Problem 109 - Chemistry and Chemical Reactivity

Question

When palladium metal is exposed to \(\mathrm{H}_{2}\) gas, the metal become brittle because \(\mathrm{H}_{2}\) molecules dissociate and \(\mathrm{H}\) atoms fill some of the octahedral holes in the face-centered cubic lattice. To find the value of \(x\) in the formula \(\mathrm{PdH}_{x}\), you perform the following experiment: \(\mathrm{H}_{2}\) gas in a 2.25 -L flask has a pressure of \(113 \mathrm{mm}\) at \(23.0^{\circ} \mathrm{C}\). After exposing the gas to \(0.192 \mathrm{g}\) of \(\mathrm{Pd},\) the pressure is now \(108 \mathrm{mm}\) at \(23^{\circ} \mathrm{C} .\) What is the value of \(\mathrm{x}\) in \(\mathrm{PdH}_{x} ?\)

Step-by-Step Solution

Answer

The value of \( x \) in \( \mathrm{PdH}_x \) is approximately 0.67.

Step 1: Calculate the initial moles of \( \mathrm{H}_{2} \) gas

First, we use the ideal gas law to calculate the initial number of moles of \( \mathrm{H}_{2} \). The formula is \( PV = nRT \). Convert the given pressure from mm Hg to atm: \( 113 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.1487 \text{ atm} \). The volume \( V \) is 2.25 L, and the temperature \( T \) is 23.0°C which is 296 K. The gas constant \( R \) is 0.0821 L·atm/mol·K. Using \( n = \frac{PV}{RT} \), we get: \[ n = \frac{(0.1487 \text{ atm}) \times (2.25 \text{ L})}{(0.0821 \text{ L·atm/mol·K}) \times (296 \text{ K})} = 0.0138 \text{ moles} \]

Step 2: Calculate the moles of \( \mathrm{H}_{2} \) gas after exposure

After the exposure of \( \mathrm{Pd} \) to \( \mathrm{H}_{2} \), the pressure is reduced to 108 mm Hg. Convert this pressure to atm: \( 108 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.1421 \text{ atm} \). Using the ideal gas law to calculate the new moles of \( \mathrm{H}_{2} \): \[ n = \frac{(0.1421 \text{ atm}) \times (2.25 \text{ L})}{(0.0821 \text{ L·atm/mol·K}) \times (296 \text{ K})} = 0.0132 \text{ moles} \]

Step 3: Determine the moles of \( \mathrm{H}_{2} \) reacted

Subtract the moles of \( \mathrm{H}_{2} \) after exposure from the initial moles to find out how much \( \mathrm{H}_{2} \) reacted with \( \mathrm{Pd} \): \[ \text{Moles of } \mathrm{H}_{2} \text{ reacted} = 0.0138 - 0.0132 = 0.0006 \text{ moles} \]

Step 4: Determine the moles of Palladium

Given the mass of \( \mathrm{Pd} \) is 0.192 g, find the moles of \( \mathrm{Pd} \) using its molar mass, approximately 106.42 g/mol: \[ \text{Moles of } \mathrm{Pd} = \frac{0.192 \text{ g}}{106.42 \text{ g/mol}} = 0.0018 \text{ moles} \]

Step 5: Determine the value of \( x \) in \( \mathrm{PdH}_x \)

Assume each mole of \( \mathrm{Pd} \) reacts with \( x \) moles of H (or \( \frac{x}{2} \) moles of \( \mathrm{H}_{2} \) because \( \mathrm{H}_{2} \) molecules split into atoms). Using the moles of \( \mathrm{H}_{2} \) reacted, calculate \( x \): \[ 0.0018 \times \frac{x}{2} = 0.0006 \]Solving for \( x \), we find: \[ x = 2 \times \frac{0.0006}{0.0018} = 0.6667 \approx 0.67 \]

Step 6: Final result

Thus, the value of \( x \) in the formula \( \mathrm{PdH}_x \) is approximately 0.67.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation connecting several important properties of gases. It relates the pressure, volume, temperature, and number of moles of a gas with the formula: \( PV = nRT \).
Here, \( P \) stands for pressure, \( V \) is volume, \( n \) denotes the number of moles, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin.
The Ideal Gas Law is quite versatile and can be used to calculate any of these variables if the others are known.
To use this law effectively, always remember to:

• Convert pressure to atmospheres if it’s given in another unit, such as mm Hg.
• Convert temperature to Kelvin by adding 273.15 to the Celsius value.

These steps ensure that all units match the gas constant \( R \), allowing accurate calculations.

Mole Calculation

The concept of moles is central to chemistry as it allows us to count atoms and molecules in a practical way. A mole is Avogadro's number, \( 6.022 \times 10^{23} \), of atoms, ions, or molecules.
Finding the number of moles involves using the molar mass of a substance, which is the mass of one mole of that substance in grams.
For instance, to find the moles of a gas, such as \( \mathrm{H}_{2} \), involved in a reaction, the molar mass of \( \mathrm{H}_{2} \) (approximately 2.02 g/mol) and the ideal gas law can be applied.

• For solids like Palladium (\( \mathrm{Pd} \)), its atomic weight (around 106.42 g/mol) is used to find moles from a given mass.
• Use the formula \( n = \frac{m}{M} \), where \( m \) is mass in grams and \( M \) is molar mass in g/mol.

This calculation is essential not only in determining proportions in reactions but also in understanding the stoichiometry of the whole process.

Chemical Reaction Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It relies on the balanced chemical equation to determine the relationships between different reactants and products.
This principle is crucial when dealing with compounds like Palladium Hydride (\( \mathrm{PdH}_{x} \)). Here, understanding stoichiometry helps in calculating the value of \( x \) in \( \mathrm{PdH}_x \).
The idea is quite straightforward: the number of moles of each reactant and product are conserved based on their coefficients in the balanced equation.

• Reactants are converted into products in specific ratios, determined by their coefficients.
• If \( \mathrm{Pd} \) reacts with \( \mathrm{H} \) atoms, it does so in a fixed stoichiometric ratio, allowing for the calculation of unknown values like \( x \) in \( \mathrm{PdH}_{x} \).

This stoichiometric relationship helps predict quantities of products, manage resources efficiently, and understand the underlying chemistry of reactions.