Question

When 145 mL of 0.204 M NaCl(aq) and 145 mL of 0.204 M AgNO3(aq), both at 21.1°C, are mixed in a coffee cup calorimeter, the temperature of the mixture increases to 23.6°C as solid AgCl forms. NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) This precipitation reaction produces 3.02 ✕ 103 J of heat, assuming no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and density of the solutions are the same as those for water (4.18 J/g·°C, and 0.997 g/mL, respectively). Using this data, calculate ΔH in kJ/mol of AgNO3(aq) for the given reaction.

Answer 1

ΔH of AgNO3(aq) is - 10.21 kJ/ mol

Step-by-step explanation:

Step 1: Data given

Volume of 0.204 M NaCl = 145 mL = 0.145 L

Volume of 0.204 M AgNO3 = 145 mL = 0.145 L

Initial temperature of both = 21.1 °C

Final temperature = 23.6 °C

This precipitation reaction produces 3.02*10³ J of heat

Specific heat of the solution = 4.18 J/g°C

Density = 0.997 g/mL

Step 2: The balanced equation

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

In a coffee cup calorimeter, pressure is constant → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ with Q = heat = 3.02 * 10³ J

⇒ with m = mass

⇒ Mass of the solution = density * volume

Mass of solution = 0.99 J/g°C7 g/mL * 290 mL = 289.13 grams

⇒ with c is heat capacity = 4.18 J/g°C

⇒ with ΔT = temperature change. = 23.6 -21.1 = 2.5

Step 3: Calculate moles of AgNO3

Moles of AgNO3 = Molarity * volume

Moles of AgNO3 = 0.204 * 0.145 L = 0.02958 moles

Step 4: Calculate ΔH of AgNO3

ΔH = 3.02 kJ / 0.02958 moles = 10.21 kJ /mol (negative because it's exothermic)

ΔH of AgNO3(aq) is - 10.21 kJ/ mol