Question

\[\ce{2SO2 (g) + O2 (g) <=> 2SO3 (g)}\]

In an equilibrium mixture, the partial pressures are

`"P"_("SO"_2)` = 43 kPa; `"P"_("O"_2)` = 530 Pa and

`"P"_("SO"_2)` = 45 kPA. The equilibrium constant K_p = _____ × 10^-2. (Nearest integer)

Options

• 17228

• 25000

• 15889

• 10000

Answer 1

\[\ce{2SO2 (g) + O2 (g) <=> 2SO3 (g)}\]

In an equilibrium mixture, the partial pressures are

`"P"_("SO"_2)` = 43 kPa; `"P"_("O"_2)` = 530 Pa and

`"P"_("SO"_2)` = 45 kPA. The equilibrium constant K_p = 17228 × 10^-2.

Explanation:

Given: `"P"_("SO"_2)` = 43 kPa; `"P"_("O"_2)` = 530 Pa and `"P"_("SO"_2)` = 45 kPA

The given chemical reaction is,

\[\ce{2SO2 (g) + O2 (g) <=> 2SO3 (g)}\]

For-tne given chemical reaction, the equilibrium constant K_p can be expressed as,

`"K"_"p" = (["P"_("SO"_3)]^2)/(["P"_("SO"_2)]^2 xx ["P"_("O"_2)])`

Here, K_p denotes the equilibrium constant.

The partial pressures of SO₃, SO₂ and O₂ are denoted by `"P"_("SO"_3), "P"_("SO"_2)` and `"P"_("O"_2)` respectively.

Substituting respective values in the above equation,

`"K"_"p" = ([43]^2)/([45]^2 xx [530])`

`therefore "K"_"p" = 1849/1073250`

`therefore "K"_"p" = 172.28 xx 10^-5 " Pa"^-1`

= 172.28 atm^-1 

= 17228 × 10^-2 atm^-1

The equilibrium constant K_p = 17228 × 10^-2 atm^-1