5. At 1024°C, the pressure of oxygen gas from the decomposition of copper(I)oxid

Question

5. At 1024°C, the pressure of oxygen gas from the decomposition of copper(I)oxide is 0.490 bar ecomposition of copper(II)oxide is 0.490 bar: 4 CuO(s)2 Cu20(s) +029) (a) What is the value of K for this reaction? (Hint: Use I atm as your standard pressure) (b) Calculate the fraction of CuO that will decompose if 0.160 mole of it is placed in a 2.00 L flask at 1024°C (c) What would the fraction be if a 1.00 mole sample of CuO were used? (d) What is the smallest amount of CuO in moles that is needed in order to establish equilibrium?

Explanation / Answer

5.

a ) kp = 0.484

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

0.484 atm (2.00 L) = n(0.082)(1024 + 273)

n = 0.009083 moles O2

0.009083 moles O2 x 4 moles CuO/ 1 mole O2 = 0.036332 mole CuO decomposed

Fraction CuO decomposed = 0.036332 moles CuO / 0.160 moles CuO = 0.227 = 22.7 %

c.) If 1.00 mole CuO used then:

Fraction CuO decomposed = 0.036332 / 1 = 0.036332

fraction will be 0.036332 = 3.63 %

d.) Stoichiometry gives you the moles of CuO that reacted to give you that oxygen product:
0.009083 mol O2
0.009083 x (1 mol O2) / (4 mol CuO)
= 0.036332 mol CuO

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