Calculate the conventional molar Gibbs energy of [latex]\left( a \right)[/latex] Ar[latex]\left( g\right)[/latex] and [latex]\left( b \right) H_{2}O\left( l \right)[/latex] at 1 bar and 298.15 K.

a. Because Ar [latex]\left( g\right)[/latex] is an element in its standard state for these conditions, [latex] H°_{m}=0[/latex] and [latex]G°_{m}=H°_{m}-TS°_{m}=-TS°_{m}=-298.15 K ×154.8 J K^{-1} mol^{-1} =-4.617 KJ mol^{-1}.[/latex] b. The formation reaction for [latex]H_{2}O\left( l \right)[/latex] is [latex]H_{2} \left( g \right)+1/2 O_{2} \left( g\right)\to H_{2}O\left( l \right)[/latex] [latex]G°_{m,product}=\Delta G°_{f} \left( product \right)+\sum_{i}^{elements only} \nu_{i} TS°_{m.i}[/latex] [latex]G°_{m}\left( H_{2} O,l\right)=\Delta G°_{f}\left( H_{2} O,l\right)-\left[ TS°_{m}\left( H_{2},g \right)+\frac{1}{2}TS°_{m}\left( O_{2} ,g\right) \right][/latex] [latex]=-237.1 KJ mol^{-1} -298.15 K × \left[ ^{130.7 J K^{-1} mol^{-1}+\frac{1}{2}× }_{205.2 J K^{-1} mol^{-1}} \right][/latex] [latex]=-306.6 KJ mol^{-1}[/latex] These calculations give [latex]G°_{m}[/latex] at 298.15 K. To calculate the conventional molar Gibbs energy at another temperature, use the Gibbs Helmholtz equation as discussed next.

Question 6.17: Calculate the conventional molar Gibbs energy of (a) Ar(g) a...

Thermodynamics, Statistical Thermodynamics, & Kinetics

Calculate the conventional molar Gibbs energy of $\left( a \right)$ Ar $\left( g\right)$ and $\left( b \right) H_{2}O\left( l \right)$ at 1 bar and 298.15 K.

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a. Because Ar $\left( g\right)$ is an element in its standard state for these conditions, $H°_{m}=0$ and $G°_{m}=H°_{m}-TS°_{m}=-TS°_{m}=-298.15 K ×154.8 J K^{-1} mol^{-1} =-4.617 KJ mol^{-1}.$

b. The formation reaction for $H_{2}O\left( l \right)$ is $H_{2} \left( g \right)+1/2 O_{2} \left( g\right)\to H_{2}O\left( l \right)$

G°_{m,product}=\Delta G°_{f} \left( product \right)+\sum_{i}^{elements  only} \nu_{i} TS°_{m.i}

G°_{m}\left( H_{2} O,l\right)=\Delta G°_{f}\left( H_{2} O,l\right)-\left[ TS°_{m}\left( H_{2},g \right)+\frac{1}{2}TS°_{m}\left( O_{2} ,g\right) \right]

=-237.1 KJ mol^{-1} -298.15 K × \left[ ^{130.7 J K^{-1} mol^{-1}+\frac{1}{2}× }_{205.2 J K^{-1} mol^{-1}} \right]

=-306.6 KJ mol^{-1}

These calculations give $G°_{m}$ at 298.15 K. To calculate the conventional molar Gibbs energy at another temperature, use the Gibbs Helmholtz equation as discussed next.