Question

108. Using Appendix 2 in your textbook, at what temperature (in K) will the reaction, 2H₂O(l) --> 2H₂(g) + O₂(g), change in spontaneity?

Assume standard change in enthalpy and standard change in entropy do not change with temperature.

Answer 1

we have:

Hof(H2O(l)) = -285.83 KJ/mol

Hof(H2(g)) = 0.0 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

we have the Balanced chemical equation as:

2 H2O(l) ---> 2 H2(g) + O2(g)

deltaHo rxn = 2*Hof(H2(g)) + 1*Hof(O2(g)) - 2*Hof( H2O(l))

deltaHo rxn = 2*(0.0) + 1*(0.0) - 2*(-285.83)

deltaHo rxn = 571.66 KJ

we have:

Sof(H2O(l)) = 69.91 J/mol.K

Sof(H2(g)) = 130.684 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

we have the Balanced chemical equation as:

2 H2O(l) ---> 2 H2(g) + O2(g)

deltaSo rxn = 2*Sof(H2(g)) + 1*Sof(O2(g)) - 2*Sof( H2O(l))

deltaSo rxn = 2*(130.684) + 1*(205.138) - 2*(69.91)

deltaSo rxn = 326.686 J/K

deltaHo = 571.66 KJ/mol

deltaSo = 326.686 J/mol.K

= 0.32669 KJ/mol.K

we have below equation to be used:

deltaGo = deltaHo - T*deltaSo

for reaction to be spontaneous, deltaGo should be negative

that is deltaGo<0

since deltaGo = deltaHo - T*deltaSo

so, deltaHo - T*deltaSo < 0

571.66- T *0.326686 < 0

T *0.326686 > 571.66

T > 1750 K

Reaction is spontaneous above 1750 K and non spontaneous below 1750 K

Change occur at 1750 K

Answer: 1750 K