Calculate ΔH for formation of 0.480mol of AgCl by this reaction.?
Calculate ΔH for the formation of 0.480 mol of AgCl using the following reaction:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s) ΔH = -65.5 kJ.
Please explain how to perform this calculation, and additionally, complete the following:
- Calculate ΔH for the formation of 8.00 g of AgCl.
- Calculate ΔH when 9.21 × 10⁻⁴ mol of AgCl dissolves in water.
1 Answers
(0.480 mol AgCl) × (−65.5 kJ/mol AgCl) = −31.44 kJ
(8.00 g AgCl) / (143.3212 g AgCl/mol) × (−65.5 kJ/mol AgCl) = −3.66 kJ
(9.21 × 10^−4 mol AgCl) × (+65.5 kJ/mol) = 0.0603255 kJ = +60.3 J
[Note that the sign on ΔH changed because the dissolving of AgCl in water is represented by the backwards form of the given equation.]
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