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The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. The standard free energy of each substance are ΔG∘fS(s) = 0 kJ/mol, ΔG∘fH2O(g) = -228.57 kJ/mol, ΔG∘fSO2(g) = -300.4 kJ/mol, ΔG∘fH2S(g) = -33.01 kJ/mol. If PSO2 = PH2S and the vapor pressure of water is 22 torr , calculate the equilibrium SO2 pressure in the system at 298 K.

Respuesta :

Answer : The equilibrium [tex]SO_2[/tex] pressure is, [tex]3.93\times 10^{-5}torr[/tex]

Explanation :

The given balanced chemical reaction is,

[tex]SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)[/tex]

First we have to calculate the standard free energy of reaction [tex](\Delta G^o)[/tex].

[tex]\Delta G^o=G_f_{product}-G_f_{reactant}[/tex]

[tex]\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}][/tex]

where,

[tex]\Delta G^o[/tex] = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

[tex]\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)][/tex]

[tex]\Delta G^o=-90.72kJ/mol[/tex]

Now we have to calculate the value of [tex]K_p[/tex]

[tex]\Delta G^o=-RT\ln K_p[/tex]

where,

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

[tex]K_p[/tex] = equilibrium constant  = ?

Now put all the given values in this expression, we get:

[tex]-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p[/tex]

[tex]K_p=7.98\times 10^{15}[/tex]

Now we have to calculate the value of [tex]K_p[/tex].

The given balanced chemical reaction is,

[tex]SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)[/tex]

The expression for equilibrium constant will be :

[tex]K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium [tex]SO_2[/tex] pressure be, x

Pressure of [tex]SO_2[/tex] = Pressure of [tex]H_2S[/tex] = x

Now put all the given values in this expression, we get

[tex]7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}[/tex]

[tex]x=3.93\times 10^{-5}torr[/tex]

Thus, the equilibrium [tex]SO_2[/tex] pressure is, [tex]3.93\times 10^{-5}torr[/tex]