Question

41.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 0.560 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?

40.A quantity of 0.29 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: C(s) + CO2(g) ⇌ 2 CO(g) Under these conditions, the average molar mass of the gases was 35.9 g/mol. (a) Calculate the mole fractions of CO and CO2. The mole fraction of CO is The mole fraction of CO2 is (b)What is KP if the total pressure is 11.1 atm?

Answer 1

41)

a)

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

Kp = P_O2

Kp = 0.490

KP for reaction = 0.490

b)

P = 0.490 atm

T = 1024 deg C + 273.15 = 1297.15 K

V = 2.0 L , R = 0.08206 L atm per K per mol

Lets plug all the values to calculate moles of O2

PV= nRT

n = PV / RT

= 0.490 atm * 2.0 L / (0.08206 L atm per K per mol * 1297.15 K )

= 9.21 x 10^-3 mol O2

moles of CuO required to produce 9.21 x 10^-3 mol oxygen = 4 x 9.21 x 10^-3

moles of CuO required = 0.037

fraction of CuO decomposed = 0.037 / 0.560

fraction of CuO decomposed = 0.066      

c)

fraction of CuO decomposed = 0.037 / 1 = 0.037