The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

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Answer 1

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Students have asked these similar questions

There is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS). Ca, ppm V, ppm SCa, arb. units SV, arb. units 20.0 10.0 14375.11 14261.02 40.0 10.0 36182.15 17997.10 60.0 10.0 39275.74 12988.01 80.0 10.0 57530.75 14268.54 100.0…

A mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C. H₂O(g) + C₁₂O(g) = 2 HOCl(g) K = 0.0900 at 25°C с Calculate the equilibrium concentrations of each gas at 25 °C. [H₂O]= [C₁₂O]= [HOCI]= M Σ M

What units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?

Answer 2

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 3

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 4

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 5

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 6

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

Answer 7

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

Videos

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Explanation of Solution

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

Explanation of Solution

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Chapter 5 Solutions

The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

Videos

Interpretation: The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Explanation of Solution

Interpretation: The equilibrium concentrations of CO2 and H2O has to be calculated.

Explanation of Solution

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Chapter 5 Solutions

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.