Question

2XY3(g) -> 2XY2(g) +Y2(g)

(delta)Hf(kj/mol) S(J/K*mol)

XY2(g) -296.3 262.3

Y2(g) 0 205.4

XY3(g) -403.4 275.2

Given the following data, assuming (delta)H and (delta)S are temperature independent, detemrine the best response for the temperature/temperature range for which this reaction is spontaneous.

Answer 1

we have the Balanced chemical equation as:

2 XY3(g) ---> 2 XY2(g) + Y2(g)

we have:

Hof(XY3(g)) = -403.4 KJ/mol

Hof(XY2(g)) = -296.3 KJ/mol

Hof(Y2(g)) = 0.0 KJ/mol

deltaHo rxn = 2*Hof(XY2(g)) + 1*Hof(Y2(g)) - 2*Hof( XY3(g))

deltaHo rxn = 2*Hof(XY2(g)) + 1*Hof(Y2(g)) - 2*Hof( XY3(g))

deltaHo rxn = 2*(-296.3) + 1*(0.0) - 2*(-403.4)

deltaHo rxn = 214.2 KJ/mol

we have:

Sof(XY3(g)) = 275.2 J/mol.K

Sof(XY2(g)) = 262.3 J/mol.K

Sof(Y2(g)) = 205.4 J/mol.K

deltaSo rxn = 2*Sof(XY2(g)) + 1*Sof(Y2(g)) - 2*Sof( XY3(g))

deltaSo rxn = 2*Sof(XY2(g)) + 1*Sof(Y2(g)) - 2*Sof( XY3(g))

deltaSo rxn = 2*(262.3) + 1*(205.4) - 2*(275.2)

deltaSo rxn = 179.6 J/mol.K

deltaSo rxn = 0.1796 KJ/mol.K

we have below equation to be used:

deltaG = deltaH - T*deltaS

for reaction to be spontaneous, deltaG should be negative

that is deltaG<0

since deltaG = deltaH - T*deltaS

so, deltaH - T*deltaS < 0

214.2- T *0.18 < 0

T *0.18 > 214.2

T > 1193 K

It will be spontaneous for temperatures greater than 1193 K