Problem 84 - Chemistry

Question

Consider the reaction: \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) When 2 moles of Na react with water at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), the volume of \(\mathrm{H}_{2}\) formed is \(24.5 \mathrm{~L}\). Calculate the work done in joules when \(0.34 \mathrm{~g}\) of Na reacts with water under the same conditions. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)

Step-by-Step Solution

Answer

The work done is -9.19 joules.

Step 1: Determine Moles of Na

To find the moles of Na, use the formula: \text{moles} = \frac{\text{mass}}{\text{molar mass}}. The molar mass of Na is 23.0 g/mol.\[ \text{moles of } \mathrm{Na} = \frac{0.34 \ \mathrm{g}}{23.0 \ \mathrm{g/mol}} \approx 0.01478 \ \mathrm{mol} \]

Step 2: Calculate Moles of H2 Gas

According to the balanced chemical equation, 2 moles of Na produce 1 mole of \(\mathrm{H}_{2}\). Thus, divide the moles of Na by 2 to find the moles of \(\mathrm{H}_{2}\).\[ \text{moles of } \mathrm{H}_{2} = \frac{0.01478 \ \mathrm{mol}}{2} \approx 0.00739 \ \mathrm{mol} \]

Step 3: Calculate Volume of H2

Under standard conditions, 2 moles of Na produce 24.5 L of \(\mathrm{H}_{2}\). Therefore, the volume of \(\mathrm{H}_{2}\) can be calculated using the moles determined.\[ \text{volume of } \mathrm{H}_{2} = 0.00739 \ \mathrm{mol} \times \frac{24.5 \ \mathrm{L}}{1 \ \mathrm{mol}} \approx 0.0907 \ \mathrm{L} \]

Step 4: Calculate Work Done

The work done is calculated using the formula \( W = -P\Delta V \), where \(P = 1\, \mathrm{atm}\) and \(\Delta V = \text{volume of } \mathrm{H}_{2} = 0.0907\, \mathrm{L}\).\[ W = -(1\, \mathrm{atm}) \times (0.0907\, \mathrm{L}) \times \frac{101.3\, \mathrm{J}}{1\, \mathrm{L} \cdot \mathrm{atm}} = -9.19\, \mathrm{J} \]

Key Concepts

Moles Calculation

In chemistry, calculating the number of moles in a substance is crucial for understanding how chemical reactions proceed. A mole is a unit that measures the number of particles, like atoms or molecules, in a given sample. It is based on Avogadro's number, which is approximately \(6.022 imes 10^{23}\).

Whenever you need to calculate the moles of a substance, the formula you will use is:

• moles = \( \frac{\text{mass of the substance}}{\text{molar mass of the substance}} \)

Let's apply this to sodium (Na): If you have a mass of 0.34 g of Na and its molar mass is 23.0 g/mol, then the moles of Na are:\[ \text{moles of Na} = \frac{0.34 \, \text{g}}{23.0 \, \text{g/mol}} \approx 0.01478 \, \text{mol} \]This calculation allows us to proceed in finding out how much product can be expected or how it participates in a reaction.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. This exercise revolves around a simple chemical reaction:\[ \text{2 Na (s) + 2 H}_{2}\text{O (l) } \longrightarrow \text{2 NaOH (aq) + H}_{2}\text{(g)} \]This equation tells us that when sodium (Na) reacts with water \(\text{H}_2\text{O} \), it forms sodium hydroxide (NaOH) and hydrogen gas \(\text{H}_2\).

• The equation is balanced, implying that the number of atoms for each element is the same on both sides of the equation.
• 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.

Understanding this transformation is key to predicting the volumes and moles of the products formed.

Work Done

In a chemical reaction, especially those involving gases, work can be done by or on the system. In this reaction, as hydrogen gas \(\text{H}_2\) is produced, work is done in expanding against the atmospheric pressure. The formula used to calculate the work done is:\[ W = -P\Delta V \]

Understanding the Variables

- \( P \) is the pressure, which in this case is 1 atm (standard pressure).- \( \Delta V \) is the change in volume, equivalent to the volume of the product gas in liters.By putting in the given values: - \( P = 1 \, \text{atm} \)- \( \Delta V = 0.0907 \, \text{L} \)Let's calculate the work done:\[ W = - (1 \, \text{atm}) \times (0.0907 \, \text{L}) \times \frac{101.3 \, \text{J}}{1 \, \text{L} \times \text{atm}} \approx -9.19 \, \text{J} \]Therefore, the reaction performs about \(-9.19\) joules of work by expanding the gas.

Stoichiometry

Stoichiometry is the field of chemistry that pertains to the quantitative relationships between reactants and products in a chemical reaction. This information comes from the balanced chemical equation. The reaction provided:\[ \text{2 Na (s) + 2 H}_{2}\text{O (l) } \longrightarrow \text{2 NaOH (aq) + H}_{2}\text{(g)} \]helps us understand these relationships clearly.

Understanding the Ratios

• Ratios indicate how much of each reactant is needed to produce a desired amount of product.
• In this case, 2 moles of Na yield 1 mole of \(\text{H}_2\), so Na is consumed by half the moles as \(\text{H}_2\) is produced.

By knowing the ratio, you can determine unknown quantities when given certain masses or moles of reactants by scaling the reaction up or down. For this specific exercise, knowing that 0.01478 moles of Na will result in 0.00739 moles of \(\text{H}_2\) keeps the calculations grounded and based on known principles.