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Chapter 13: Problem 46

A sample of \(\mathrm{S}_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$ \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) $$ At equilibrium, the partial pressure of \(S_{8}\) is \(0.25\) atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{~K}\).

Short Answer

Expert verified
The equilibrium constant K_p for the given reaction at 1325 K is 324.

Step by step solution

01

Write down the balanced equation and equilibrium constant expression

We are given the balanced equation: \( \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) \) The equilibrium constant expression for this reaction will be: \[ K_{p} = \frac{(\mathrm{[S_{2}]})^4}{\mathrm{[S_{8}]}} \]
02

Set up the I.C.E. table

To set up the I.C.E. table, we'll write the initial partial pressure of S₈ and S₂, the pressure change during the reaction, and the equilibrium partial pressures. I: Initial (atm): S₈: 1.00 S₂: 0 C: Change (atm): S₈: -x S₂: +4x E: Equilibrium (atm): S₈: 1-x S₂: 4x We are also given that the equilibrium pressure of S₈ is 0.25 atm. This means that: \(1-x = 0.25\)
03

Find the change in pressures and the equilibrium pressures

To find the change in pressures, we'll solve for x: \(1-0.25 = x\) \(x = 0.75\) Now, we can find the equilibrium pressure of S₂: S₂: 4x = 4(0.75) = 3.0 atm Hence, the equilibria partial pressures are: S₈: 0.25 atm, S₂: 3.0 atm
04

Calculate K_p using the equilibrium pressures

Now that we have the equilibrium pressures of S₈ and S₂, we can plug these values into the equilibrium constant expression: \[ K_{p} = \frac{(3.0 \thinspace \mathrm{atm})^4}{(0.25 \thinspace \mathrm{atm})} \] Calculate the value of K_p: \[ K_{p} = \frac{81.0 \thinspace \mathrm{atm^3}}{0.25 \thinspace \mathrm{atm}} = 324 \] Finally, the equilibrium constant K_p for this reaction at 1325 K is 324.

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Most popular questions from this chapter

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