Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 5, Problem 5.139AP

(a)

Interpretation Introduction

Interpretation: Hydrogen is given to be an attractive fuel as it has high value and produces no CO2 . However, the production and storage of hydrogen fuel remain problematic. On the basis of this information the given questions are to be answered.

Concept introduction: The fuel density of hydrogen is calculated by the formula,

Fueldensityofhydrogen=Fuelvalueofhydrogendensityofhydrogenliquid/gas

The enthalpy change for a balanced reaction (ΔHbalanced rxn) is calculated by the formula,

ΔHbalanced rxn=n×ΔHfο(Products)m×ΔHfο(Reactants)

The amount of ammonia borane (mH3NBH3) needed is calculated by the formula,

mH3NBH3=mH2×MH3NBH3MH2

To determine: The fuel density of hydrogen in gaseous and liquid phase.

(a)

Expert Solution
Check Mark

Answer to Problem 5.139AP

Solution

The fuel densities of hydrogen in gaseous and liquid phases are 1,668.52kJ/L_ and 2.118kJ/L_ .

Explanation of Solution

Explanation

Given

The density of hydrogen gas is 0.0899g/L .

The density of hydrogen liquid is 70.8g/L .

The fuel value of hydrogen is 150kJ/g .

The fuel density of hydrogen is calculated by the formula,

Fueldensityofhydrogen=Fuelvalueofhydrogendensityofhydrogenliquid/gas

So, the fuel density of hydrogen in gaseous phase is calculated by the formula,

Fueldensityofhydrogen(g)=Fuelvalueofhydrogendensityofhydrogengas (1)

Where,

  • Fuel density of hydrogen (g) represents fuel density of hydrogen in gaseous phase.

Substitute the fuel value of hydrogen and density of hydrogen gas in equation (1).

Fueldensityofhydrogen(g)=150kJ/g0.0899g/L=1,668.52kJ/L

Similarly, the fuel density of hydrogen in liquid phase is calculated by the formula,

Fueldensityofhydrogen(l)=Fuelvalueofhydrogendensityofhydrogenliquid (2)

Where,

  • Fuel density of hydrogen (l) represents fuel density of hydrogen in liquid phase.

Substitute the fuel value of hydrogen and density of hydrogen liquid in equation (1).

Fueldensityofhydrogen(l)=150kJ/g70.8g/L=2.118kJ/L

Hence, the fuel densities of hydrogen in gaseous and liquid phases are 1,668.52kJ/L_ and 2.118kJ/L_ .

(b)

Interpretation Introduction

To determine: The enthalpy change (ΔH) for the given reactions.

(b)

Expert Solution
Check Mark

Answer to Problem 5.139AP

Solution

The enthalpy change (ΔH) for the given reactions is 197.2kJ_ .

Explanation of Solution

Explanation

Given

The value of ΔHf,H3NBH3ο is 38.1kJ/mol .

The value of ΔHf,BH3ο is +110.2kJ/mol .

The value of ΔHf,NH3ο is 46.1kJ/mol .

The value of ΔHf,HNBHο is +56.9kJ/mol .

The stated reactions are,

H3NBH3(g)NH3(g)+BH3(g)H3NBH3(g)H2(g)+H2NBH2(g)H2NBH2(g)H2(g)+HNBH(g)

The overall balanced reaction is the summation of above stated reaction.

H3NBH3(g)NH3(g)+BH3(g)H3NBH3(g)H2(g)+H2NBH2(g)H2NBH2(g)H2(g)+HNBH(g)2H3NBH3(g)2H2(g)+NH3(g)+BH3(g)+HNBH(g)¯_ (3)

The enthalpy change for a balanced reaction (ΔHbalanced rxn) is calculated by the formula,

ΔHbalanced rxn=n×ΔHfο(Products)m×ΔHfο(Reactants) (4)

Where,

  • ΔHfο(Products) is standard enthalpy of formation of the products.
  • ΔHfο(Reactants) is of standard enthalpy of formation of the reactants.
  • n is number of moles of products.
  • m is number of moles of reactants.

