Final answer:
The enthalpy change (ΔH) in kJ/mol of AgNO3(aq) for the precipitation reaction NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) is -33.35 kJ/mol, calculated by using the provided calorimetric data and the concept of specific heat capacity.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) in kJ/mol of AgNO3(aq) for the reaction NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq), we use the information from the provided data. The reaction involves mixing 100 mL of 0.200 M NaCl(aq) with 100 mL of 0.200 M AgNO3(aq), with the temperature increasing from 21.9 °C to 23.5 °C.
The total volume of the solution is 200 mL, which is 0.2 L, and since the concentration of both reactants is 0.200 M, we are dealing with a reaction involving 0.0400 mol of AgNO3.
The specific heat capacity of water (which we'll assume represents the solution well) is 4.18 J/g°C, and the density of water is approximately 1 g/mL.
Therefore, the mass of the solution is 200 g. To find the amount of heat (q) produced, we use the formula: q = mcΔT.
Here, m = 200 g, c = 4.18 J/g°C, and ΔT = (23.5 °C - 21.9 °C) = 1.6 °C. So, q = 200 g * 4.18 J/g°C * 1.6 °C = 1334.4 J.
Since the reaction is exothermic, the heat produced is released to the surroundings, which is why we use a negative sign for q. Thus, q = -1334.4 J = -1.334 kJ (since 1 kJ = 1000 J).
To find ΔH per mole of AgNO3, we divide the total heat by the moles of AgNO3:
ΔH = q/mol = -1.334 kJ / 0.0400 mol = -33.35 kJ/mol.