   Chapter 13, Problem 38E

Chapter
Section
Textbook Problem

# In a study of the reaction 3 Fe ( s ) + 4 H 2 O ( g ) ⇌ Fe 3 O 4 ( s ) + 4 H 2 ( g ) at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of Kp for this reaction at 1200 K. (Hint: Apply Dalton's law of partial pressures.)

Interpretation Introduction

Interpretation: A reaction between Fe and H2O at 1200K is given. At the given partial pressure value for water, the total pressure at equilibrium is 36.3torr . The value of the equilibrium constant (Kp) for the reaction is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of pressure, it is represented as Kp .

To determine: The equilibrium constant (Kp) value for the given reaction.

Explanation

Given

The stated reaction is,

3Fe(s)+4H2O(g)Fe3O4(s)+4H2(g)

At equilibrium, the equilibrium ratio is expressed by the formula,

Kp=PartialpressureofproductsPartialpressureofreactants

Where,

• Kp is the equilibrium constant in terms of partial pressure.

The solid components do not affect the value of the equilibrium constant.

The equilibrium ratio for the given reaction is,

Kp=(PH2)4(PH2O)4 (1)

The total pressure at equilibrium is given to be 36.3torr .

Partial pressure of water vapor (H2O) is given to be 15.0torr .

The partial pressure of hydrogen (H2) is calculated by the formula,

Ptotal=PH2O+PH2

Substitute the given values of PH2O and PH2 in the above expression.

Ptotal=PH2O+PH236.3torr=15.0torr+PH2

Simplify the above expression.

36.3torr=15.0torr+PH2PH2=36

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