Calculation of the Isothermal Enthalpy Change of Mixing

One mole of sulfuric acid at 65.6°C is added to 1000 moles of water at the same temperature. If the mixing is done isothermally, estimate the change in enthalpy of the mixture.

Question

Calculation of the Isothermal Enthalpy Change of Mixing

One mole of sulfuric acid at 65.6°C is added to 1000 moles of water at the same temperature. If the mixing is done isothermally, estimate the change in enthalpy of the mixture.

Accepted Answer

From Eq. 8.6-14,

[latex] heta\left(T, P, x_{1} \sim 1 ight)=N_{1} heta(T, P)+\bar{ heta}_{2}\left(T, P, x_{2} \sim 0 ight)[/latex] (8.6-14)

[latex]H\left(T=65.6^{\circ} C , N_{ H _{2} O }=1000, N_{ H _{2} SO _{4}}=1 ight)-H\left(T=65.6^{\circ} C , N_{ H _{2} O }=1000, N_{ H _{2} SO _{4}}=0 ight)[/latex]

[latex]=\bar{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C , x_{ H _{2} SO _{4}}=0.001 \sim 0 ight)=-73.80 \frac{ kJ }{ mol }[/latex]

[latex]=\Delta_{\operatorname{mix}} H\left(1000 mol H_ 2 O +1 mol H _{2} SO _{4} ight)[/latex]

[The numerical value for [latex]\bar{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C , x_{ H _{2} SO _{4}} \sim 0 ight)[/latex] was obtained from the previous illustration.]

Question 8.6.3: Calculation of the Isothermal Enthalpy Change of Mixing One ...

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