Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of solid silver chloride. When 10.0 mL of 1.00M AgNO3 solution is added to 10.0mL of 1.00 M NaCl solution at 25oC in a calorimeter a white precipitate of AgCl forms and the temperature of the aqueous mixture increases to 32.6oC. Assuming that the specific heat of the aqueous mixture is 4.18 J/goC, that the density of the mixture is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate the amount of heat absorbed in kJ/mol of Ag+.

Question

Chemistry

Emanuel Shea

5 June, 02:42

Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of solid silver chloride. When 10.0 mL of 1.00M AgNO3 solution is added to 10.0mL of 1.00 M NaCl solution at 25oC in a calorimeter a white precipitate of AgCl forms and the temperature of the aqueous mixture increases to 32.6oC. Assuming that the specific heat of the aqueous mixture is 4.18 J/goC, that the density of the mixture is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate the amount of heat absorbed in kJ/mol of Ag+.

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Alan Hays · Answer 1

AgNO3 + NaCl - -> AgCl + NaNO3

Number of moles

AgNO3

= molar concentration * volume

= 1 * 0.01

= 0.01 mol

NaCl

= 0.01 * 1

= 0.01 mol.

By stoichiometry, 1 mole of silver nitrate reacted with 1 mole of NaCl. Therefore,

Number of moles of AgCl formed = 0.01 * 1

= 0.01 mol AgCl formed.

Heat absorbed by solution during precipitation:

Mass of solution = density * total volume

= 1 * 20

= 20 g.

q = m * Cp * (T2 - T1)

= 20 * 4.18 * (32.6 - 25.0)

= 635 J

Since 635 J was absorbed by the solution, the reaction released - 635 J

So, Delta H = - 635 J/0.01 mol

= - 63500 J/mol

= - 63.5 kJ/mol.