Question

An engineer examining the oxidation of SO2 in the manufacture of sulfuric acid determines that Kc = 1.7 multiplied by 108 at 600. K:

2 SO2(g) + O2(g) reverse reaction arrow 2 SO3(g).
(a) At equilibrium, PSO3 = 300. atm and PO2 = 100. atm. Calculate PSO2.
PSO2 = _________. atm
(b) The engineer places a mixture of 0.0040 mol of SO2(g) and 0.0028 mol of O2(g) in a 1.0-L container and raises the temperature to 1000 K. At equilibrium, 0.0020 mol of SO3(g)
K_c =

Answer 1

PSO2 at equilibrium can be calculated as 25 atm using the provided equilibrium constant and partial pressures of SO3 and O2. Additional information is needed to calculate the new Kc at 1000 K.

Step-by-step explanation:

To calculate PSO2 using the given equilibrium constant (Kc) and partial pressures of SO3 and O2, we can use the expression for the equilibrium constant for the reaction 2 SO2(g) + O2(g) ⇒ 2 SO3(g). The expression for Kc at equilibrium is:

Kc = (PSO3)^2 / (PSO2)^2 * PO2

Given that Kc = 1.7 × 10^8, PSO3 = 300 atm, and PO2 = 100 atm, we can substitute these values into the equation and solve for PSO2:

1.7 × 10^8 = (300)^2 / (PSO2)^2 * 100

After simplifying and solving for PSO2, we find that:

PSO2 = 25 atm

For part (b), we cannot directly calculate the new equilibrium constant Kc at 1000 K with the given information. Instead, the expression for Kc must be applied using the concentrations at equilibrium. For this example, we would need additional information such as the change in concentrations of reactants and products due to the reaction to determine the equilibrium concentrations and calculate the new Kc value.

Answer 2

a. At equilibrium when PSO₃ = 300. atm and PO₂ = 100. atm, PSO₂ is 23.0 atm.

b. Kc is 1250 and PSO₂ is 0.0020 atm.

How to calculate PSO₂ at equilibrium

(a) To calculate PSO₂ at equilibrium, use the equilibrium constant expression and the given equilibrium concentrations of PSO₃ and PO₂.

The equilibrium constant expression for the reaction is:

`Kc = [PSO_2]^2 / ([PSO_2]^2 * [PO_2])`

Given:

Kc = 1.7 ×
`10^8`

PSO₃ = 300 atm

PO₂ = 100 atm

Substituting these values into the equilibrium constant expression, we get:

1.7 ×
`10^8` =
`(300^2) / ([PSO_2]^2` * 100)

Rearranging the equation to solve for PSO₂:

PSO₂ = √529.41

PSO₂ = 23.0 atm

Therefore, PSO₂ is 23.0 atm.

(b) To calculate Kc at the new equilibrium and PSO₂, use the given initial amounts of SO₂, O₂, and SO₃.

The initial moles of SO₂ = 0.0040 mol

The initial moles of O₂ = 0.0028 mol

The initial moles of SO₃ = 0 mol (since it is not given)

The change in moles for SO₂ = -0.0020 mol (since 0.0020 mol of SO₂ reacts to form 0.0020 mol of SO₃)

The change in moles for O₂ = -0.0020 mol (since 0.0020 mol of O₂ reacts to form 0.0020 mol of SO₃)

The change in moles for SO₃ = +0.0020 mol (since 0.0020 mol of SO3 is formed)

The equilibrium moles of SO₂ = initial moles of SO₂ + change in moles of SO₂

= 0.0040 mol + (-0.0020 mol)

= 0.0020 mol

The equilibrium moles of O₂ = initial moles of O₂ + change in moles of O₂

= 0.0028 mol + (-0.0020 mol)

= 0.0008 mol

The equilibrium moles of SO₃ = initial moles of SO₃ + change in moles of SO₃

= 0 mol + 0.0020 mol

= 0.0020 mol

Now, calculate Kc using the equilibrium concentrations:

`Kc = [SO_3]^2 / ([SO_2]^2 * [O_2])`

Substituting the values:

Kc =
`(0.0020^2) / ((0.0020^2)` * 0.0008)

Kc = 1 / 0.0008

Kc = 1250

The equilibrium concentration of SO₂, PSO₂, is given by the moles of SO₂ divided by the volume:

PSO₂ = moles of SO₂ / volume

= 0.0020 mol / 1.0 L

= 0.0020 M

Therefore, Kc is 1250 and PSO₂ is 0.0020 atm.

Complete question

An engineer examining the oxidation of SO2 in the manufacture of sulfuric acid determines that Kc = 1.7 multiplied by 108 at 600. K:

2 SO2(g) + O2(g) reverse reaction arrow 2 SO3(g).

(a) At equilibrium, PSO3 = 300. atm and PO2 = 100. atm. Calculate PSO2.

PSO2 = _________. atm

(b) The engineer places a mixture of 0.0040 mol of SO2(g) and 0.0028 mol of O2(g) in a 1.0-L container and raises the temperature to 1000 K. At equilibrium, 0.0020 mol of SO3(g)

K_c =

PSO2 = _________. atm