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

Solution Steps

First, we need to calculate the number of moles of sodium (Na) used in the reaction. The molar mass of sodium is approximately 22.99 g/mol.

\[ \text{Moles of Na} = \frac{0.617 \, \text{g}}{22.99 \, \text{g/mol}} = 0.02684 \, \text{mol} \]

The given reaction is:

\[ 2 \mathrm{Na}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \]

The heat produced is 6410 J for 0.02684 moles of Na. Since the reaction involves 2 moles of Na, we need to find the enthalpy change per mole of reaction (which involves 2 moles of Na).

First, calculate the heat produced per mole of Na:

\[ \text{Heat per mole of Na} = \frac{6410 \, \text{J}}{0.02684 \, \text{mol}} = 238,800 \, \text{J/mol} \]

Since the reaction involves 2 moles of Na, the enthalpy change for the reaction as written (per 2 moles of Na) is:

\[ \Delta H = 238,800 \, \text{J/mol} \times 2 = 477,600 \, \text{J/mol} \]

Convert the enthalpy change from joules to kilojoules:

\[ \Delta H = \frac{477,600 \, \text{J}}{1000} = 477.6 \, \text{kJ} \]

Final Answer