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) 2 Na ( s ) + 2 HCl ( aq ) ⟶ 2 NaCl ( aq ) + H 2 ⁡ ( g )

Answer 1

To calculate the enthalpy of reaction of sodium with hydrochloric acid, the mass of sodium was converted to moles and then related to the heat released. The result is an enthalpy change of -119.4 kJ/mol, indicating an exothermic reaction.

Step-by-step explanation:

Calculating the Enthalpy of Reaction

When sodium (Na) reacts with hydrochloric acid (HCl) to form sodium chloride (NaCl) and hydrogen gas (H2), the enthalpy of the reaction can be determined from the amount of heat released. Given that 0.617 g of sodium reacts with an excess of HCl and produces 6410 J of heat, we must first convert this mass of sodium into moles using the molar mass of sodium (22.99 g/mol). This gives us 0.617 g / 22.99 g/mol = 0.02685 mol of sodium. Since the balanced chemical equation is

2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)

each mole of sodium corresponds to half a mole of reaction as per the stoichiometry. Hence, we need to multiply the moles of sodium by 2 to get the moles of reaction, yielding 0.0537 moles of reaction. The enthalpy change (ΔH) for the amount of reaction that took place is obtained by dividing the heat produced by the moles of reaction:

6410 J / 0.0537 mol = 119402.6 J/mol = 119.4 kJ/mol

Since the reaction is exothermic (releasing heat), the enthalpy change should have a negative sign. Therefore, the enthalpy change for the reaction is -119.4 kJ/mol. This represents the heat released when one mole of reaction occurs as per the stoichiometry of the balanced equation provided.