equilibrium chem

1.
ZnO(s) + CO(g) ⇌ Zn(g) + CO2(g)

At eqm:
Let no. of moles of Zn = y mol
Then, no. of moles of CO = (1 - y) mol
and no. of moles of (1 + y) mol
Total no. of moles = y + (1 - y) + (1 + y) = (2 + y) mol

PCO = 1 x [(1 - y)/(2 + y)] = (1 - y)/(2 + y) atm
PZn = 1 x [y/(2 + y)] = y/(2 + y) atm
PCO2 = 1 x [(1 + y)/(2 + y)] = (1 + y)/(2 + y) atm

Kp = (PZn)(PCO2)/(PCO)
[y/(2 + y)] x [(1 + y)/(2 + y)] / [(1 - y)/(2 + y)] = 1
y(1 + y) / (1 - y)(2 + y) = 1
y(1 + y) = (1 - y)(2 + y)
y + y^2 = 2 - y - y^2
2y^2 + 2y - 2 = 0
y^2 + y - 1 = 0
y = {-1 + √[1² - 4x1x(-1)]}/2
y = 0.618

Eqm partial pressure of Zn = 0.618/(2 + 0.618) = 0.236 atm


= = = = =
2.
The answer depends on the initial mole ratio of N2­ and H2.
Let 1 mol and 3 mol be the initial number of moles of N2 and H2­ respectively.

N2 + 3H2 ⇌ 2NH3

At eqm:
Let no. of moles of NH3 = 2y mol
Then no. of moles of N2 = (1 - y) mol
and no. of moles of H2 = (3 - 3y) mol
Total no. of moles = 2y + (1 - y) + (3 - 3y) = (4 - 2y) mol

2y/(4 - 2y) = 8%
2y = 0.32 - 0.16y
2.16y = 0.32
y = 0.148

PNH3 = 45 x [2(0.148)/(4 - 2x0.148)] = 3.60 atm
PN2 = 45 x [(1 - 0.148)/(4 - 2x0.148)] = 10.35 atm
PH2 = 45 x [(3 - 3x0.148)/(4 - 2x0.148)] = 31.05 atm

Kp = (PNH3)^2/(PN2)(PH2)^3 = (3.6)^2 /(10.35)(31.05)^3 = 4.18 x 10^-5 atm^-2


= = = = =
3.
NH4HS(s) ⇌ NH3(g) + H2S(g)

At eqm:
PNH3 = PH2S = 0.77/2 = 0.385 atm

Kp = (PNH3) x (PH2S) = (0.385)^2 = 0.148 atm^2

After 0.1 atm of NH3 was added,
let y atm of partial pressure of NH3 is decreased in the new eqm.

At the new eqm:
PNH3 = (0.385 + 0.1 - y) atm = (0.485 - y) atm
PH2S = (0.385 - y) atm

Kp = (PNH3) x (PH2S)
(0.485 - y)(0.385 - y) = 0.148
y^2 - 0.87y + 0.0387 = 0
y = [0.87 - √(0.87^2 - 4x0.0387x1)/]/2
y = 0.047

Partial pressure of NH3 at eqm = (0.485 - 0.047) = 0.438 atm