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Chapter 5: Problem 112

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 \(2.50\) -L vessel. $$\mathrm{KCHO}_{2}(s)+\mathrm{KOH}(s) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2}(g)$$ If \(50.0 \mathrm{~g}\) of potassium formate reacts in a \(2.50\) -L vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to \(21^{\circ} \mathrm{C}\) ? Use the preceding chemical equation and ignore the volume of solid product.

Short Answer

Expert verified
The pressure of hydrogen gas attained is approximately 5.73 atm.

Step by step solution

01

Identify the Reactant and Equation

The reaction described is between potassium formate, \( \mathrm{KCHO}_2 \), and potassium hydroxide, \( \mathrm{KOH} \), producing potassium carbonate \( \mathrm{K}_2\mathrm{CO}_3 \) and hydrogen gas \( \mathrm{H}_2 \). The balanced chemical equation is provided as: \[ \mathrm{KCHO}_2(s) + \mathrm{KOH}(s) \rightarrow \mathrm{K}_2\mathrm{CO}_3(s) + \mathrm{H}_2(g) \]
02

Determine Moles of Potassium Formate

First, calculate the number of moles of potassium formate (\( \mathrm{KCHO}_2 \)) used in the reaction. The molar mass of \( \mathrm{KCHO}_2 \) is calculated as follows:- K: 39.10 g/mol- C: 12.01 g/mol- H: 1.01 g/mol- O (x2): 16.00 g/mol * 2Adding these gives the molar mass of \( \mathrm{KCHO}_2 \) as \( 39.10 + 12.01 + 1.01 + 32.00 = 84.12 \text{ g/mol} \).Given 50.0 g of \( \mathrm{KCHO}_2 \), the number of moles is calculated:\[ n(\mathrm{KCHO}_2) = \frac{50.0 \text{ g}}{84.12 \text{ g/mol}} \approx 0.5947 \text{ moles} \]
03

Moles of Hydrogen Produced

According to the balanced chemical equation, each mole of \( \mathrm{KCHO}_2 \) produces one mole of \( \mathrm{H}_2 \). Therefore, the moles of hydrogen produced in the reaction is the same as the moles of \( \mathrm{KCHO}_2 \):\[ n(\mathrm{H}_2) = 0.5947 \text{ moles} \]
04

Calculate the Pressure of Hydrogen Gas

To find the pressure of the hydrogen gas, use the Ideal Gas Law \( PV = nRT \), where:- \( P \) is the pressure,- \( V \) is the volume of the vessel (2.50 L),- \( n \) is the moles of gas (0.5947 moles),- \( R \) is the ideal gas constant (0.0821 L路atm/mol路K),- \( T \) is the temperature in Kelvin (21掳C + 273 = 294 K).Rearranging the Ideal Gas Law for pressure, we get:\[ P = \frac{nRT}{V} \]Substituting the values:\[ P = \frac{0.5947 \times 0.0821 \times 294}{2.50} \approx 5.732 \text{ atm} \]
05

Conclusion

The pressure of hydrogen gas, \( \mathrm{H}_2 \), at 21掳C in a 2.50-L vessel after the reaction has completed is approximately 5.732 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equation
Understanding a chemical equation is crucial for any chemical reaction process. A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants on the left side and the products on the right side, separated by a reaction arrow. For the exercise given, the reaction equation is:
  • Reactants: Potassium formate (\(\mathrm{KCHO}_2\)) and potassium hydroxide (\(\mathrm{KOH}\))
  • Products: Potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) and hydrogen gas (\(\mathrm{H}_2\))
This equation must be balanced, meaning that the number of each type of atom is equal on both sides of the equation. Balancing a chemical equation is based on the Law of Conservation of Mass, which states that matter cannot be created or destroyed.
Moles of Gas
The concept of moles is fundamental in chemistry, allowing chemists to count particles by weighing them. By calculating moles, we can determine the amount of substance involved in a reaction. To find the moles of a reactant or product, you divide the mass of the substance by its molar mass.
In our exercise, for potassium formate (\(\mathrm{KCHO}_2\)), we calculated the moles using its molar mass:
  • Molar mass of \(\mathrm{KCHO}_2\) = 84.12 g/mol
  • Mass of \(\mathrm{KCHO}_2\) = 50.0 g
Formula: \[ n(\mathrm{KCHO}_2) = \frac{50.0}{84.12} \approx 0.5947 \text{ moles} \]According to the balanced chemical equation, 1 mole of \(\mathrm{KCHO}_2\) produces 1 mole of \(\mathrm{H}_2\), hence, 0.5947 moles of hydrogen gas are produced.
Pressure Calculation
To calculate gas pressure, we use the Ideal Gas Law: \(PV = nRT\), where:
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume of gas (2.50 L in this case),
  • \(n\) is the number of moles of the gas (0.5947 moles of hydrogen),
  • \(R\) is the ideal gas constant (0.0821 L路atm/mol路K),
  • \(T\) is the temperature in Kelvin (\(21掳C + 273 = 294 \text{ K}\))
To find the pressure \(P\), we rearrange the formula:\[ P = \frac{nRT}{V} \]Substitute the values:\[ P = \frac{0.5947 \times 0.0821 \times 294}{2.50} \approx 5.732 \text{ atm} \]Thus, the pressure of hydrogen gas is approximately 5.732 atm.
Molar Mass Calculation
Molar mass is a measure of the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating molar mass is essential because it helps in converting between mass and number of moles, which is crucial for stoichiometry in chemical reactions.
For this exercise, the molar mass of potassium formate \(\mathrm{KCHO}_2\) was determined using the periodic table:
  • Potassium (K) = 39.10 g/mol
  • Carbon (C) = 12.01 g/mol
  • Hydrogen (H) = 1.01 g/mol
  • Oxygen (O): 16.00 g/mol, and remember it's multiplied by 2 since there are two oxygen atoms
Adding together gives:\[ 39.10 + 12.01 + 1.01 + (16.00 \times 2) = 84.12 \text{ g/mol} \]Understanding this calculation allows you to translate between the mass of a compound and the amount (in moles) that will react.

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