Question

1.

A sample of S8 (8) is placed in an otherwise empty rigid container at 1325 K at an initial pressure of 1.00 atm, where it

decomposes to S2 by the reaction: S8() 4 S2(2). |

At equilibrium, the partial pressure of S, is 0.25 atm. Calculate K, for this reaction at 1325 K

Answer 1

Answer: The value of
`K_p` is 324

Step-by-step explanation:

We are given:

Initial pressure of
`S_8(g)` = 1.00 atm

The chemical equation for the conversion of
`S_8(g)` to
`S_2(g)` follows:

`S_8(g)\rightleftharpoons 4S_2(g)`

Initial: 1

At eqllm: 1-x 4x

We are given:

Equilibrium partial pressure of
`S_8(g)=0.25atm`

Evaluating the value of 'x', we get:

`\Rightarrow (1-x)=0.25\\\\\Rightarrow x=1-0.25=0.75`

The expression of
`K_p` for the above reaction follows:

`K_p=((p_(S_2))^4)/(p_(S_8))`

`p_(S_2)=4x=(4* 0.75)=3atm`

`p_(S_8)=0.25atm`

Putting values in above expression, we get:

`K_p=(3^4)/(0.25)\\\\K_p=324`

Hence, the value of
`K_p` is 324