   # A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: 2 FeSO 4 ( s ) ⇌ Fe 2 O 3 ( s ) + SO 3 ( g ) + SO 2 ( g ) SO 3 ( g ) ⇌ SO 2 ( g ) + 1 2 O 2 ( g ) After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate K p for each of these reactions. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 110CP
Textbook Problem
24 views

## A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: 2 FeSO 4 ( s ) ⇌ Fe 2 O 3 ( s ) + SO 3 ( g ) + SO 2 ( g )                 SO 3 ( g ) ⇌ SO 2 ( g ) + 1 2 O 2 ( g ) After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate Kp for each of these reactions.

Interpretation Introduction

Interpretation: The equilibrium constants for the given reactions are to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

Partial pressure of a gas is defined as the pressure applied by the each gas to the surroundings.

The equilibrium constant depends upon temperature.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

Law of mass action states that at a given temperature the equilibrium constant is equal to the partial pressure of the products to the reactants raised the power of stoichiometric coefficient.

To determine: The equilibrium constants for the given reactions.

### Explanation of Solution

Explanation

Given

The first reaction is given as,

2FeSO4(s)Fe2O3(s)+SO3(g)+SO2(g)

In this reaction equal amounts of SO3 and SO2 are produced.

The second reaction is given as’

SO3(g)SO2(g)+12O2(g)

Partial pressure of O2=0.0275atm

In order to solve the equilibrium constant (ICE-chart) that is the initial pressure, change in pressure and equilibrium pressure is determined.

The initial partial pressure is supposed to be P0 and the change in partial pressure is x . The ICE table for the given reaction is,

SO3(g)SO2(g)1/2O2(g)InitialP0P0P0Changex+x+x/2EquilibriumP0xP0+xx/2

At equilibrium the partial pressure is equal to the sum of the individual pressure of the all gases. Therefore,

Ptotal=Ppartial

Where,

• Ptotal is the total pressure at equilibrium.
• Ppartial is the partial pressure of the gases.
• represents the sum of all partial pressures.

Substitute the given value of total pressure and value of partial pressure of all gases from the ICE-chart.

Ptotal=Ppartial0.836atm=P0+x+P0+x+x/20.836atm=2P0+x/2 (1)

At equilibrium the given partial pressure of O2=0.0275atm .

Therefore, from the ICE table the value of x calculated as,

PO2=x/20.0275atm=x/2x=0.0550atm

Substitute the value of x in equation (1).

0.836atm=2P0+x/20.836atm=2P0+0.0550atm2P0=0.405atm

Therefore, from the ICE table the partial pressure of SO2 and SO3 is as follows,

PSO2=P0x=0.4050.0550=0.350atm_

PSO3=P0+x=0

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