Question

When 0.617 0.617 g of sodium metal is added to an excess of hydrochloric acid, 6410 6410 J of heat are produced. What is the enthalpy of the reaction as written? 2Na(s)+2HCl(aq)⟶2NaCl(aq)+H₂(g) 2Na( s ) + 2 HCl ( aq ) ⟶ 2 NaCl ( aq ) + H₂ ( g )

Answer 1

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.