Question

An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:

2SO2(g) + O2(g) ⇌ 2SO3(g)

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

Answer 1

Answer:
`p_(SO_2)=0.017atm`

Step-by-step explanation:

We are given:

`K_c=1.7* 10^8`

Relation of
`K_p` with
`K_c` is given by the formula:

`K_p=K_c(RT)^(\Delta ng)`

Where,

`K_p` = equilibrium constant in terms of partial pressure = ?

`K_c` = equilibrium constant in terms of concentration

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature =
`600K`

`2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)`

`\Delta ng` = change in number of moles of gas particles =
`n_(products)-n_(reactants)=2-3=-1`

Putting values in above equation, we get:

`K_p=1.7* 10^8* (0.0821* 600)^(-1)\\\\K_p=3.4* 10^6`

The chemical reaction follows the equation:

`2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)`

The expression for
`K_p` for the given reaction follows:

`K_p=\frac{(p_(SO_3))^2}{ p_(O_2)* {(p_(SO_2))^2}}`

We are given:

`p_(SO_3)=300atm`
`p_(O_2)=100atm`

Putting values in above equation, we get:

`3.4* 10^6=\frac{(300)^2}{100* {(p_(SO_2))^2}}`

`p_(SO_2)=0.017atm`

Hence, the partial pressure of the
`SO_2` at equilibrium is 0.017 atm.