Given: Gof(WO3(s)) = -763.1 KJ/mol Gof(H2(g)) = 0.0 KJ/mol Gof(W(s)) = 0.0 KJ/mo

Question

Given:
Gof(WO3(s)) = -763.1 KJ/mol
Gof(H2(g)) = 0.0 KJ/mol
Gof(W(s)) = 0.0 KJ/mol
Gof(H2O(g)) = -228.6 KJ/mol

Balanced chemical equation is:
WO3(s) + 3 H2(g) ---> W(s) + 3 H2O(g)

Go rxn = 1*Gof(W(s)) + 3*Gof(H2O(g)) - 1*Gof( WO3(s)) - 3*Gof(H2(g))
Go rxn = 1*(0.0) + 3*(-228.6) - 1*(-763.1) - 3*(0.0)
Go rxn = 77.3 KJ

Given:
Hof(WO3(s)) = -839.9 KJ/mol
Hof(H2(g)) = 0.0 KJ/mol
Hof(W(s)) = 0.0 KJ/mol
Hof(H2O(g)) = -241.8 KJ/mol

Balanced chemical equation is:
WO3(s) + 3 H2(g) ---> W(s) + 3 H2O(g)

Ho rxn = 1*Hof(W(s)) + 3*Hof(H2O(g)) - 1*Hof( WO3(s)) - 3*Hof(H2(g))
Ho rxn = 1*(0.0) + 3*(-241.8) - 1*(-839.9) - 3*(0.0)
Ho rxn = 114.5 KJ

Go = 77.3 KJ/mol
Ho = 114.5 KJ/mol
T = 298 K

use:
Go = Ho - T*So
77.3 = 114.5 - 298.0 *So
So = 0.1248 KJ/mol.K
So = 124.8322 J/mol.K

Now we have:
Go = 0.0 KJ/mol
Ho = 114.5 KJ/mol
So = 124.8322 J/mol.K
= 0.12483 KJ/mol.K

use:
Go = Ho - T*So
0.0 = 114.5 - T *0.1248322
T = 917 K
Answer: A

Explanation / Answer

20. Tungsten, W(s), is usually produced by the reduction of WO(s) with hydrogen wo,(s) + 3 H2(g) w(s) + 3 H2O(g) Based on the data at 298 K be considered independent of the temperature, estimate the temperature at which the standard Gibbs free energy change AG of the reaction vanishes inthe table below a dass mingthat randSoforthereactioncan AH (KI mol ) -839.9 acr (kJ mol i) 763.1 -241.8 I-228.6 A) 917 K B) 125 K C) 276 K D) 298 K E) 425 K

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