Question

5.118 Raoul Pictet, the Swiss physicist who first liquefied oxygen, attempted to liquefy hydrogen. He heated potassium formate, \(\mathrm{KCHO}_{2}\), with \(\mathrm{KOH}\) in a closed 2.50-Lvessel. \(\mathrm{KCHO}_{2}(s)+\mathrm{KOH}(s) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2}(g)\) If \(75.0 \mathrm{~g}\) of potassium formate reacts in a \(2.50-\mathrm{L}\) vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.

Short answer

The pressure of hydrogen gas is approximately 10.81 atm.

Step-by-step solution

Determine Molar Mass of Potassium Formate

The first step is to calculate the molar mass of potassium formate, \( \text{KCHO}_2 \). By adding the atomic masses of K (39.1 g/mol), C (12.0 g/mol), H (1.0 g/mol), and O (2 \times 16.0 g/mol), we find the molar mass is 68.1 g/mol.

Calculate Moles of Potassium Formate

Next, convert the mass of potassium formate to moles. Using the mass of \( 75.0 \) g, divide by the molar mass: \( n = \frac{75.0 \text{ g}}{68.1 \text{ g/mol}} \approx 1.101 \text{ moles}. \)

Identify Reaction Stoichiometry

From the balanced reaction \( \text{KCHO}_2(s) + \text{KOH(s)} \rightarrow \text{K}_2\text{CO}_3(s) + \text{H}_2(g) \), we see that 1 mole of \( \text{KCHO}_2 \) produces 1 mole of \( \text{H}_2(g) \). Thus, \( 1.101 \) moles of \( \text{KCHO}_2 \) will produce \( 1.101 \) moles of \( \text{H}_2 \).

Use Ideal Gas Law to Calculate Pressure

Apply the ideal gas law \( PV = nRT \) to find the pressure. Substitute \( n = 1.101 \text{ moles} \), \( V = 2.50 \text{ L} \), \( R = 0.0821 \text{ L atm/mol K} \), and temperature \( T = 25 + 273 = 298 \text{ K} \). Solve for \( P \): \[ P = \frac{nRT}{V} = \frac{1.101 \times 0.0821 \times 298}{2.50} \approx 10.81 \text{ atm}. \]

Key concepts

Chemical Stoichiometry

Chemical stoichiometry helps us understand the quantities of reactants and products in a chemical reaction. It is important because it allows chemists to predict how much of each substance will be needed or produced in a reaction.
In our exercise, the equation is simplified to show the transformation of potassium formate and potassium hydroxide into potassium carbonate and hydrogen gas. This equation reveals a key stoichiometric detail: for every mole of potassium formate (\(\text{KCHO}_2\)), one mole of hydrogen gas (\(\text{H}_2\)) is created.This one-to-one relationship is crucial for stoichiometric calculations. It tells us that the moles of a reactant will directly translate into the moles of a product when the reaction fully goes to completion.

• 1 mole of \(\text{KCHO}_2\) produces 1 mole of \(\text{H}_2\).

Understanding this allows us to apply the data from the exercise, knowing that after converting the mass of reactants to moles, we can directly relate it to the amount of product formed. This step is key before proceeding with further calculations like molar mass and gas laws.

Molar Mass Calculation

Molar mass calculation involves adding up the atomic masses of all the atoms in a compound to find its mass per mole. This is essential when converting between the mass of a substance and the number of moles.For potassium formate (\(\text{KCHO}_2\)), the molar mass is calculated by considering the individual atomic weights:

• Potassium (\(\text{K}\)): 39.1 g/mol
• Carbon (\(\text{C}\)): 12.0 g/mol
• Hydrogen (\(\text{H}\)): 1.0 g/mol
• Oxygen (\(\text{O}\)): 16.0 g/mol (and there are two oxygen atoms)

Adding these together gives:\[39.1 + 12.0 + 1.0 + (2 \times 16.0) = 68.1 \text{ g/mol}\]This molar mass is vital in converting the given mass of 75.0 g of potassium formate into moles, enabling further stoichiometric calculations. By dividing the mass by the molar mass, we find that the substance approximately equals 1.101 moles, thus facilitating the calculation of resulting products from the reaction.

Gas Pressure Calculation

Gas pressure calculation is often accomplished using the Ideal Gas Law equation, \(PV = nRT\). This equation relates the pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), ideal gas constant (\(R\)), and temperature (\(T\)) of an ideal gas.For this exercise:

• \(n\) (moles of \(\text{H}_2\)) is 1.101, as calculated.
• \(V\) (volume) is given as 2.50 L.
• \(R\) is a constant: 0.0821 L atm/mol K.
• Temperature \(T\) is 298 K (converting 25°C to Kelvin by adding 273).

Plug these values into the formula to find the pressure (\(P\)):\[P = \frac{nRT}{V} = \frac{1.101 \times 0.0821 \times 298}{2.50} \approx 10.81 \text{ atm}\]The gas pressure computed reflects how much force the hydrogen gas exerts within the confines of the reaction vessel. Understanding this relationship is essential for students managing chemical reactions involving gases at defined conditions.