a. Find Δ(PV) and ΔU° for the reaction of $Eq. (2.7-1): 2H_{2}(g)+O_{2}(g)\longrightarrow 2H_{2}O(l)$ at 298.15 K.

b. Using the fact that the molar volume of liquid water is $18 cm^{3} mol^{−1}$ at 298.15 K, make a more accurate calculation of Δ(PV) for the reaction of the previous example.

Question

a. Find Δ(PV) and ΔU° for the reaction of [latex]Eq. (2.7-1): 2H_{2}(g)+O_{2}(g)\longrightarrow 2H_{2}O(l)[/latex] at 298.15 K.

b. Using the fact that the molar volume of liquid water is [latex]18 cm^{3} mol^{−1}[/latex] at 298.15 K, make a more accurate calculation of Δ(PV) for the reaction of the previous example.

Accepted Answer

a. The reaction is

[latex]2H_{2}(g)+O_{2}(g)\longrightarrow 2H_{2}O(l)[/latex]

Since the single product, [latex]H_{2}O[/latex], is liquid, [latex]\Delta u (g)=-3 and \Delta n(g)=-3.000mol[/latex]

[latex]\Delta (PV)=\Delta u (g)RT=(-3)(8.3145JK^{-1}mol^{-1})(298.15K)[/latex]

[latex]=-7437Jmol^{-1}=-7.437kJmol^{-1}[/latex]

[latex]\Delta H°=2\Delta _{f}H°(H_{2}O)-0-0=-571.660kJmol^{-1}[/latex]

[latex]\Delta U°=-571.660kJmol^{-1}-(-7.437kJmol^{-1})=-564.223kJmol^{-1}[/latex]

b. [latex]V_{2}=(2.000mol)(18cm^{3}mol^{-1})=36cm^{3}=3.6×10^{-5}m^{3}[/latex]

[latex]V_{1}=\frac{(3.000mol)(8.3145JK^{-1}mol^{-1})(298.15K)}{100000Pa} =0.0743690m^{3}[/latex]

[latex]\Delta V=3.6×10^{-5}m^{3}-0.0743690m^{3}=-0.07433m^{3}[/latex]

Since the pressure is constant at P°=100000 Pa,

[latex]\Delta (PV)=P\Delta V=(100000Pa)(-0.07340m^{3})=-7433Jmol^{-1}[/latex]

This value compares with the value of [latex]-7437 Jmol^{-1}[/latex] in part a.

Question 2.33: a. Find Δ(PV) and ΔU° for the reaction of Eq. (2.7-1) at 298...

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