Question

1. 2O3(g)↔ 3O2(g) ; if 10.0g of O2 is at equilibrium with 7.50g of O3 calculate Kp if the total pressure is 1.10atm.

A. 2.94

B. 0.499

C. 0.339

^ I submitted this question already but the answer was wrong, I worked through it and still couldn't find one of these options. Suggestions?

2. For: N2(g) + O2(g) ↔ 2NO(g) 0.500M O2 and 0.750M N2 is allowed to reach equilibrium. Kc = 1.00 x 10-1. Calculate equilibrium concentration of N2

A. 0.667

B. 0.167

C. 0.083

*Hint from professor: Are the given concentrations equilibrium concentrations or initial? Please read carefully.* Please show how to set up given the hint.

Answer 1

1.

The molar masses of O2 and O3 are 32 g/mol and 48 g/mol respectively.
The number of moles are obtained by dividing mass with molar mass.
The number of moles of $O2 = \dfrac {10.0}{32} = 0.3125$
The number of moles of $O3 = \dfrac {7.50}{48} = 0.15625$
The mole fraction of O2 $= \dfrac {0.3125}{0.3125 + 0.15625} = 0.667.$
The mole fraction of O3 = 1- 0.667 = 0.334
The partial pressure of O2 $= 0.667 \times 1.10 = 0.733$ atm
The partial pressure of O3 $= 0.334 \times 1.10 = 0.366$ atm
$K_p = \dfrac {P^3_{O2}}{P^2_{O3}} = \dfrac {(0.733)^3}{(0.366)^2}= 2.94$
Hence, the option (A) is the answer.