   # At 125°C, K P = 0.25 for the reaction 2 NaHCO 3 ( s ) ⇌ Na 2 CO 3 ( s ) + CO 2 ( g ) + H 2 O ( g ) A 1.00-L flask containing 10.0 g NaHCO 3 is evacuated and heated to 125°C. a . Calculate the par1ial pressures of CO 2 and H 2 O after equilibrium is established. b . Calculate the masses of NaHCO 3 and Na 2 CO 3 present at equilibrium. c . Calculate the minimum container volume necessary for all of the NaHCO 3 to decompose. ### 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 106CP
Textbook Problem
223 views

## At 125°C, KP = 0.25 for the reaction 2 NaHCO 3 ( s ) ⇌ Na 2 CO 3 ( s ) + CO 2 ( g ) + H 2 O ( g ) A 1.00-L flask containing 10.0 g NaHCO3 is evacuated and heated to 125°C.a. Calculate the par1ial pressures of CO2 and H2O after equilibrium is established.b. Calculate the masses of NaHCO3 and Na2CO3 present at equilibrium.c. Calculate the minimum container volume necessary for all of the NaHCO3 to decompose.

(a)

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of CO2 and H2O is to be calculated. At equilibrium the masses of NaHCO3 and Na2CO3   is to be calculated. For all the decomposition of NaHCO3 the minimum container volume is 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.

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

Law of mass action is applicable on the equilibrium reactions.

To determine: The partial pressure of CO2 and H2O

### Explanation of Solution

Explanation

The partial pressure of CO2 and H2O is 0.50atm_ .

Given

The reaction is as,

2NaHCO3(s)Na2CO3(s)+CO2(g)+H2O(g)

The equilibrium constant Kp is 0.25 at 125°C .

Mass of NaHCO3 is 10g .

Volume of the flask is 1.00L .

To calculate the partial pressure of CO2 and H2O after equilibrium established, the (ICE-table) is given as,

2NaHCO3(s)Na2CO3(s)+CO2(g)+H2O(g)Initialpressure0.00.0changeinpressure+x+xEquilibriumpressurexx

Solids are not considered in ICE-chart because solids do not affect the pressure

(b)

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of CO2 and H2O is to be calculated. At equilibrium the masses of NaHCO3 and Na2CO3   is to be calculated. For all the decomposition of NaHCO3 the minimum container volume is 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.

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

Law of mass action is applicable on the equilibrium reactions.

To determine: The masses of NaHCO3 and Na2CO3 at equilibrium.

(c)

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of CO2 and H2O is to be calculated. At equilibrium the masses of NaHCO3 and Na2CO3   is to be calculated. For all the decomposition of NaHCO3 the minimum container volume is 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.

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

Law of mass action is applicable on the equilibrium reactions.

To determine: The minimum container volume to decompose the all amount of NaHCO3 .

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