Concept explainers
Interpretation:
The standard enthalpy of combustion of glucose has to be determined.
Concept Introduction:
The standard enthalpy of the reaction is the difference of the sum of standard enthalpy of formation of all the products and the sum of standard enthalpy of formation of all the reactants. The mathematical equation for the calculation of standard enthalpy of the reaction is shown below.
ΔH° =∑nΔH∘f(products)− ∑nΔH∘f(reactants)
Explanation of Solution
The balanced chemical equation for the combustion of glucose is shown below.
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
The standard enthalpy of formation of products that are CO2(g) and H2O(l) is −393.51 kJ⋅mol−1 and −285.83 kJ⋅mol−1 respectively.
The total enthalpy of formation of products is calculated by the expression shown below.
∑nΔH∘f(products)=nCO2ΔH∘f(CO2, g)+nH2OΔH∘f(H2O, l) (1)
Where,
- ∑nΔH∘f(products) is total enthalpy of formation of products.
- nCO2 is the number of moles of CO2(g).
- ΔH∘f(CO2, g) is the standard enthalpy of formation of CO2(g).
- nH2O is the number of moles of H2O(l).
- ΔH∘f(H2O, l) is the standard enthalpy of formation of H2O(l).
The value of nCO2 is 6 mol.
The value of ΔH∘f(CO2, g) is −393.51 kJ⋅mol−1.
The value of nH2O is 6 mol.
The value of ΔH∘f(H2O, l) is −285.83 kJ⋅mol−1.
Substitute the value of nCO2, ΔH∘f(CO2, g), nH2O and ΔH∘f(H2O, l) in equation (1).
∑nΔH∘f(products)=(6 mol)×(−393.51 kJ⋅mol−1)+(6 mol)×(−285.83 kJ⋅mol−1)=−2361.06 kJ+(−1714.98 kJ)=−2361.06 kJ−1714.98 kJ=−4076.04 kJ
The standard enthalpy of formation of reactants that are C6H12O6(s) and O2(g) is −1268.0 kJ⋅mol−1 and 0.0 kJ⋅mol−1 respectively.
The total enthalpy of formation of reactants is calculated by the expression shown below.
∑nΔH∘f(reactants)=nC6H12O6ΔH∘f(C6H12O6, s)+nO2ΔH∘f(O2, g) (2)
Where,
- ∑nΔH∘f(reactants) is total enthalpy of formation of reactants.
- nC6H12O6 is the number of moles of C6H12O6(s).
- ΔH∘f(C6H12O6, s) is the standard enthalpy of formation of C6H12O6(s).
- nO2 is the number of moles of O2(g).
- ΔH∘f(O2, g) is the standard enthalpy of formation of O2(g).
The value of nC6H12O6 is 1 mol.
The value of ΔH∘f(C6H12O6, s) is −1268.0 kJ⋅mol−1.
The value of nO2 is 6 mol.
The value of ΔH∘f(O2, g) is 0.0 kJ⋅mol−1.
Substitute the value of nC6H12O6, ΔH∘f(C6H12O6, s), nO2 and ΔH∘f(O2, g) in equation (2).
∑nΔH∘f(reactants)=(1 mol)×(−1268.0 kJ⋅mol−1)+(6 mol)×(0.0 kJ⋅mol−1)=−1268.0 kJ+0.0 kJ=−1268.0 kJ
The standard enthalpy of combustion of glucose is calculated by the expression shown below.
ΔH°=∑nΔH∘f(products)− ∑nΔH∘f(reactants) (3)
Where,
- ΔH° is standard enthalpy of reaction.
- ∑nΔH∘f(products) is total enthalpy of formation of products.
- ∑nΔH∘f(reactants) is total enthalpy of formation of reactants.
The value of ∑nΔH∘f(products) is −4076.04 kJ.
The value of ∑nΔH∘f(reactants) is −1268.0 kJ.
Substitute the value of ∑nΔH∘f(products) and ∑nΔH∘f(reactants) in equation (3).
ΔH° =(−4076.04 kJ)−(−1268.0 kJ)=−4076.04 kJ+1268.0 kJ=−2808.04 kJ
Thus, after rounding to four significant figures, the standard enthalpy of combustion of glucose is -2808 kJ⋅mol_.
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Chapter 4 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
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