Silver ions react with chloride ions in aqueous solution to form an insoluble silver chloride precipitate:
Ag^+ (aq) + Cl^-(aq) -----> AgCl(s)
When 10.0 mL of a 1.00 M AgNO3 solution is mixed with 10 mL of a 1.00 M NaCl solution at 25 degrees celsius in a calorimeter, a white precipitate is formed and the temperature increased to 32.6 degrees celsius. Calculate the enthalpy (delta H) for the production of 1 mol of AgCL in the reaction shown above. (Cs of solution = 4.184J/g celsius; density of solution = 1.00g/mL)
Ag^+ (aq) + Cl^-(aq) -----> AgCl(s)
When 10.0 mL of a 1.00 M AgNO3 solution is mixed with 10 mL of a 1.00 M NaCl solution at 25 degrees celsius in a calorimeter, a white precipitate is formed and the temperature increased to 32.6 degrees celsius. Calculate the enthalpy (delta H) for the production of 1 mol of AgCL in the reaction shown above. (Cs of solution = 4.184J/g celsius; density of solution = 1.00g/mL)
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moles Ag+ = M Ag+ x L Ag+ = (1.00)(0.0100) = 0.0100 moles Ag+
moles Cl- = M Cl- x L Cl- = (1.00)(0.0100) = 0.0100 moles Cl-
So 0.0100 moles of Ag+ react with the same amount of Cl- to form 0.0100 moles of AgCl.
The final solution volume = 10.0 mL + 10.0 mL = 20.0 mL (x 1.00 g/mL) = 20.0 g H2O
Heat evolved = (mass of solution)(Cs of solution)(Tf - Ti) = (20.0 g)(4.184 J/g C)(32.6 C - 25.0 C)
= -636 J evolved to make 0.0100 moles of AgCl
-636 J / 0.0100 moles AgCl = -63600 J/mole = -63.6 kJ/mole
moles Cl- = M Cl- x L Cl- = (1.00)(0.0100) = 0.0100 moles Cl-
So 0.0100 moles of Ag+ react with the same amount of Cl- to form 0.0100 moles of AgCl.
The final solution volume = 10.0 mL + 10.0 mL = 20.0 mL (x 1.00 g/mL) = 20.0 g H2O
Heat evolved = (mass of solution)(Cs of solution)(Tf - Ti) = (20.0 g)(4.184 J/g C)(32.6 C - 25.0 C)
= -636 J evolved to make 0.0100 moles of AgCl
-636 J / 0.0100 moles AgCl = -63600 J/mole = -63.6 kJ/mole