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?

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, \(K_P = 0.49 atm\).

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 \(0.49/1.96 = 0.25\)

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

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.