Concept explainers
Interpretation: The value of ΔrH° at 750 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.
Answer to Problem 2C.8BE
The value of ΔrH° at 750 K is −23115.911 kJ mol−1_.
Explanation of Solution
The given reaction is,
N2(g) + 3H2(g) → 2NH3(g)
The standard enthalpy of the reaction is, ΔrH° is calculated by the expression,
ΔrHο=∑m⋅ΔHοf(products)−∑n⋅ΔHοf(reactants) (1)
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.
The temperature dependent heat capacity is given by the expression,
ΔrCοp=Δa+ΔbT+Δc/T2 (2)
Where,
- • Δa, Δb, Δc are the temperature dependent variables.
- • T is the temperature.
For the reaction, Δa is calculated by the expression,
Δa=∑va(products)−∑va(reactants) (3)
Where,
- • ∑va(products) is the sum of a of products.
- • ∑va(reactants) is the sum of a of reactants.
- • v is the stoichiometry.
For the reaction, Δb is calculated by the expression,
Δb=∑vb(products)−∑vb(reactants) (4)
Where,
- • ∑vb(products) is the sum of b of products.
- • ∑vb(reactants) is the sum of b of reactants.
- • v is the stoichiometry.
For the reaction, Δc is calculated by the expression,
Δc=∑vc(products)−∑vc(reactants) (5)
Where,
- • ∑vc(products) is the sum of c of products.
- • ∑vc(reactants) is the sum of c of reactants.
- • v is the stoichiometry.
Enthalpy at a temperature T2= 750 K is calculated by the expression,
ΔrH°(T2)= ΔrH°(T1)+T2∫T1ΔrCp° dT=ΔrH°(T1)+T2∫T1(a+bT+c/T2)dT
ΔrH°(T2)=ΔrH°(T1)+Δa(T2−T1)+12Δb(T22−T21)−Δc(1T2−1T1) (6)
The value of Δa is calculated for the given reaction by equation (3)
Δa=∑2×a(NH3, g)−∑[3×a(H2,g)+a(N2,g)]
The value of a for NH3, N2 and H2 is 29.75 J K−1mol−1, 28.58 J K−1mol−1 and 27.28 J K−1mol−1, respectively. Substitute the values in the above equation to calculate the value of Δa.
Δa=2×29.75 J K−1mol−1−(28.58 J K−1mol−1+3×27.28 J K−1mol−1)=−50.92 J K−1mol−1=−50.92 kJ K−1mol−1
The value of Δb is calculated for the given reaction by equation (4)
Δb=∑2×b(NH3, g)−∑[3×b(H2,g)+b(N2,g)]
The value of b for NH3, N2 and H2 is 25.1×10−3 J K−2mol−1, 3.77×10−3 J K−2mol−1 and 3.26×10−3 J K−2mol−1 respectively. Substitute the values in the above equation to calculate the value of Δb.
Δb=2×25.1×10−3 J K−2mol−1−(3.77×10−3 J K−2mol−1+3×3.26×10−3 J K−2mol−1)=−0.03665 J K−2mol−1=−0.03665×10−3 kJ K−2mol−1
The value of Δc is calculated for the given reaction by equation (5)
Δc=∑2×c(NH3, g)−∑[3×c(H2,g)+c(N2,g)]
The value of c for NH3, N2 and H2 is −1.55×105 J K mol−1, −0.50×105 J K mol−1 and 0.50×105 J K mol−1, respectively. Substitute the values in the above equation to calculate the value of Δb.
Δc=2×−1.55×105 J K mol−1−(−0.50×105 J K mol−1+3×0.50×105 J K mol−1)=4.1 ×105 J K mol−1=4.1 ×102 kJ K mol−1
To calculate the value of ΔrH° at 298 K, use equation (1).
ΔrHο=∑2×ΔHοf(NH3, g)−[3×ΔHοf(H2,g)+ΔHοf(N2,g)]
Substitute the values of ΔHοf(NH3, g)=-46.11 kJ mol-1, ΔHοf(N2,g)=0 kJ mol−1 and ΔHοf(H2,g)=0 kJ mol−1 in the above equation to calculate ΔrH°.
ΔrHο=−2×46.11 kJ mol−1−(0 +0) kJ mol−1=−92.22 kJ mol−1
To calculate the value of ΔrH° at 750 K, use equation (6).
ΔrH°(750 K)=ΔrH°(298 K)+Δa(750 K−298 K)+12Δb((750 K)2−(298 K)2)−Δc(1750 K−1298 K)=−92.22 kJ mol−1−50.92 kJ K−1mol−1(750 K−298 K)+12×(−0.03665×10−3 kJ K−2mol−1)((750 K)2−(298 K)2)−4.1 ×102 kJ K mol−1(1750 K−1298 K)= −23115.911 kJ mol−1_
Hence, the value of ΔrH° at 750 K is −23115.911 kJ mol−1_.
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Chapter 2 Solutions
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