CHEMISTRY

V. I. LAVRENT'EV, Corresponding Member of the Academy of Sciences of the USSR Ya. I. GERASIMOV
and T. N. REZUKHINA

EQUILIBRIUM WITH HYDROGEN AND THERMODYNAMIC CHARACTERISTICS OF BaMoO$_4$ AND BaMoO$_3$

In the present work, which is a continuation of a series of studies on the thermodynamics of tungstates and molybdates ($^{1-4}$), the equilibrium of barium molybdate with hydrogen was investigated. BaMoO$_4$ was prepared by precipitating sodium molybdate with an equivalent amount of Ba(NO$_3$)$_2$ solution; the precipitate was thoroughly washed, dried, and calcined. The BaMoO$_4$ obtained has a tetragonal structure; the lattice constants found by us agree well with literature data ($^5$).

Table 1

The equilibrium of BaMoO$_4$ with hydrogen in the temperature range 1200–1400°C was studied by the circulation method in the apparatus described in ($^4$). An X-ray diffraction study showed that the reduction of BaMoO$_4$ in this temperature range proceeds with the formation of the intermediate compound BaMoO$_3$ (perovskite type, $a = 4.03$ Å) ($^6$); the final products of reduction are BaO and Mo. BaO remains at these temperatures in the solid phase; at 1400°C, according to ($^{7,8}$), the saturated vapor pressure of BaO does not exceed 0.001 mm Hg.

Thus, analogously to the reductions of magnesium ($^9$), calcium ($^1$), and strontium ($^4$) molybdates studied earlier by us, the reduction of BaMoO$_4$ by hydrogen proceeds in two stages:

$$ \mathrm{BaMoO}_4 + \mathrm{H}_2 \to \mathrm{BaMoO}_3 + \mathrm{H}_2\mathrm{O}, \tag{I} $$

$$ \frac{1}{2}\mathrm{BaMoO}_3 + \mathrm{H}_2 \to \frac{1}{2}\mathrm{BaO} + \frac{1}{2}\mathrm{Mo} + \mathrm{H}_2\mathrm{O}. \tag{II} $$

Table 1 gives the mean values of the equilibrium constants $K_{p\mathrm{I}} = p_{\mathrm{I}H_2O}/p_{\mathrm{I}H_2}$ and $K_{p\mathrm{II}} = p_{\mathrm{II}H_2O}/p_{\mathrm{II}H_2}$ for a number of temperatures of the I and II stages of reduction of BaMoO$_4$; in Fig. 1 the dependence of $\lg K_{p\mathrm{I}}$ and $\lg K_{p\mathrm{II}}$ on $\frac{1}{T}$ is presented graphically.

The logarithmic polynomials of the equilibrium constants for the two stages of reduction of BaMoO$_4$ are described (with an accuracy of $\pm 0.25\%$ for the first and $\pm 0.2\%$ for the second) by the following equations:

$$ \lg K_{p\mathrm{I}} = -\frac{11155}{4.575T} + 0.5708, \qquad \lg K_{p\mathrm{II}} = -\frac{26610}{4.575T} + 2.3639. $$

Combining reactions (I) and (II), we obtain the equation for the reaction of complete reduction of BaMoO$_4$:

$$ \mathrm{BaMoO}_4 + 3\mathrm{H}_2 \to \mathrm{BaO} + \mathrm{Mo} + 3\mathrm{H}_2\mathrm{O}. \tag{III} $$

For it

\[ \lg K_{p\mathrm{III}}=\lg \frac{p_{\mathrm{III}\ \mathrm{H_2O}}^{3}}{p_{\mathrm{III}\ \mathrm{H_2}}^{3}} =\lg K_{p\mathrm{I}}+2\lg K_{p\mathrm{II}} =-\frac{64370}{4.575T}+5.299,\quad \Delta Z_{\mathrm{III}}^{0}= \]

\[ =-RT\ln K_{p\mathrm{III}}=64370-24.24T . \]

For the reaction of reduction of \(\mathrm{BaMoO_3}\) by hydrogen

\[ \mathrm{BaMoO_3}+2\mathrm{H_2}\to \mathrm{BaO}+\mathrm{Mo}+2\mathrm{H_2O}, \tag{IIa} \]

\[ K_{p\mathrm{IIa}}=K_{p\mathrm{II}}^{2};\qquad \Delta Z_{\mathrm{IIa}}^{0}\,(\mathrm{kcal})=-4.575T\lg K_{p\mathrm{II}}^{2}=5321Q-21.63T . \]

The values of \(\Delta Z_{\mathrm{IIa}}^{0}\), \(\Delta Z_{\mathrm{III}}^{0}\), as well as \(\Delta H_{\mathrm{IIa}}\) and \(\Delta H_{\mathrm{III}}\) for a number of temperatures are given in Table 2.

