Thermodynamic Modeling of Nitric Acid Speciation Using eUNIQUAC Activity Coefficient Model

Abstract

The speciation of nitric acid was modeled using eUNIQUAC activity coefficient model and the thermodynamic dissociation constant was estimated by solving the dissociation reaction equilibrium. eUNIQUAC model was used for the estimation of activity of species during the estimation of dissociation constant. The eUNIQUAC model parameter was reported for the determination of activity coefficients of ionic species, neutral HNO₃ molecule and water activity up to nitric acid concentration of 15 mol·L⁻¹. The estimated dissociation constant at molar scale is 23.89.

Appendix A

The partial Gibbs energy of un dissociated HNO₃ can be represented as follows

$$\overline{G}_{{{\text{HNO}}_{{3}} }} = \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right) + RT\ln a_{{{\text{HNO}}_{{3}} }} \left( x \right)$$

(A1)

$$\overline{G}_{{{\text{HNO}}_{{3}} }} = \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) + RT\ln a_{{{\text{HNO}}_{{3}} }} \left( m \right)$$

(A2)

where $\overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right)$ and $\overline{G}_{{{\text{HNO}}_{{3}} }} \left( m \right)$ are the standard state partial Gibbs energy of un dissociated HNO₃ in mole fraction and molal scale respectively. The activity of un dissociated HNO₃ is given as follows

$$a_{{{\text{HNO}}_{{3}} }} \left( x \right) = x_{{{\text{HNO}}_{{3}} }} f_{{{\text{HNO}}_{{3}} }}$$

(A3)

$$a_{{{\text{HNO}}_{{3}} }} \left( m \right) = m_{{{\text{HNO}}_{{3}} }} \gamma_{{{\text{HNO}}_{{3}} }}$$

(A4)

where $x_{{{\text{HNO}}_{{3}} }}$ and $m_{{{\text{HNO}}_{{3}} }}$ are the mole fraction and molality of undissociated HNO₃, respectively $f_{{{\text{HNO}}_{{3}} }}$ and $\gamma_{{{\text{HNO}}_{{3}} }}$ is the activity coefficient in mole fraction and molal scale respectively,

Now substituting Eqs. A3 and A4 in A1 and A2 gives

$$\overline{G}_{{{\text{HNO}}_{{3}} }} = \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right) + RT\ln x_{{{\text{HNO}}_{{3}} }} f_{{{\text{HNO}}_{{3}} }}$$

(A5)

$$\overline{G}_{{{\text{HNO}}_{{3}} }} = \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) + RT\ln m_{{{\text{HNO}}_{{3}} }} \gamma_{{{\text{HNO}}_{{3}} }}$$

(A6)

Partial Gibbs energy is constant irrespective of the concentration scale used, So now equating the Eqs. A5 and A6 we get

$$\overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right) + RT\ln x_{{{\text{HNO}}_{{3}} }} f_{{{\text{HNO}}_{{3}} }} = \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) + RT\ln m_{{{\text{HNO}}_{{3}} }} \gamma_{{{\text{HNO}}_{{3}} }}$$

(A7)

So by rearranging the above equation

$$\ln f_{{{\text{HNO}}_{{3}} }} = \frac{{\overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) - \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right)}}{RT} + \ln \frac{{m_{{{\text{HNO}}_{{3}} }} }}{{x_{{{\text{HNO}}_{{3}} }} }} + \ln \gamma_{{{\text{HNO}}_{{3}} }}$$

(A8)

The relation between Mole fraction and molality can be written as

$$x_{{{\text{HNO}}_{{3}} }} = \frac{{m_{{{\text{HNO}}_{{3}} ,st\left( {1 - \alpha } \right)}} }}{{m_{{{\text{HNO}}_{3} ,st\left( {1 - \alpha } \right)}} + \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + {{1000} \mathord{\left/ {\vphantom {{1000} {W_{{\text{A}}} }}} \right. \kern-\nulldelimiterspace} {W_{{\text{A}}} }}}}$$

(A9)

where $m_{{{\text{HNO}}_{{3}} }} - m_{{{\text{HNO}}_{3} ,st}} \left( {1 - \alpha } \right)$, α is the degree of dissociation, ν is the number of ions produced during dissociation and W_A is the molecular weight of the solvent.

