Calculate ΔH for formation of 0.480mol of AgCl by this reaction.?

Ag+(aq)+Cl−(aq)→AgCl(s)ΔH=−65.5kJ

Please explain how to do this as well as..

Calculate ΔH for the formation of 8.00g of AgCl.

and

Calculate ΔH when 9.21×10−4mol of AgCl dissolves in water.

1 Answer

  • (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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