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 K_P for reaction?

(b) Calculate the fraction of CuO decomposed if 0.310 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?

Answer 1

4CuO(s) <---------------> 2Cu2O(s) + O2(g)

Kp = PO2 Kp = 0.490

KP for reaction = 0.490

(b) determine the number of moles of O2 produced using the Ideal Gas Law
PV = nRT

0.490atm(2.00 L) = n(0.082)(1024 + 273)
n = 0.92 / (0.082)(1297)
n = 0.00921 moles O2

0.00921 moles O2 x 4 moles CuO/ 1 mole O2 = 0.0368 mole CuO decomposed

Fraction CuO decomposed = 0.0368 moles CuO/ 0.310 moles CuO = 0.1187

(c) If 1.00 mole CuO used then:

Fraction CuO decomposed = 0.0368/ 1 = 0.0368