For the following reaction, the percent yield of H 2 when an equimolar mixture of CH 4 and CO 2 with a total pressure of 20 .0 atm reaches equilibrium at 1200 .K , at which K p = 3 .548×10 6 has to be found. Concept Introduction: Equilibrium constant using partial pressure: The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure. For a reaction, aA (g) + bB (g) ⇌ cC (g) + dD (g) The expression of K p can be given as K p = (P C ) c (P D ) d (P A ) a (P B ) b

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Answer 1

Question

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

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Steps and explanations please

Answer 2

Question

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

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Answer 3

Question

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

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Answer 4

Question

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

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Answer 5

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

Answer 6

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

Answer 7

Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1200.K$ , at which $Kp = 3.548×106$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

Answer 8

(a)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 3.548×106Taking the square root on both side(2x)2(10.0-x) = 1.8836135×103Quadratic equation:ax2+ bx + c = 04x2 + (1.8836135×103x) -1.8836135×104= 0a = 4, b = 1.8836135×103, c = -1.8836135×104x = -b±b2-4ac2a = -1.8836135×103±(1.8836135×103)2-4(4)(-1.883615×104)2(4)x = 9.796209PH2= 2x = 2(9.796209) = 19.592419 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.592418atm20.0atm(100%) = 97096209 = 98.0%$

The percent yield = $98.0%$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

Answer 13

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of $H2$ when an equimolar mixture of $CH4 and CO2$ with a total pressure of $20.0 atm$ reaches equilibrium at $1300.K$ , at which $Kp = 2.626 ×107$ has to be found.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

$aA(g)+ bB(g)⇌cC(g)+ dD(g)$

The expression of $Kp$ can be given as

$Kp = (PC)c(PD)d(PA)a(PB)b$

Answer 14

(b)

Expert Solution

Explanation of Solution

Given,

$PCH4(init) = PCO2(init) = 10.0atm$

$Pressure(atm) CH4(g) + CO2(g) ⇌ 2CO(g) + 2H2(g)Initial 10.0 10.0 0 0Change -x -x +2x +2xEquilibrium 10.0-x 10.0-x 2x 2xKp=PCO2PH22PCH4PCO2 = (2x)2(2x)2(10.0-x)(10.0-x)= 2.626×107Taking the square root on both side(2x)2(10.0-x) = 5.124451×103Quadratic equation:ax2+ bx + c = 04x2 + (5.124451×103x) - 5.124451×104= 0a = 4, b = 5.124451×103, c = -5.124451×104x = -b±b2-4ac2a = -5.124451×103±(5.124451×103)2-4(4)(-5.124451×104)2(4)x = 9.923144PH2= 2x = 2(9.923144) = 19.84629 atmIf the reaction proceeded to completion, the partial pressure of H2wouldbe20.0atmThe percent yield = 19.84629atm20.0atm(100%) = 99.23144 = 99.0%$

The percent yield = $99.0%$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$

$ΔHrxno$ for the reaction is $2.60×105J/mol$

Answer 19

(c)

Interpretation Introduction

Interpretation:

$ΔHrxno$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

$lnK2K1 = -ΔHr×n°R(1T2-1T1)where, K = Rate constant R = Universal gas constant = 8.314 J/mol.K T = Temperature ΔHr×n°= enthalpy of the reaction$

Answer 20

(c)

Expert Solution

Explanation of Solution

$K1 = 3.548×106 T1=1200.K ΔHrxno=?K2 = 2.626×107 T2=1300.K R=8.314J/mol.KlnK2K1 = -ΔHrxnoR(1T2-1T1)ln2.626×1073.548×106 = ΔHrxno(8.314JmolK)(11200.K-11300.K)ΔHrxno = 2.00166287.710195×10-6 = 2.5961247×105 = 2.60×105J/mol$