Table 2

Combining reactions IIa and III with the reaction of formation of water vapor

\[ \mathrm{H_2}+{}^{1}/{}_{2}\mathrm{O_2}\rightleftharpoons \mathrm{H_2O}\,(\mathrm{g}), \tag{IV} \]

whose isobaric potential and enthalpy according to Chipman \({}^{(10)}\) are expressed by the equations

\[ \Delta Z_{\mathrm{IV}}^{0}\,(\mathrm{kcal})=-59251+2.006T\lg T-7.5\cdot 10^{-5}T^{2} +\frac{408000}{T}+6.8085T, \]

\[ \Delta H_{\mathrm{IV}}\,(\mathrm{kcal})=-59251-0.871T+7.5\cdot 10^{-5}T^{2}, \]

one can calculate* \(\Delta Z_{\mathrm{V}}^{0}\) (respectively \(\Delta H_{\mathrm{V}}\)) and \(\Delta Z_{\mathrm{VI}}^{0}\) (respectively \(\Delta H_{\mathrm{VI}}\)) for the formation of \(\mathrm{BaMoO_3}\) and \(\mathrm{BaMoO_4}\) according to the reactions:

\[ \mathrm{BaO}+\mathrm{Mo}+\mathrm{O_2}\to \mathrm{BaMoO_3} \tag{V} \]

\[ \mathrm{BaO}+\mathrm{Mo}+{}^{3}/{}_{2}\mathrm{O_2}\to \mathrm{BaMoO_4}. \tag{VI} \]

The numerical values of \(\Delta Z_{\mathrm{V}}^{0}\) and \(\Delta Z_{\mathrm{VI}}^{0}\) are given in Table 2, and their temperature dependence in the temperature interval we studied is expressed by the equations:

Fig. 1

\[ \Delta Z_{\mathrm{V}}^{0}\,(\mathrm{kcal})=-\Delta Z_{\mathrm{IIa}}^{0}+2\Delta Z_{\mathrm{IV}}^{0} =-173020+49.19T, \]

\[ \Delta Z_{\mathrm{VI}}^{0}\,(\mathrm{kcal})=-\Delta Z_{\mathrm{III}}^{0}+3\Delta Z_{\mathrm{IV}}^{0} =-244070+65.58T . \]

It is not possible to calculate \(\Delta Z_{\mathrm{V}}^{0}\) for 298°K because of the absence of heat-capacity data for \(\mathrm{BaMoO_3}\). For \(\mathrm{BaO}\), \(\mathrm{Mo}\), and \(\mathrm{O_2}\), according to \({}^{(11)}\), and for

* In the calculation we used the simpler equation \(\Delta Z_{\mathrm{IV}}^{0}=-59000+13.782T\), which is obtained from Chipman’s equation for the temperature interval 1473–1643°K.

For $\mathrm{BaMoO_4}$, according to (12), the following equations are available for the dependence of the true molar heat capacities of these compounds on temperature:

\[ \mathrm{Mo}: \quad c_p = 5.48 + 1.30 \cdot 10^{-3}T; \]

\[ \mathrm{O_2}: \quad c_p = 8.27 + 0.258 \cdot 10^{-3}T - 1.877 \cdot 10^5 T^{-2}; \]

\[ \mathrm{BaO}: \quad c_p = 12.74 + 1.04 \cdot 10^{-3}T - 1.984 \cdot 10^5 T^{-2}; \]

\[ \mathrm{BaMoO_4}: \quad c_p = 25.37 + 13.38 \cdot 10^{-3}T. \]

Using these equations and our data, we obtain the following equation for the dependence of $\Delta Z^0_{\mathrm{VI}}$ in the temperature interval 298—1643°K:

\[ \Delta Z^0_{\mathrm{VI}} = -248520 + 5.255T \ln T - 5.325 \cdot 10^{-3}T^2 - \frac{2.4 \cdot 10^5}{T} + 38.21T, \]

whence, for 298.2°K: $\Delta Z^0_{\mathrm{VI}} = -229.5$ kcal/mole; $\Delta H^0_{\mathrm{VI}} = -251.2$ kcal/mole; $\Delta S^0_{\mathrm{VI}} = -72.9$ e.u. Taking, according to (13), the standard molar entropies of $\mathrm{BaO}$, $\mathrm{Mo}$, and $\mathrm{O_2}$ as equal, respectively, to 16.8, 6.83, and 49.00 e.u., we obtain $(\mathrm{BaMoO_4})\, S_{298.2} = 24.2$ e.u.

Moscow State University
named after M. V. Lomonosov

Received
5 IV 1960

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