Question

Explain how it is possible to reduce tungsten (VI) oxide (WO₂)to metal with hydrogen at an elevated temperature. Over what temperature range is this reaction spontaneous? Use the data of Appendix D.

Short answer

$W{O}_3\left(s\right)+3H_2\left(g\right)\to W\left(s\right)+3H_{2}O\left(g\right)$

The standard enthalpy change for the reaction is +117.41 kJ.

The standard entropy change for the reaction is +131.19 J/K.mol.

Hence, the reaction will be spontaneous over the temperature range 0<T>895 K.

Step-by-step solution

The reaction and the Standard enthalpy of formation of different components

The chemical reaction representing the reduction of tungsten oxide is as follows:

$W{O}_3\left(s\right)+3H_2\left(g\right)\to W\left(s\right)+3H_{2}O\left(g\right)$

The enthalpies of formation are as given:

Standard enthalpy of formation of tungsten oxide isrole="math" localid="1663850139713" $△H_F^0\left(W{O}_3\left(s\right)\right)=-842.87kJ/mol$

Standard enthalpy of formation of hydrogen,role="math" localid="1663850077861" $△H_F^0\left(H_2\left(g\right)\right)=0kJ/mol$

Standard enthalpy of formation of tungsten,role="math" localid="1663850201254" $△H_f^0\left(W\left(s\right)\right)=0kJ/mol$

Standard enthalpy of formation of water vapor,role="math" localid="1663850305796" $△H_f^0\left(H_{2}O\left(g\right)\right)=-241.82kJ/mol$

The enthalpies change for the reaction.

It is calculated by using the formula;

By substituting and simplifying the equation;

Thus, the standard enthalpy change for the reaction is +117.41 kJ.

The reaction and the Standard entropy of different components.

Standard entropy of tungsten oxide,$△H_F^0\left(W{O}_3\left(s\right)\right)=75.90J{K}^{-1}mo{l}^{-1}$

Standard entropy of hydrogen,$△H_F^0\left(H_2\left(g\right)\right)=130.57k{J}^{-1}mo{l}^{-1}$

Standard entropy of tungsten,$△H_f^0\left(W\left(s\right)\right)=32.64J{K}^{-1}mo{l}^{-1}$

Standard entropy of water vapor, $△H_f^0\left(H_{2}O\left(g\right)\right)=188.72J{K}^{-1}mo{l}^{-1}$

The entropies change for the reaction is calculated as:

By substituting and simplifying the equation;

Thus, the standard entropy change for the reaction is +131.19 J/K.mol.

The spontaneity.

As both standard enthalpy and entropy are positive, from this we can conclude that the reaction will be spontaneous at a temperature greater than T.

The change in Gibbs free energy is given as;

At equilibrium, the change in free energy is always zero. Then the equation becomes.$△H°=T△S°$

The temperature is calculated as;

$\begin{array}{c}T=\frac{△H°}{△S°}=\frac{+117.41 kJ}{+131.19J{K}^{-1}}\\ \\ =\frac{+117.41\times {10}^{3}J}{+131.19J{K}^{-1}}\\ =895K\end{array}$

Hence, the reaction will be spontaneous over the temperature range 0<T>895 K.