Question 8.3: Calculating ΔH in a Calorimetry Experiment Aqueous silver io......
Calculating ΔH in a Calorimetry Experiment
Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of solid silver chloride:
Ag^+(aq) + Cl^-(aq) → AgCl(s)
When 10.0 mL of 1.00 M AgNO_3 solution is added to 10.0 mL of 1.00 M NaCl solution at 25.0 °C in a calorimeter, a white precipitate of AgCl forms and the temperature of the aqueous mixture increases to 32.6 ºC. Assuming that the specific heat of the aqueous mixture is 4.18 J/(g . °C), that the density of the mixture is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction.
STRATEGY
Because the temperature rises during the reaction, heat must be released and ΔH must be negative. The amount of heat evolved during the reaction is equal to the amount of heat absorbed by the mixture:
Heat evolved = Specific heat × Mass of mixture× Temperature change
Calculating the heat evolved on a per-mole basis then gives the enthalpy change ΔH.
According to the balanced equation, the number of moles of AgCl produced equals the number of moles of Ag^+ (or Cl^-) reacted:
\text { Moles of } \mathrm{Ag}^{+}=(10.0 \mathrm{~mL})\left(\frac{1.00 \mathrm{~mol} \mathrm{Ag}^{+}}{1000 \mathrm{~mL}}\right)=1.00 \times 10^{-2} \mathrm{~mol} \mathrm{Ag}^{+} \\ \text { Moles of } \mathrm{AgCl}=1.00 \times 10^{-2} \mathrm{~mol} \mathrm{AgCl} \\ \text { Heat evolved per mole of } \mathrm{AgCl}=\frac{6.4 \times 10^2 \mathrm{~J}}{1.00 \times 10^{-2} \mathrm{~mol} \mathrm{AgCl}}=64 \mathrm{~kJ} / \mathrm{mol} \mathrm{AgCl}Therefore, ΔH = -64 kJ (negative because heat is released)