Question

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}\) $$ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a 2.0 - \(L\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

Short answer

\[K_P = 0.49\], \(fraction_{decomposed} = 0.25\) for both 0.16 mole and 1 mole of CuO, and \(min_{amount} = 0.080 moles\]

Step-by-step solution

Calculate \(K_P\)

Given that the pressure of O2 is 0.49 atm, We can use the balanced chemical equation and the definition of \(K_P\) to solve for it. \(K_P\) is equal to the partial pressure of the products raised to their stoichiometric coefficients, divided by the partial pressure of the reactants raised to their stoichiometric coefficients. For this reaction, there are no gaseous reactants, so \(K_P = P_{O2}\). Therefore, \(K_P = 0.49 atm\).

Calculate the fraction of CuO decomposed

To find the fraction of CuO decomposed, we calculate the initial moles of CuO through \(n = \frac{PV}{RT}\) where n is the number of moles, P is the pressure, V is the volume, R is the ideal gas constant (0.08206 L·atm/K·mol), and T is the temperature in Kelvin. Since the pressure of oxygen arises solely from the decomposition of CuO, the moles of O2 gas formed can be calculated via stoichiometry from the pressure given. Thus, \(moles_{input} - moles_{decomposed} = moles_{remaining}\) and the fraction decomposed is \(fraction_{decomposed} = \frac{moles_{decomposed}}{moles_{initial}}\). It's necessary to change the temperature to Kelvin (from Celsius) by adding 273 to the temperature in Celsius (1024°C + 273 = 1297K). Hence, we can assume that since 1 mole of O2 is produced from 4 moles of CuO, then 0.49 atm of O2 would result from the decomposition of four times that amount of CuO. Therefore, the fraction decomposed is \(0.49/1.96 = 0.25\)

Calculate the fraction for 1 mole of CuO

If 1 mole of CuO was used, we can use the same ratio as derived in Step 2. Thus, the fraction decomposed would still be 0.25

Calculate the minimum amount of CuO to establish the equilibrium

The smallest number of moles of CuO that can be decomposed to establish this equilibrium would be enough to generate 0.49 atm of pressure within the 2.0-L flask at 1024°C (1297 K). Therefore, using the ideal gas law equation, we can find this minimum amount to be 0.080 moles, \(n = \frac{PV}{RT} = \frac{0.49*2}{0.08206*1297} = 0.080 moles\) therefore indicating that 0.080 moles would be the smallest amount of CuO that could establish this equilibrium under the given conditions.

Key concepts

Equilibrium Constant (Kp)

In the context of chemical equilibrium, the equilibrium constant denoted as \( K_p \) is pivotal in understanding the balance between reactants and products in a gaseous reaction. In this particular decomposition reaction of copper(II) oxide (\( CuO \)), \( K_p \) represents the concentration of products compared to that of the reactants at equilibrium. Since this system only involves a gaseous product and solid reactants (copper oxides), the constant is determined solely from the partial pressures of the products. In solid-gas equilibria, the partial pressure of solids, like \( CuO \), does not affect \( K_p \) as solids are not included in the calculation. Thus, the equation simplifies to

• \( K_p = P_{O_2} \), where \( P_{O_2} \) is the pressure of \( O_2 \) gas in atmospheres.

The solution identifies that \( P_{O_2} = 0.49 \) atm, leading directly to \( K_p = 0.49 \) atm. This shows that at equilibrium, the room occupied by oxygen gas is characterized entirely by its own partial pressure, linking the concept of \( K_p \) to product pressures found in reaction equilibria.

Decomposition Reaction

A decomposition reaction involves the breakdown of a compound into simpler substances. In the exercise, copper(II) oxide (\( CuO \)) decomposes to form copper(I) oxide (\( Cu_2O \)) and oxygen gas (\( O_2 \)). This specific reaction is represented as: \[ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_2\mathrm{O}(s) + \mathrm{O}_2(g) \]
A few important points to keep in mind about decomposition reactions include:

• They often require energy input, such as heat, to proceed, as seen with \( CuO \).
• Decomposition reactions typically result in the production of simpler substances from a single complex compound.

In this reaction, oxygen gas is the only product that is gaseous, which simplifies calculations for partial pressures and places emphasis on stoichiometry for understanding the decomposition extent. Applying stoichiometry, every mole of \( O_2 \) is produced from 4 moles of \( CuO \), demonstrating a key stoichiometric relationship that helps calculate the fraction decomposed.

Ideal Gas Law

The ideal gas law is fundamental in linking the state variables of a gas: pressure, volume, and temperature. The formula is expressed as: \[ PV = nRT \]
where \( P \) is the pressure in atmospheres, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant (approximately 0.08206 L·atm/K·mol), and \( T \) is the temperature in Kelvin. This law is particularly useful in the decomposition reaction of \( CuO \), as it allows for determining the number of moles of gas produced from a given pressure and temperature condition.

• To find the number of moles (\( n \)) of \( O_2 \) produced, rearrange the formula to \( n = \frac{PV}{RT} \).
• Changing temperature from Celsius to Kelvin involves adding 273 to the Celsius temperature, essential for using \( T \) in the gas law equation.

For example, to calculate the minimum moles of \( CuO \) needed to establish equilibrium, \( n \) was calculated, offering practical insights into the application of the ideal gas law in determining gas-phase quantities in chemical reactions.