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Q.

Tungsten is obtained commercially by the reduction of ${WO}_{3}$ with $H_{2}$ according to the equation
${WO}_{3} (s) + 3 H_{2} (g) ⟶ W (s) + 3 H_{2} O (g)$
The following data related to this reaction at 25° C are available.
${WO}_{3} (s) H_{2} O (g)$
$\begin{array}{ccc} Δ H^{\circ} kJ / mol & - 840.3 & - 241.8 \\ Δ G^{\circ} kJ / mol & - 763.5 & - 228.5 \end{array}$
The temperature at which this reaction is at equilibrium at 1 $atm$ is closest to which of the following?

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a

213 K

b

2810 K

c

928 K

d

124 K

answer is C.

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Detailed Solution

$\begin{array}{l} Δ G_{R x^{n}}^{\circ} = - 3 \times 228.5 + 763.5 = 78 \\ Δ H_{R x^{n}}^{\circ} = - 3 \times 241.5 + 840.3 = 115.8 \\ Δ S^{\circ} = \frac{115.8 - 78}{298} = \frac{37.8}{298} \\ T = \frac{115.8}{37.8} \times 298 = 913 K \end{array}$

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Tungsten is obtained commercially by the reduction of WO3 with H2 according to the equationWO3(s)+3H2(g)⟶W(s)+3H2O(g)The following data related to this reaction at 25° C are available. WO3(s) H2O(g)ΔH∘kJ/mol−840.3−241.8ΔG∘kJ/mol−763.5−228.5The temperature at which this reaction is at equilibrium at 1 atm is closest to which of the following?