Final answer:
To determine the enthalpy of the reaction where 0.617 g of sodium reacts with hydrochloric acid, the number of moles of sodium was first calculated. The heat released was then converted to kJ per mole of reaction, yielding an enthalpy of -239.18 kJ/mol reaction.
Step-by-step explanation:
When 0.617 g of sodium metal is added to hydrochloric acid, producing 6410 J of heat, we need to determine the enthalpy of the reaction for the balanced equation:
2Na(s) + 2HCl(aq) ⟶ 2NaCl(aq) + H₂(g)
First, we find the amount in moles of sodium. The molar mass of sodium (Na) is approximately 23.0 g/mol. Therefore, the amount of Na used in the reaction is:
0.617 g Na × (1 mol Na / 23.0 g Na) = 0.0268 mol Na
Since the balanced equation shows 2 moles of Na reacting, we divide the moles of Na used by 2 to find the enthalpy change per mole of reaction:
6410 J × (1 mole reaction / 0.0268 mol Na) = 239179.1 J/mol reaction
Since 1 kJ = 1000 J, we can convert this to kJ:
239179.1 J/mol reaction × (1 kJ / 1000 J) = 239.18 kJ/mol reaction
The enthalpy of the reaction is then -239.18 kJ/mol reaction, with the negative sign indicating that heat is released by the reaction.