Final answer:
To calculate the enthalpy change ΔH for 9.25×10⁻⁴ mol AgCl dissolving, we use the relation ΔH = q/n with q = -2.9 kJ. The result is ΔH ≈ -20.9 kJ/mol. Multiplying by the given moles and converting to J gives a final answer close to option a) -20.9 kJ.
Step-by-step explanation:
To calculate ΔH when 9.25×10⁻⁴ mol of AgCl dissolves in water, we use the given data that the reaction produces 2.9 kJ of heat. The reaction for the formation of AgCl(s) from NaCl(aq) and AgNO3(aq) is exothermic, as indicated by the negative sign. Therefore, the enthalpy change for dissolving AgCl in water will also be negative. The equation relating ΔH (enthalpy change) to the amount of heat produced (q) and the amount of substance (n) in moles is ΔH = q/n.
Using the given amount of heat, q = -2.9×10³ J, and the number of moles, n = 9.25×10⁻⁴ mol, we calculate: ΔH = (-2.9×10³ J) / (9.25×10⁻⁴ mol), which results in a value close to -20.9 kJ/mol. Hence, since we are only given the amount for 9.25×10⁻⁴ mol, we multiply -20.9 kJ/mol by this amount, converting kJ to J (as 1 kJ = 1000 J) to match the units.