   # Given the following equilibrium constants at 427°C, Na 2 O ( s ) ⇌ 2 Na ( l ) + 1 2 O 2 ( g ) K 1 = 2 × 10 − 25 NaO ( g ) ⇌ Na ( l ) + 1 2 O 2 ( g ) K 2 = 2 × 10 − 5 Na 2 O 2 ( s ) ⇌ 2 Na ( l ) + O 2 ( g ) K 3 = 5 × 10 − 29 NaO 2 ( s ) ⇌ Na ( l ) + O 2 ( g ) K 4 = 3 × 10 − 14 determine the values for the equilibrium constants for the following reactions: a. Na 2 O ( s ) + 1 2 O 2 ( g ) ⇌ Na 2 O 2 ( s ) b. NaO ( g ) + Na 2 O ( s ) ⇌ Na 2 O 2 ( s ) + Na ( l ) c. 2 NaO ( g ) ⇌ Na 2 O 2 ( s ) ( Hint: When reaction equations are added, the equilibrium expressions are multiplied.) ### 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 72AE
Textbook Problem
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## Given the following equilibrium constants at 427°C, Na 2 O ( s ) ⇌ 2 Na ( l ) + 1 2 O 2 ( g )         K 1 = 2 × 10 − 25 NaO ( g ) ⇌ Na ( l ) + 1 2 O 2 ( g )               K 2 = 2 × 10 − 5 Na 2 O 2 ( s ) ⇌ 2 Na ( l ) + O 2 ( g )           K 3 = 5 × 10 − 29 NaO 2 ( s ) ⇌ Na ( l ) + O 2 ( g )                     K 4 = 3 × 10 − 14 determine the values for the equilibrium constants for the following reactions:a. Na 2 O ( s ) + 1 2 O 2 ( g ) ⇌ Na 2 O 2 ( s ) b. NaO ( g ) + Na 2 O ( s ) ⇌ Na 2 O 2 ( s ) + Na ( l )   c.   2 NaO ( g ) ⇌ Na 2 O 2 ( s ) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

Interpretation Introduction

Interpretation: The equilibrium constants for the decomposition of Na2O(g) , NaO(g) , Na2O2(g) and NaO2(g) is given. The equilibrium constant of given reactions are 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 concentration, it is represented as K .

### Explanation of Solution

EXPLANATION

(a)

To determine: The equilibrium constant of given reaction.

The stated reaction is,

Na2O(s)+12O2(g)Na2O2(s)

Given

The rate constant of decomposition of Na2O(g) is 2×1025 .

The rate constant of decomposition reaction of Na2O2(g) is 5×1029 .

The stated reaction is,

Na2O(s)+12O2(g)Na2O2(s)

The reaction of decomposition of Na2O(g) is,

Na2O(s)2Na(l)+12O2(g) (1)

The reaction of decomposition Na2O2(g) is,

Na2O2(s)2Na(l)+O2(g)

The reverse reaction of decomposition Na2O2(g) is,

2Na(l)+O2(g)Na2O2(s) (2)

Na2O(s)+2Na(l)+O2(g)2Na(l)+12O2(g)+Na2O2(s)

The common terms on both sides is canceled. The required reaction is,

Na2O(s)+12O2(g)Na2O2(s)

It is noted that,

• When reactions are added together, the equilibrium expressions are multiplied.
• The rate constant of any reverse reaction is 1K .

Hence, its rate constant is calculated by using the formula,

K=K1×1K3

Where,

• K is the equilibrium constant.
• K1 is rate constant of decomposition of Na2O(g) .
• K3 is rate constant of reaction of decomposition reaction of Na2O2(g) .

Substitute the values of K1 and K3 in above equation.

K=K1×1K3=2×1025×15×1029=4×103_

(b)

To determine: The equilibrium constant of given reaction.

The stated reaction is,

NaO(g)+Na2O(s)Na2O2(s)+Na(l)

Given

The rate constant of decomposition of Na2O(g) is 2×1025 .

The rate constant of decomposition reaction of Na2O2(g) is 5×1029 .

The rate constant of decomposition of NaO(g) is 2×105 .

The stated reaction is,

NaO(g)+Na2O(s)Na2O2(s)+Na(l)

The reaction of decomposition of Na2O(g) is,

Na2O(s)2Na(l)+12O2(g) (1)

The reaction of decomposition Na2O2(g) is,

Na2O2(s)2Na(l)+O2(g)

The reverse reaction of decomposition Na2O2(g) is,

2Na(l)+O2(g)Na2O2(s) (2)

The reaction of decomposition of NaO(g) is,

NaO(g)Na(l)+12O2(g) (3)

Adding equation (1) and (2) and (3),

Na2O(s)+2Na(l)+O2(g)+NaO(g)2Na(l)+12O2(g)+Na2O2(s)+Na(l)+12O2(g)

The common terms on both sides is canceled

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