Question

38. The formation of SO₃ from SO₂ and O₂ is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant K_P for the reaction

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

is 0.13 at 803

°

C. In one experiment, 2.00 mol SO₂ and 2.00 mol O₂ were initially present in a flask.

What must

be the total pressure at equilibrium in order to have an 76.0% yield of SO₃?

b.The equilibrium constant K_P for the following reaction is found to be 4.31

×

10⁻⁴ at 375

°

C.

N₂(g) + 3H₂(g) ⇆ 2NH₃(g)

In a certain experiment a student starts with

0.942 atm of N₂

and

0.433 atm of H₂

in a

constant-volume

vessel at 375

°

C. Calculate the partial pressures of all species when equilibrium is reached.

P_N2 = atm

P_H2 = atm

P_NH3 = atm

Answer 1

38)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

2.0          2.0               0

2-2x         2-x               2x

at equilibrium 76% of SO3 obtained

moles of SO3 = 0.76 x 2 = 1.52 mol

at equilibrium :

moles of SO3 = 1.52

moles of SO2 = 0.48

moles of O2 = 1.24

total moles = 1.52 + 0.48 + 1.24 = 3.24

mole fraction of SO2 = 0.48 / 3.24 = 0.148

mole fraction of O2 = 1.24 / 3.24 = 0.383

mole fraction of SO3 = 1.52 / 3.24 = 0.469

let us assume :

total pressure = P atm

PSO2 = mole fraction x total pressure

PSO2 = 0.148 x P

PO2 = 0.383 x P

PSO3 = 0.469 x P

Kp = (PSO3)^2 / (PSO2)^2 x PO2

0.13 = (0.469 P)^2 / (0.148P)^2 x (0.383P)

P = 201.7

total pressure = 201.7 atm