In the balanced chemical equation (3) the,

  • Number of moles of product H2(g) is 2 .
  • Number of moles of product NH3(g) is 1 .
  • Number of moles of product BH3(g) is 1 .
  • Number of moles of product HNBH(g) is 1 .
  • Number of moles of reactant H3NBH3(g) is 2 .

The n×ΔHfο(Products) for the balanced chemical reaction is calculated by the formula,

n×ΔHfο(Products)=2mol×ΔHf,H2(g)ο+1mol×ΔHf,NH3ο+1mol×ΔHf,BH3ο+1mol×ΔHf,HNBHο

Substitute the value of ΔHf,H2(g)ο , ΔHf,NH3ο , ΔHf,BH3ο and ΔHf,HNBHο in above expression.

n×ΔHfο(Products)=(2mol×0kJ/mol+1mol×46.1kJ/mol+1mol×(+110.2kJ/mol)+1mol×(+56.9kJ/mol))=46.1kJ+110.2kJ+56.9kJ=121kJ

The m×ΔHfο(Reactants) for the balanced chemical reaction (1) is calculated by the formula,

m×ΔHfο(Reactants)=2mol×ΔHf,H3NBH3ο (5)

Substitute the value of ΔHf,H3NBH3ο in equation (5).

m×ΔHfο(Reactants)=2mol×38.1kJ/mol=76.2kJ

Substitute the values of n×ΔHfο(Products) and m×ΔHfο(Reactants) in equation (4).

ΔHbalanced rxn=121kJ(76.2kJ)=121kJ+76.2kJ=197.2kJ_

Thus, the enthalpy change (ΔH) for the given reactions is 197.2kJ_ .

(c)

Interpretation Introduction

To determine: The amount of ammonia borane needed to supply 10.0kg of hydrogen.

(c)

Expert Solution
Check Mark

Answer to Problem 5.139AP

Solution

The amount of ammonia borane needed to supply 10.0kg of hydrogen is 153.100kg_ .

Explanation of Solution

Explanation

In the above balanced chemical equation, 2 moles of ammonia borane are utilized to produce 2 moles of hydrogen. That means number of moles of ammonia borane and hydrogen are same.

The amount of hydrogen needed is 10.0kg . To change this in to grams use conversion factor.

1kg=103g10kg=104g

Thus, in grams the amount of hydrogen is 104g .

The amount of ammonia borane (mH3NBH3) needed is calculated by the formula,

mH3NBH3=mH2×MH3NBH3MH2 (6)

Where,

  • mH2 is the amount of hydrogen in grams.
  • MH2 is the molar mass of hydrogen.
  • MH3NBH3 is the molar mass of ammonia borane.

The molar mass of hydrogen is =2×H=2×1.008g/mol=2.016g/mol

The molar mass of H3NBH3 is =1×N+1×B+6×H=1×14.007g/mol+1×10.81g/mol+6×1.008g/mol=14.007g/mol+10.81g/mol+6.048=30.865g/mol

Substitute the values of mH2 , MH2 and MH3NBH3 in equation (6).

mH3NBH3=104g×30.865g/mol2.016g/mol=153,100.19g

To, change the value of mH3NBH3 in kg use conversion factor.

1g=103kg153,100.19g=153,100.19×103kg=153.100kg_

Thus, the amount of ammonia borane needed to supply 10.0kg of hydrogen is 153.100kg_ .

Conclusion

  1. a. The fuel densities of hydrogen in gaseous and liquid phases are 1,668.52kJ/L_ and 2.118kJ/L_ .
  2. b. The enthalpy change (ΔH) for the given reactions is 197.2kJ_ .
  3. c. The amount of ammonia borane needed to supply 10.0kg of hydrogen is 153.100kg_

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Chapter 5 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 5.6 - Prob. 11PECh. 5.7 - Prob. 12PECh. 5.7 - Prob. 13PECh. 5.8 - Prob. 14PECh. 5.8 - Prob. 15PECh. 5.9 - Prob. 16PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.63QPCh. 5 - Prob. 5.64QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143APCh. 5 - Prob. 5.144AP
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