a)
Interpretation:
Burning of 1.00 gal gasoline to produce carbon-dioxide gas and water vapour, value of ΔH∘ has to be calculated.
Concept Introduction:
The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as ΔH°rxn.
ΔH°rxn= ΔH°p− ΔH°rΔH°p = standard enthalpy of productΔH°r = standard enthalpy of reactant
a)
Explanation of Solution
The given chemical equation:
2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)
As known, heat of reaction is calculated using heat of formation as follows,
ΔH°rxn=∑m ΔH°f(poducts)−∑m ΔH°f(reactants)
ΔH°rxn=[(16 mol CO2)(ΔH°f of CO2) + (18 mol H2O)(ΔH°f of H2O) ]- [(2 mol C8H18)(ΔH°f of C8H18) + (25 mol O2)(ΔH°f of O2) ]ΔH°rxn=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔH°rxn= -10148.868 = -10148.9 kJ.
The energy from 1.00 gal of gasoline as follows,
Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.
Hence, the calculated value of ΔH∘ is -1.18 × 105 kJ.
b)
Interpretation:
Liters of H2 required to burn the given quantity of energy has to be calculated.
Concept introduction:
Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the
Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.
PV = nRT
The unit of each term in the equation is,
Where, R is proportionality constant called gas constant (atm.L/mol.K)
V is volume (L)
N is number of moles (moles)
P is Pressure (atm)
T is temperature (K)
Standard temperature and pressure: The conditions 0oC and 1 atm are called standard atmospheric conditions.
b)
Explanation of Solution
The reaction as follows,
H2(g) + 12 O2 (g) → H2O (g)
The heat of formation of water vapour is ΔHo = -241.826 kJ.
The moles of hydrogen needed to produce the energy from given part a) is,
Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.
In order to determine the volume, the ideal gas equation is used as shown below,
V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.
Therefore, the volume required calculated as 1.20 × 104 L.
c)
Interpretation:
For the amount of H2 to produce the time required by
c)
Explanation of Solution
The reaction as follows,
2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)
Using the known unit conversion; 1A = 1 C/s
Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.
Therefore, the required time calculated as 9.43 × 104 seconds.
d)
Interpretation:
Power required generating the amount of H2 in kilowatt-hours has to be calculated.
d)
Explanation of Solution
The coulombs involved in electrolysis of 488.6 mol of H2.
Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.
Conversion of Coulomb into Joules:
Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.
Conversion of Joules into Power:
Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.
Therefore, the Power required in order of generating the amount of H2 in kilowatt-hours calculated as 157 kW.h.
e)
Interpretation:
The cost of producing the amount of H2 equivalent to 1.00 gal of gasoline has to be calculated.
e)
Explanation of Solution
Given:
Cell has efficiency of 88.0 %; electricity costs of Per kW.h is $ 0.123.
Additional electricity is necessary to produce sufficient hydrogen; the purpose of (100 %80 %) factor.
Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.
Therefore, the cost of producing the amount of H2 equivalent to 1.00 gal of gasoline calculated as 22.0 cents.
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Chapter 21 Solutions
Chemistry: The Molecular Nature of Matter and Change
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