Question

Calculate the heat released when \(2.00 \mathrm{~L}\) of \(\mathrm{Cl}_{2}(g)\) with a density of \(1.88 \mathrm{~g} / \mathrm{L}\) reacts with an excess of sodium metal at \(25^{\circ} \mathrm{C}\) and 1 atm to form sodium chloride.

Short answer

The heat released is approximately \(-21.8 \, \text{kJ}\).

Step-by-step solution

Understanding the chemical reaction

The chemical equation for the reaction between chlorine gas and sodium to form sodium chloride is:\[2 ext{Na}_{(s)} + ext{Cl}_2{(g)} ightarrow 2 ext{NaCl}_{(s)}\]This equation tells us that 1 mole of \( \text{Cl}_2 \) reacts with 2 moles of sodium to form 2 moles of sodium chloride.

Determine the amount of \( \text{Cl}_2 \) in moles

Given the density of \( \text{Cl}_2 \) is \( 1.88 \, \text{g/L} \), we first calculate the mass of \( \text{Cl}_2 \) in 2 L:\[\text{Mass} = \text{Density} \times \text{Volume} = 1.88 \times 2.00 = 3.76 \, \text{g}\]Now, calculate the moles of \( \text{Cl}_2 \). The molar mass of \( \text{Cl}_2 \) is \( 70.9 \, \text{g/mol} \):\[\text{Moles of } \text{Cl}_2 = \frac{3.76}{70.9} \approx 0.0531 \, \text{mol}\]

Find the heat released per mole of \( \text{NaCl} \) formed

The standard enthalpy of formation of \( \text{NaCl} \) from sodium and chlorine gas is \(-411 \, \text{kJ/mol}\). This value means that 411 kJ of heat is released when 2 moles of \( \text{NaCl} \) are formed.

Calculate the total heat released

From Step 2, we derived that \( 0.0531 \, \text{mol} \) of \( \text{Cl}_2 \) results in \( 2 \times 0.0531 = 0.1062 \, \text{mol} \) of \( \text{NaCl} \). The total heat released is given as:\[\text{Total heat released} = 0.1062 \, \text{mol} \times \left( \frac{-411 \, \text{kJ/mol}}{2} \right) \approx -21.8 \, \text{kJ}\]Note that we divide by 2 because the enthalpy corresponds to 2 moles of \( \text{NaCl} \).

Key concepts

Chemical Reaction Balancing

Balancing chemical equations is all about ensuring that the number of atoms for each element is the same on both sides of the equation. This is crucial for accurately representing chemical reactions. In the given exercise, the reaction is between chlorine gas (\( \text{Cl}_2(g) \)) and solid sodium (\( \text{Na}_{(s)} \)) to form sodium chloride (\( \text{NaCl}_{(s)} \)).

• The balanced equation is: \[2 \text{Na}_{(s)} + \text{Cl}_2(g) \rightarrow 2 \text{NaCl}_{(s)}\]
• This equation indicates that 1 mole of chlorine gas reacts with 2 moles of sodium to produce 2 moles of sodium chloride.
• Balancing the equation ensures the conservation of mass during the reaction, meaning that all atoms you start with are accounted for in the products.

This balance allows us to use the stoichiometry of the reaction to calculate the amount of heat released, as shown in the exercise solution.

Moles Calculation

Calculating moles is a fundamental step in stoichiometry, helping us relate mass to the entities involved in the reaction. In the problem, we started with chlorine gas, which has a density of 1.88 g/L.

• First, compute the mass of \( \text{Cl}_2 \) by multiplying its density by the volume: \( \text{Mass} = \text{Density} \times \text{Volume} = 1.88 \, \text{g/L} \times 2.00 \, \text{L} = 3.76 \, \text{g} \).
• Knowing the molar mass of \( \text{Cl}_2 \) is 70.9 g/mol, we find moles by: \( \frac{3.76 \, \text{g}}{70.9 \, \text{g/mol}} \approx 0.0531 \, \text{mol} \).

By calculating the moles of reactants, we can determine the proportion of products formed, crucial for determining the reaction's heat output.

Enthalpy of Formation

The enthalpy of formation (\( \Delta H_{f}^{\circ} \)) of a compound is the heat change when 1 mole of a compound forms from its elements in their standard states. Most enthalpy values, like that of sodium chloride in this exercise (-411 kJ/mol), are negative, indicating that the reaction releases heat, or is exothermic.

• The standard enthalpy of formation tells us how much energy is released or absorbed, providing insight into the stability of the compound formed.
• In our case, for every 2 moles of \( \text{NaCl} \) produced, the reaction releases 411 kJ of energy.

This value helps convert moles of product into energy change, aligning perfectly with our stoichiometric calculations.

Chemical Stoichiometry

Stoichiometry provides a quantitative relationship between reactants and products in a chemical reaction, guiding us through calculations. It relates the moles of chlorine used to the moles of sodium chloride produced.

• With stability provided by a balanced equation, we use stoichiometry to find the total heat released.
• From the exercise, we computed 0.0531 mol of \( \text{Cl}_2 \) produces \( 2 \times 0.0531 \approx 0.1062 \) mol of \( \text{NaCl} \).
• The total heat release calculation involves multiplying the moles of \( \text{NaCl} \) by the enthalpy change per mole, adjusted for the 2 moles in the enthalpy data: \( 0.1062 \, \text{mol} \times \frac{-411 \, \text{kJ/mol}}{2} \approx -21.8 \, \text{kJ} \).

Together, these concepts allow us to confidently determine the energy change associated with the reaction, mirroring physical processes in chemical laboratories.