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(i)
Interpretation: The values of ΔrH° and ΔrU° at 298 K has to be calculated for the given reaction.
Concept introduction: Enthalpy is the internal heat content of the system. Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system. Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system. The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.
(i)
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Answer to Problem 2C.7AE
The value of ΔrH° for the reaction is 131.29kJ mol_−1 . The value of ΔrU° for the reaction is 128.81kJ mol-1_.
Explanation of Solution
The given reaction is:
C(graphite) + H2O(g) → CO(g) + H2(g) (1)
The standard enthalpy for the reaction is calculated by the formula,
ΔrHο=∑m⋅ΔHοf(products)−∑n⋅ΔHοf(reactants) (2)
Where,
- • ΔHοf(products) is the standard enthalpy of formation of products.
- • ΔHοf(reactants) is the standard enthalpy of formation of reactants.
- • m is the
stoichiometric coefficient of products. - • n is the stoichiometric coefficient of reactants
For standard enthalpy of the reaction, equation (2) is written as,
ΔrHο=(1×ΔHοf(CO)+1×ΔHοf(H2)−(1×ΔHοf(C(graphite))+1×ΔHοf(H2O)))
The standard enthalpy of formation of CO is −110.53 kJ mol−1.
The standard enthalpy of formation of H2 is 0 kJ mol−1.
The standard enthalpy of formation of H2O is −241.82 kJ mol−1.
The standard enthalpy of formation of C(graphite) is 0 kJ mol−1.
Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.
ΔrHο=(1×(−110.53 kJ mol−1)+1×(0 kJ mol−1)−(1×(0 kJ mol−1)+1×(−241.82 kJ mol−1)))=−110.53 kJ mol−1+241.82 kJ mol−1=131.29kJ mol_−1
Hence, the value of ΔrH° for the reaction is 131.29kJ mol_−1 .
The change in number of moles of the reaction is calculated by the formula,
Number of moles, Δng= Moles of gaseous product− Moles of gaseous reactant
Hence, the number of moles for the reaction is,
Δng= (1+1)−(1)=1
The expression to calculate the value of ΔrU° for the reaction is,
ΔrU°= ΔrH°−ΔngRT
Substitute the values of ΔrH°, R, Δng and T= 298 K in the above equation to calculate the value of ΔrU°.
ΔrU°= 131.29 kJ mol−1 −(1×8.314× 10−3 kJ K−1 mol−1×298 K)= 131.29 kJ mol−1 − (2477.572× 10−3kJ mol−1)= 128.81kJ mol-1_
Hence, the value of ΔrU° for the reaction is 128.81kJ mol-1_.
(ii)
Interpretation: The values of ΔrH° and ΔrU° has to be calculated for the given reaction at 478 K.
Concept introduction: Enthalpy is the internal heat content of the system. Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system. Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system. The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.
(ii)
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Answer to Problem 2C.7AE
The value of ΔrH° for the reaction is 131.29kJ mol_−1 . The value of ΔrU° for the reaction is 127.31kJ mol-1_.
Explanation of Solution
The given reaction is:
C(graphite) + H2O(g) → CO(g) + H2(g) (1)
The standard enthalpy for the reaction is calculated by the formula,
ΔrHο=∑m⋅ΔHοf(products)−∑n⋅ΔHοf(reactants) (2)
Where,
- • ΔHοf(products) is the standard enthalpy of formation of products.
- • ΔHοf(reactants) is the standard enthalpy of formation of reactants.
- • m is the stoichiometric coefficient of products.
- • n is the stoichiometric coefficient of reactants
For standard enthalpy of the reaction, equation (2) is written as,
ΔrHο=(1×ΔHοf(CO)+1×ΔHοf(H2)−(1×ΔHοf(C(graphite))+1×ΔHοf(H2O)))
The standard enthalpy of formation of CO is −110.53 kJ mol−1.
The standard enthalpy of formation of H2 is 0 kJ mol−1.
The standard enthalpy of formation of H2O is −241.82 kJ mol−1.
The standard enthalpy of formation of C(graphite) is 0 kJ mol−1.
Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.
ΔrHο=(1×(−110.53 kJ mol−1)+1×(0 kJ mol−1)−(1×(0 kJ mol−1)+1×(−241.82 kJ mol−1)))=−110.53 kJ mol−1+241.82 kJ mol−1=131.29kJ mol_−1
Hence, the value of ΔrH° for the reaction is 131.29kJ mol_−1 .
The change in number of moles of the reaction is calculated by the formula,
Number of moles, Δng= Moles of gaseous product− Moles of gaseous reactant
Hence, the number of moles for the reaction is,
Δng= (1+1)−(+1)=1
The expression to calculate the value of ΔrU° for the reaction is,
ΔrU°= ΔrH°−ΔngRT
Substitute the values of ΔrH°, R, Δng and T= 478 K in the above equation to calculate the value of ΔrU°.
ΔrU°= 131.29 kJ mol−1 −(1×8.314× 10−3 kJ K−1 mol−1×478 K)= 131.29 kJ mol−1 − (3.97 kJ mol−1)= 127.31kJ mol-1_
Hence, the value of ΔrU° for the reaction is 127.31kJ mol-1_.
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Chapter 2 Solutions
Atkins' Physical Chemistry
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