Ag+(aq) + Cl -(aq) AgCl(s) ΔH = -65.5 kJ?

2016-10-18 1:50 am

(a) Calculate ΔH for formation of 0.400 mol of AgCl by this reaction.

(b) Calculate ΔH for the formation of 3.80 g of AgCl.

(c) Calculate ΔH when 0.450 mmol of AgCl dissolves in water.

回答 (1)

micatkie

2016-10-18 2:13 am

✔ 最佳答案

(a)
Ag⁺(aq) + Cl⁻(aq) → AgCl(s) …. ΔH = -65.5 kJ
ΔH for the formation of 1 mol of AgCl = -65.5 kJ
ΔH for the formation of 0.400 mol of AgCl = (-65.5 kJ) × 0.400 = -26.2 kJ

(b)
Molar mass of AgCl = (107.9 + 35.5) g/mol = 143.4 g/mol
No. of moles of 3.80 g of AgCl = (3.80 g) / (143.4 g/mol) = 0.0265 mol
ΔH for the formation of 3.80 g of AgCl = (-65.5 kJ) × 0.0265 = -1.74 kJ

(c)
Reverse the given thermochemical equation :
AgCl(s) → Ag⁺(aq) + Cl⁻(aq) …. ΔH = +65.5 kJ
ΔH for dissolving 1 mol of AgCl in water = +65.5 kJ
ΔH for dissolving 0.450 mmol of AgCl in water = (+65.5 × 10³ J) × (0.450 × 10⁻³) = +29.5 J

👍 3 👎 0