Question

15 g sample of $Ba{O}_2$ is heatedto ${794}^{o}C$ in a closed evacuated vessel of 5 litres capacity . How many g of peroxide are converted to $BaO\left(s\right).$
$2Ba{O}_2\left(s\right)\rightleftharpoons 2BaO\left(s\right)+O_2\left(g\right),K_p=0.5atm.$

Answer 1

2BaO2(s) 2BaO(s) + O2(g)

Initially 0.109 0 0

At equi 0.109-2x 2x x

Number of moles of BaO2 are calculated

Using formula weight of BaO2 (given)/ M.Mass of BaO2 = 15/169 = 0.089 moles.

Now, Kp = pO2 as others are in solid state so are not taken in account for finding equilibrium constant.

p(O2) = nRT/V = (x)(0.0821)(1067K)/ 5

( because T = 7940C = 794 + 273 K = 1067 K and V= 5 L)

Kp = 0.5 atm (given)

Therefore 0.5 =(x)(0.0821)(1067 K)/ 5

x = 0.0285 moles

as BaO = 2x = 2 x 0.0285 = 0.057 moles.

Now as moles = given wt/ M.Mass

0.057 = wt/153

Wt = 153 X 0.057 = 8.73 g