Final answer:
To determine the enthalpy change for the reaction of sodium metal with hydrochloric acid, the moles of sodium are calculated and then used to compute the enthalpy change per mole. The reaction yields -238.9 kJ/mol of sodium, and since the reaction is for 2 moles of sodium, the total enthalpy change is -477.8 kJ.
Step-by-step explanation:
To determine the enthalpy of the reaction for the reaction between sodium metal and hydrochloric acid, we use the information provided about the amount of heat produced and the mass of sodium reacted. From the equation 2Na(s) + 2HCl(aq) → 2NaCl(aq) + H₂(g), we can calculate the moles of sodium reacted and then compute the enthalpy change per mole of sodium.
First, we calculate the moles of sodium:
- Molar mass of Na = 22.99 g/mol
- Moles of Na = 0.703 g / 22.99 g/mol = 0.0306 mol
The given reaction involves 2 moles of sodium, so we need to find the enthalpy for half the reaction that would be for 1 mole of sodium. Since 7310 J of heat were produced for 0.0306 mol of sodium, we use the following calculation:
Enthalpy (ΔH) per mole = 7310 J / 0.0306 mol = 238889 J/mol or 238.9 kJ/mol
The reaction releases energy, so the enthalpy change is negative:
ΔH = -238.9 kJ/mol for the reaction as written for 2 moles of Na:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H₂(g), ΔH = -477.8 kJ