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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

Respuesta :

Answer: The value of [tex]K_p[/tex] is 324

Explanation:

We are given:

Initial pressure of [tex]S_8(g)[/tex] = 1.00 atm

The chemical equation for the conversion of [tex]S_8(g)[/tex] to [tex]S_2(g)[/tex] follows:

                  [tex]S_8(g)\rightleftharpoons 4S_2(g)[/tex]

Initial:          1

At eqllm:    1-x       4x

We are given:

Equilibrium partial pressure of [tex]S_8(g)=0.25atm[/tex]

Evaluating the value of 'x', we get:

[tex]\Rightarrow (1-x)=0.25\\\\\Rightarrow x=1-0.25=0.75[/tex]

The expression of [tex]K_p[/tex] for the above reaction follows:

[tex]K_p=\frac{(p_{S_2})^4}{p_{S_8}}[/tex]

[tex]p_{S_2}=4x=(4\times 0.75)=3atm[/tex]

[tex]p_{S_8}=0.25atm[/tex]

Putting values in above expression, we get:

[tex]K_p=\frac{3^4}{0.25}\\\\K_p=324[/tex]

Hence, the value of [tex]K_p[/tex] is 324

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