Question

At 1024°C, the pressure of oxygen gas from the decomposition of copper(II) oxide (CuO) is 0.490 atm: 4CuO(s) ⇆ 2Cu2O(s) + O2(g)

(a) What is KP for reaction?

(b) Calculate the fraction of CuO decomposed if 0.860 mole of it is placed in a 2.00 L flask at 1024°C.

(c) What would be the fraction if a 1.00 mole sample of CuO were used?

Can someone please help me with part B and C

Answer 1

a) For the calculation of K_p, Ordinarily you'd have to calculate the partial pressures of the other gases from the given partial pressure and stoichiometry.

4CuO(s) ⇆ 2Cu₂O(s) + O₂(g)

Here, in this reaction O₂ is the only gas present, so the K_p of the reaction is, K_p = P_O2 = 0.49 atm

b)

Now, from the ideal gas law, PV = nRT or, P_O2V = n_o2RT

or, n_o2 = P_O2V /RT = (0.49 atm)(2.0 L)/ (0.082 L atm K⁻¹ mol⁻¹) (1297 K) = 9.21 10^-3 mol

So, 9.21 10^-3 mol oxygen produced and for that CuO decomposed 49.21 10^-3 mol = 0.03686 mol

So, the fraction of CuO decompose is = (0.03686/0.86) = 0.04286

c) If 1.00 mole sample of CuO were used, then the fraction would be = (0.03686/1.00) = 0.03686