Now substituting Eqs. A9 in A8 gives as

$$\ln f_{{{\text{HNO}}_{3} }} = \frac{{\overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) - \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right)}}{RT} + \ln \left( {m_{{{\text{HNO}}_{3} ,st\left( {1 - \alpha } \right)}} + \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + {{1000} \mathord{\left/ {\vphantom {{1000} {W_{{\text{A}}} }}} \right. \kern-\nulldelimiterspace} {W_{{\text{A}}} }}} \right) + \ln \gamma_{{{\text{HNO}}_{{3}} }}$$

(A10)

when $m_{{{\text{HNO}}_{3} {,}st}} \to 0$, $f_{{{\text{HNO}}_{3} }}$ and $\gamma_{{{\text{HNO}}_{3} }}$ tend to one, so Eq. A10 becomes

$$\frac{{\overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( m \right) - \overline{G}_{{{\text{HNO}}_{3} }}^{0} \left( x \right)}}{RT} = - \ln \left( {\frac{1000}{{W_{{\text{A}}} }}} \right)$$

(A11)

Substituting Eq. A11 into A10

$$\ln f_{{{\text{HNO}}_{3} }} = - \ln \left( {\frac{1000}{{W_{{\text{A}}} }}} \right) + \ln \left( {m_{{{\text{HNO}}_{3} ,st\left( {1 - \alpha } \right)}} + \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + {{1000} \mathord{\left/ {\vphantom {{1000} {W_{{\text{A}}} }}} \right. \kern-\nulldelimiterspace} {W_{{\text{A}}} }}} \right) + \ln \gamma_{{{\text{HNO}}_{{3}} }}$$

(A12)

After rearranging we get the expression as follows

$$\ln f_{{{\text{HNO}}_{{3}} }} = \ln \left( {0.001W_{{\text{A}}} m_{{{\text{HNO}}_{3} ,st\left( {1 - \alpha } \right)}} + 0.001W_{{\text{A}}} \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + 1} \right) + \ln \gamma_{{{\text{HNO}}_{{3}} }}$$

(A13)

and further simplification results in the following relation between mole fraction activity coefficient to molal activity coefficient

$$f_{{{\text{HNO}}_{{3}} }} = \gamma_{{{\text{HNO}}_{{3}} }} \left( {0.001W_{{\text{A}}} m_{{{\text{HNO}}_{3} ,st}} \left( {1 - \alpha } \right) + 0.001W_{{\text{A}}} \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + 1} \right)$$

(A14)

Similar to the above equation relation between the molal activity coefficient and molar activity coefficient (y_HNO3) is given as follows

$$y_{{{\text{HNO}}_{{3}} }} = \frac{{\gamma_{{{\text{HNO}}_{{3}} }} m_{{{\text{HNO}}_{{3}} }} \rho_{0} }}{{M_{{{\text{HNO}}_{{3}} }} }}$$

(A15)

where ρ₀ is the density of water at 25 °C. Now substitute expression for γ_HNO3 using equation A14 in A15 we get

$$y_{{{\text{HNO}}_{3} }} = \frac{{m_{{{\text{HNO}}_{{3}} }} \rho_{0} }}{{M_{{{\text{HNO}}_{{3}} }} }}\left( {\frac{{f_{{{\text{HNO}}_{{3}} }} }}{{\left( {0.001W_{{\text{A}}} m_{{{\text{HNO}}_{3} ,st}} \left( {1 - \alpha } \right) + 0.001W_{A} \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + 1} \right)}}} \right)$$

(A16)

The conversion from molal concentration to molar concentration is given as follows

$$m_{{{\text{HNO}}_{{3}} }} = \frac{{M_{{{\text{HNO}}_{{3}} }} }}{{\left( {\rho_{0} - 0.001M_{{{\text{HNO}}_{3} ,st}} W_{{\text{B}}} } \right)}}$$

(A17)

By substituting the Eq. A17 into A16 we get the required relation between the mole fraction activity coefficient to molar activity coefficient of un dissociated nitric acid

$$y_{{{\text{HNO}}_{{3}} }} = f_{{{\text{HNO}}_{{3}} }} \frac{{\rho_{0} }}{{\left( {\rho_{0} - 0.001M_{{{\text{HNO}}_{3} ,st}} W_{{\text{B}}} } \right)\left( {0.001W_{{\text{A}}} m_{{{\text{HNO}}_{3} ,st}} \left( {1 - \alpha } \right) + 0.001W_{{\text{A}}} \nu m_{{{\text{HNO}}_{3} ,st}} \alpha + 1} \right)}}$$

(A18)