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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. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

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

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

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

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

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

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

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

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

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

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

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

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

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