water: 2$H_2O$(l) → 2H+(aq) + 2OH-(aq), ΔH° = 2 × 57 = 114 kJ. Thus, Statement D is correct.<br><br>All four statements are correct, so the answer is A, B, C, D.